Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap17 Planar motion of an object 19 Constrained particles and rigid objects

Chapter 18 Kinematics using time-varying basis vectors

Download this chapter by section

18.0Introductionstudy PDF
18.1Polar coordinates and path coordinatesstudy PDF + solutions
18.2Rotating frames and their base vectorsstudy PDF + solutions
18.3General formulas for ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ study PDF + solutions
18.4Kinematics of 2-D mechanismsstudy PDF + solutions
18.5Advanced kinematics of planar motionstudy PDF + solutions

All chapters and individual parts

Here is a second approach to the kinematics of particle motion. Now, instead of using constant base vectors, we use time-varying base vectors. The discussion of polar coordinates started in Chapter 15 is completed here. Path coordinates, where one base vector is parallel to the velocity and the others orthogonal to that, are introduced. The challenging kinematics topic of relative motion is introduced in two stages: first using rotating base vectors connected to a moving rigid object and then using the more abstract notation associated with frame-dependent differentiation and the famous “five term acceleration formula.”

Many parts of practical machines and structures move in ways that can be idealized as straight-line motion (Chapter 14) or circular motion (Chapter 15). But often an engineer must analyze parts with more general motions as we began to study in Chapter 17.

In principle one can study all motions of all things using one fixed, say $x y z$ , coordinate system. If one knows the $x, y,$ and $z$ coordinates of all points at all times then one can evaluate the linear momentum, the angular momentum, and their rates of change. In this way one can do all of mechanics. But when a machine has various parts, each moving relative to the other, it turns out it is helpful to make use of additional base vectors besides those fixed to a Newtonian (“fixed”) reference frame. That is, the formulas for velocity and acceleration are in some senses simplified (or clarified) by using moving base vectors. Most often these moving base vectors move with the moving parts.

You have seen the time-varying base vectors $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ , the polar coordinate base vectors used to describe circular motion. These are the ideas on which we build here. Altogether we discuss 4 approaches that use time varying base vectors:

1.

Polar coordinates are generalized to include more than just circular motion;
2.

Path coordinates and path base vectors;
3.

General rotating base vectors and coordinate systems with an origin that moves are introduced; and
4.

Formulas are presented for differentiation ‘in’ moving frames that don’t depend on any particular choice of base vectors.

The basic idea is to try to use coordinate systems that most simply describe the motions of interest, even if these coordinate systems are superficially confusing because they rotate and move. Time-varying base vectors are difficult at first. Like many shortcuts, they have a cost. In the end, however, they can aid intuition and simplify calculations.

18.1 Polar coordinates and path coordinates

As you learned in Chapter 15, when a particle moves in a plane while going in circles around the origin its position, velocity, and acceleration can be described like this:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=$	$\displaystyle R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=v% _{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle-R\dot{\theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R% \ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=-{v_{\theta}^{2% }\over R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\dot{v}_{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}.$

These three equations say that the position is the distance from the origin times a unit vector towards the point; that the velocity is tangent to the circle of motion; and that the acceleration has a centripetal component proportional to the speed squared, and a tangential component, tangent to the circle of motion with magnitude equal to the rate of change of speed.

We will now generalize these results two different ways.

•

First we will use polar base vectors for non-constant $R$ .
•

Next we will use path base vectors to show that, in a sense to be explained, the 2nd and 3rd formulas above (for ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ ) apply to any wild motion in 2D or 3D.

As mentioned, in principle these new methods are not needed. We could just use one fixed coordinate system with base vectors $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{k}}$ and write the velocity and acceleration of a point at position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=x\hat{\boldsymbol{\imath}}+y\hat% {\boldsymbol{\jmath}}$ as

{\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{x}\hat{\boldsymbol{\imath}}% +\dot{y}\hat{\boldsymbol{\jmath}}+\dot{z}\hat{\boldsymbol{k}}\hbox to36.135pt{% \hfil and\hfil}{\accentset{\rightharpoonup}{\boldsymbol{a}}}=\ddot{x}\hat{% \boldsymbol{\imath}}+\ddot{y}\hat{\boldsymbol{\jmath}}+\ddot{z}\hat{% \boldsymbol{k}}

as in Chapters 9-12. But, as for the circular motion of Chapter 15, rotating base vectors are helpful for simplifying some kinematics and mechanics problems.

Polar coordinates for general (non-circular) motion

The extension of polar coordinates to 3 dimensions as cylindrical coordinates is shown in fig. 18.1.

Filename:tfigure8-rel-ang-vel

Figure 18.1: Polar coordinates.

Rather than identifying the location of a point by its $x$ , $y$ and $z$ coordinates, a point is located by its cylindrical coordinates

$R,$

the distance to the point from the $z$ axis,
$\theta,$

the angle that the most direct line from the $z$ axis to the point makes with the positive $x$ direction,
$z,$

the conventional $z$ coordinate of the particle,

and base vectors:

$\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R},$

a unit vector normal to the $z$ axis that points from the $z$ axis to the particle
(in 2-D $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}/r$ ,
(in 3-D $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=({\accentset{\rightharpoonup}{% \boldsymbol{r}}}-({\accentset{\rightharpoonup}{\boldsymbol{r}}}\cdot\hat{% \boldsymbol{k}})\hat{\boldsymbol{k}})/|{\accentset{\rightharpoonup}{% \boldsymbol{r}}}-({\accentset{\rightharpoonup}{\boldsymbol{r}}}\cdot\hat{% \boldsymbol{k}})\hat{\boldsymbol{k}}|$
(in 3-D $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ $=$ a unit vector in the direction of the shadow of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ in the $x y$ plane)
$\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta},$

a vector in the $x y$ plane normal to $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ (Formally $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=\hat{\boldsymbol{k}}\times% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ ),
$\hat{\boldsymbol{k}},$

the conventional $\hat{\boldsymbol{k}}$ base vector.

The position vector of a particle is

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+z\hat{\boldsymbol{k}}.

The $z$ component of position, velocity and acceleration is the same in cylindrical coordinates as in Cartesian coordinates. If only two-dimensional problems are being considered then

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}.

As the particle moves, the values of its coordinates $R$ , $\theta$ and $z$ change as do the base vectors $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ .

Example: Oblong path

Filename:tfigure8-ang-vel-ex

Figure 18.2: Particle

P

on an oblong path. Note that the velocity is not parallel to

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}

at most points on the path.

A particle that moves on the oblong path $R=A+B\cos(2\theta)$ with $A>B$ is shown in fig. 18.2. The position vector is

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}=(A+B\cos(2\theta))\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}.

Note that, unlike for circular motion, $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ is not tangent to the particle’s path in general. The base vector $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ is only tangent to the path at those points where the path is closest to, or furthest from, the origin (which, in this example, are also the points where the path crosses the $x$ and $y$ axes).

Velocity in polar coordinates

The velocity and acceleration are found by differentiating the position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ , taking account that the base vectors $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ also change with time just as they did for circular motion:

\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}=\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}\qquad\mbox{and}\qquad\dot{\hat{% \boldsymbol{e}}}_{\scriptscriptstyle\!\theta}=-\dot{\theta}\hat{\boldsymbol{e}% }_{\scriptscriptstyle\!R}.

We find the velocity by taking the time derivative of the position, using the product rule of differentiation:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}={d\over dt}{% \accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=$	$\displaystyle{d\over dt}[R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+z\hat{% \boldsymbol{k}}]$	(18.1)
	$\displaystyle=$	$\displaystyle{d\over dt}(R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R})+{d% \over dt}(z\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle(\dot{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R\underbrace% {\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}}_{\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}})+(\dot{z}\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle\underbrace{\dot{R}}_{v_{R}}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+\underbrace{R\dot{\theta}}_{v_{\theta}}\hat{\boldsymbol% {e}}_{\scriptscriptstyle\!\theta}+\underbrace{\dot{z}}_{v_{z}}\hat{\boldsymbol% {k}}.$

This formula is intuitive. The velocity is the sum of three vectors: one, $\dot{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ , due to moving towards or away from the $z$ axis; one, $\dot{\theta}R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ , having to do with the angle being swept; and in 3-D, one, $\dot{z}\hat{\boldsymbol{k}}$ , for motion perpendicular to the $x y$ plane. In 2-D this is shown in fig. 18.3.

Filename:tfigure8-ang-vel

Figure 18.3: Velocity in polar coordinates for general planar motion. The velocity has a radial

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}

component

\dot{R}

and a circumferential

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}

component

R\dot{\theta}

But, as has been emphasized before, this isn’t a new vector ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ but just a new way of representing the same vector:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$
	$\displaystyle v_{x}\hat{\boldsymbol{\imath}}+v_{y}\hat{\boldsymbol{\jmath}}+v_% {z}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle v_{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+v_{\theta}\hat% {\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+v_{z}\hat{\boldsymbol{k}}.$

The vector ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ can be represented in different base vector systems, in this case cartesian and polar.

Note that eqn. (18.1) adds two terms to the circular motion case from Chapter 15: one for variable $R$ and one for variable $z$ .

Acceleration in polar coordinates

Filename:tfigure8-ang-accel

Figure 18.4: Acceleration in polar coordinates for general planar motion. The acceleration has a radial

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}

component

\ddot{R}-R\dot{\theta}^{2}

and a circumferential

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}

component

R\ddot{\theta}+2\dot{R}\dot{\theta}

To find the acceleration, we differentiate once again. The resulting formula has new terms generated by the product rule of differentiation.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle{d\over dt}{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	(18.2)
	$\displaystyle=$	$\displaystyle{d\over dt}(\dot{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R% \dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+\dot{z}\hat{% \boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle(\ddot{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\dot{R}% \underbrace{\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}}_{\dot{\theta}% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}})+(\dot{R}\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}+R\ddot{\theta}\hat{\boldsymbol{e}% }_{\scriptscriptstyle\!\theta}+R\dot{\theta}\underbrace{\dot{\hat{\boldsymbol{% e}}}_{\scriptscriptstyle\!\theta}}_{-\dot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}})+(\ddot{z}\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle(\underbrace{\ddot{R}-\dot{\theta}^{2}R}_{a_{R}})\hat{\boldsymbol% {e}}_{\scriptscriptstyle\!R}+(\underbrace{2\dot{R}\dot{\theta}+R\ddot{\theta}}% _{a_{\theta}})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+\underbrace{% \ddot{z}}_{a_{z}}\hat{\boldsymbol{k}}.$

The acceleration for an arbitrary planar path is shown in fig. 18.4.

Four of the five terms comprising the polar coordinate formula for acceleration are easy to understand.

$\ddot{R}$

is just the acceleration due to the distance from the origin changing with time.
$\dot{\theta}^{2}R$

is the familiar centripetal acceleration.
$R\ddot{\theta}$

is the acceleration due to rotation proceeding at a faster and faster rate. And
$\ddot{z}$

is the same as for Cartesian coordinates.

The Coriolis term.

Filename:tfigure8-ang-accel-ex

Figure 18.5: The simplest case for understanding of the Coriolis term in the acceleration. A particle moves at constant speed

\dot{R}

out on a line which is rotating at constant rate

\dot{\theta}

. The figure shows a velocity vector as an arrow when the particle is at the origin and a time

\Delta t

later. The change in the velocity vector comes from two effects: (a) the radial line has rotated, and (b) a circumferential component is added to the velocity when the particle is away from the origin. For small

\Delta\theta

these two changes to the velocity are both approximately perpendicular to

{\accentset{\rightharpoonup}{\boldsymbol{v}}}

and parallel to each other.

The difficult term in the polar coordinate expression for the acceleration is the

$2\dot{\theta}\dot{R}$

term, called the Coriolis acceleration, after the civil engineer Gustave-Gaspard Coriolis who first wrote about it in 1835 (in a slightly more general context).

The presence of the ‘2’ in this term is due to the two effects from which it derives: 1 from the change of the $\dot{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ term in the velocity and 1 from the change of the $R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ term in the velocity ( $1+1=2$ ).

The Coriolis acceleration occurs even if both $\dot{\theta}$ and $\dot{R}$ are constant. That is, a particle that moves at constant speed ( $\dot{R}$ = constant) on a straight line that is itself rotating at constant rate ( $\dot{\theta}=$ constant) does not have a straight-line path, and thus has some acceleration. The Coriolis term catches this acceleration. Here is a situation in which the Coriolis term is the only non-zero term in the general polar-coordinate acceleration formula (eqn. (18.2)).

Example: The simplest Coriolis example

Consider, like above, a particle moving at constant speed $\dot{R}$ along a line which is itself rotating at constant $\dot{\theta}$ about the origin. Let’s look at a particle as it passes through the origin at time $t$ and a small amount of time $\Delta t$ later (see fig. 18.5). At time $t+\Delta t$ the direction of the scribed line has changed by an angle $\Delta\theta=\dot{\theta}\Delta t$ . So that, even if at that later time $\dot{\theta}=0$ the direction of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ has changed so ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ has changed by an amount $v\Delta\theta$ . But the rotation of the line does continue, so the velocity includes a part in the $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ direction with magnitude $\dot{\theta}\Delta R$ . That is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ is changed by both $\Delta\theta v$ and by $\dot{\theta}\Delta R$ so

$\displaystyle\Delta{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle\approx$	$\displaystyle\left(v\Delta\theta+\dot{\theta}\Delta R\right)\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!\theta}$
	$\displaystyle\approx$	$\displaystyle\left(\dot{R}(\dot{\theta}\Delta t)+\dot{\theta}(\dot{R}\Delta t)% \right)\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
	$\displaystyle\approx$	$\displaystyle\left(\dot{R}\dot{\theta}+\dot{\theta}\dot{R}\right)\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}\Delta t$
$\displaystyle\Rightarrow{\Delta{\accentset{\rightharpoonup}{\boldsymbol{v}}}% \over\Delta t}$	$\displaystyle\approx$	$\displaystyle\left(\dot{R}\dot{\theta}+\dot{\theta}\dot{R}\right)\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
$\displaystyle\Rightarrow{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle\approx$	$\displaystyle\left(\dot{R}\dot{\theta}+\dot{\theta}\dot{R}\right)\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}=2\dot{R}\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$

as predicted by the general polar coordinate acceleration formula.

But one need not get confounded by a desire to understand every term intuitively. Equation (18.2) is a way of describing the same acceleration we have described with cartesian coordinates. Namely,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{a}}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{x}\hat{\boldsymbol% {\imath}}+\ddot{y}\hat{\boldsymbol{\jmath}}+\ddot{z}\hat{\boldsymbol{k}}$	$\displaystyle=(\underbrace{\ddot{R}-\dot{\theta}^{2}R}_{a_{R}})\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}+(\underbrace{2\dot{R}\dot{\theta}+R% \ddot{\theta}}_{a_{\theta}})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+% \underbrace{\ddot{z}}_{a_{z}}\hat{\boldsymbol{k}}.$

Example: $R(t)$ and $\theta(t)$ are given functions.

Say $\theta$ and $R$ vary in time according to

\theta=at\hbox to36.135pt{\hfil and\hfil}R=ct^{2}

with $a$ and $c$ as given constants. Then at any $t$ the position, velocity, and acceleration are (see fig. 18.6)

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=$	$\displaystyle R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=ct^{2}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle\dot{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R\dot{\theta}% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=2ct\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+act^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta% },\quad\hbox{and}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle(\ddot{R}-\dot{\theta}^{2}R)\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(R\ddot{\theta}+2\dot{R}\dot{\theta})\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!\theta}$
	$\displaystyle=$	$\displaystyle\left(2c-a^{2}ct^{2}\right)\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+\left(0+4act\right)\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}$

with polar base vectors

	$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$	$\displaystyle=$	$\displaystyle\phantom{-}\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}}=\phantom{-}\cos(at)\hat{\boldsymbol{\imath}}+\sin(at)\hat% {\boldsymbol{\jmath}}\quad\hbox{and}$
	$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{% \jmath}}=-\sin(at)\hat{\boldsymbol{\imath}}+\cos(at)\hat{\boldsymbol{\jmath}}.$

Filename:tfigure8-ang-accel1

Figure 18.6: A particle moves with

\theta=at

and

R=ct^{2}

. In this drawing

a=.5

and

c

is anything since no scale is shown.

If we substituted these expressions for the polar base vectors into the expressions for ${\accentset{\rightharpoonup}{\boldsymbol{r}}},{\accentset{\rightharpoonup}{% \boldsymbol{v}}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ we would get the same cartesian representation (a giant mess that we don’t show) that we would get from using $x=R\cos\theta$ and $y=R\sin\theta$ with ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=x\hat{\boldsymbol{\imath}}+y\hat% {\boldsymbol{\jmath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{x}\hat{\boldsymbol{\imath}}% +\dot{y}\hat{\boldsymbol{\jmath}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=\ddot{x}\hat{\boldsymbol{\imath}% }+\ddot{y}\hat{\boldsymbol{\jmath}}$ . That is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{v}}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ even if the representation is different.

Path coordinates

Another way still to describe the velocity and acceleration is to use base vectors which are defined by the motion. In particular, the path base vectors used are:

1.

the unit tangent to the path $\hat{\boldsymbol{e}}_{t}$ , and
2.

the unit normal to the path $\hat{\boldsymbol{e}}_{n}$ .

Somewhat surprisingly at first glance, only two base vectors are needed to define the velocity and acceleration, even in three dimensions.

The base vectors can be described geometrically and analytically. Let’s start out with a geometric description.

The geometry of the path basis vectors.

As a particle moves through space it traces a path ${\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)$ . At the instant of interest the path has a unique tangent line. The unit tangent $\hat{\boldsymbol{e}}_{t}$ is along this line in the direction of motion, as shown in fig. 18.7.

Filename:sfig8-2-3

Figure 18.7: The base vectors for path coordinates are tangent and perpendicular to the path. The osculating circle is a circle that is tangent to the path and has the same curvature as the path. The circle is in the plane containing both the tangent to the path and the rate of change of that tangent. The osculating circle is not defined for a straight-line path.

Less clear is that the path has a unique ‘kissing’ plane. One line on this plane is the tangent line. The other line needed to define this plane is determined by the position of the particle just before and just after the time of interest. Just before and just after the time of interest the particle is a little off the tangent line (unless the motion happens to be a straight line and the tangent plane is not uniquely determined). Three points, the position of the particle just before, just at, and just after the instant of interest determine the tangent plane.

Another way to picture the tangent plane is to find the circle in space that is tangent to the path and which turns at the same rate and in the same direction as the path turns. This circle, which touches the path so intimately, is called the osculating or ‘kissing’ circle. The tangent plane is the plane of this circle. (See fig. 18.7).

The unit normal $\hat{\boldsymbol{e}}_{n}$ is the unit vector which is perpendicular to the unit tangent and is in the tangent plane. It is pointed in the direction from the edge of the osculating circle towards the center of the circle as shown in fig. 18.7. For 2-D motion in the $x y$ plane the osculating plane is the $x y$ plane and the osculating circle is in the $x y$ plane. The path base vectors are unit vectors that vary along the path, always tangent and normal to the path (see fig. 18.8).

Filename:sfig8-2-3a

Figure 18.8: A path in the plane with the path base vectors and osculating circle marked at three points.

Formal definition of path basis vectors

The path of a particle ${\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)$ can also be parameterized by arc length $s$ along the path, as explained in any introductory calculus text. So the path in space is also ${\accentset{\rightharpoonup}{\boldsymbol{r}}}(s)$

^†^†margin: Mathematicians often do not like this engineering sloppiness of notation. That is, the sequence of points

{\accentset{\rightharpoonup}{\boldsymbol{r}}}

that a particle traverses as both time

t

and accumulated arc length

s

increase is not the same mathematical function of both the arguments

s

and

t

. For example, when

t=6\,\mathrm{s}

the position is not the same as when

s=6\,\mathrm{cm}

. Here we don’t think it is worth the trouble of writing

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=\mathbf{f}(t)

and

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=\mathbf{g}(s)

, where the arc length $s$ is the path “coordinate”. The unit tangent is:

\hat{\boldsymbol{e}}_{t}\equiv{d{\accentset{\rightharpoonup}{\boldsymbol{r}}}(% s)\over ds}.

Using the chain rule with ${\accentset{\rightharpoonup}{\boldsymbol{r}}}(s(t))$ this is also

\hat{\boldsymbol{e}}_{t}={d{\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)% \over dt}{dt\over ds}={\,{\accentset{\rightharpoonup}{\boldsymbol{v}}}\,\over v}.

To define the unit normal let’s first define the curvature ${\accentset{\rightharpoonup}{\boldsymbol{\kappa}}}$ of the path as the rate of change of the tangent (rate in terms of arc length).

{\accentset{\rightharpoonup}{\boldsymbol{\kappa}}}\equiv\frac{d\hat{% \boldsymbol{e}}_{t}}{ds}

The unit normal $\hat{\boldsymbol{e}}_{n}$ is the unit vector in the direction of the curvature

\hat{\boldsymbol{e}}_{n}={{\accentset{\rightharpoonup}{\boldsymbol{\kappa}}}% \over|{\accentset{\rightharpoonup}{\boldsymbol{\kappa}}}|}.

Finally, the binormal $\hat{\boldsymbol{e}}_{b}$ is the unit vector perpendicular to $\hat{\boldsymbol{e}}_{t}$ and $\hat{\boldsymbol{e}}_{n}$ :

\hat{\boldsymbol{e}}_{b}\equiv\hat{\boldsymbol{e}}_{t}\times\hat{\boldsymbol{e% }}_{n}.

For 2-D motion the binormal $\hat{\boldsymbol{e}}_{b}$ is always in the $\hat{\boldsymbol{k}}$ direction. The radius of the osculating circle $\rho$ is

\rho={1\over|{\accentset{\rightharpoonup}{\boldsymbol{\kappa}}}|}.

Note that, in general, the polar and path coordinate basis vectors are not parallel; i.e., $\hat{\boldsymbol{e}}_{n}$ is not parallel to $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{t}$ is not parallel to $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ . For example, consider a particle moving on an elliptical path in the plane shown in fig. 18.9. In this case, the polar coordinate and path coordinate basis vectors are only parallel where the major and minor axes intersect the path.

Filename:sfig8-2-3b

Figure 18.9: Particle

P

on an elliptical path.

Velocity and acceleration in path coordinates.

Filename:sfig8-3-1

Figure 18.10: The acceleration of any particle can be broken into a part that is parallel to its path and a part perpendicular to its path.

Although it is not necessarily easy to compute the path basis vectors $\hat{\boldsymbol{e}}_{t}$ and $\hat{\boldsymbol{e}}_{n}$ , they lead to simple expressions for the velocity and acceleration:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=v\hat{\boldsymbol{e}}_{t}={ds\over dt}\hat{\boldsymbol{e}}_{t}$	(18.3)
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle={d\over dt}{\accentset{\rightharpoonup}{\boldsymbol{v}}}={d\over dt% }\left(v\hat{\boldsymbol{e}}_{t}\right)=\dot{v}\hat{\boldsymbol{e}}_{t}+v{% \mathrel{\mathop{\hat{\boldsymbol{e}}_{t}}\limits^{.}}}$
	$\displaystyle=\dot{v}\hat{\boldsymbol{e}}_{t}+v{d\hat{\boldsymbol{e}}_{t}\over ds% }{ds\over dt}\ =\dot{v}\hat{\boldsymbol{e}}_{t}+v{\accentset{\rightharpoonup}{% \boldsymbol{\kappa}}}v$
	$\displaystyle=\underbrace{\dot{v}\hat{\boldsymbol{e}}_{t}}_{{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{t}}+\underbrace{{{v^{2}\over\rho}\hat{% \boldsymbol{e}}_{n}}}_{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}}.$	(18.4)

This formula for velocity is intuitive: velocity is speed times a unit vector in the direction of motion. The formula for acceleration is more interesting. It says that the acceleration of any particle at any time is given by the same formula as the formula for acceleration of a particle going around in circles at non-constant rate.

•

The first term $\dot{v}\hat{\boldsymbol{e}}_{t}$ is tangent to the path (also tangent to the osculating circle), as shown in fig. 18.10. If we draw the osculating circle at the point of interest and fix it in space, then this first term is $\rho\ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ in a polar coordinate system centered at the center of that fixed circle.
•

The term $v^{2}/\rho\hat{\boldsymbol{n}}$ is directed towards the center of the osculating circle. This term is associated with change of direction and does not vanish even if the speed is constant. This normal acceleration is perpendicular to the path.

That is, the two terms in the path-coordinate acceleration formula correspond exactly to the two terms for acceleration for a particle going in circles.

Example: Estimating the direction of acceleration

By looking at the path of a particle (e.g., see fig. 18.11) and just knowing whether it is speeding up or slowing down one can estimate the direction of the acceleration.

If the particle is known to be speeding up at A then $\dot{v}>0$ so the tangential acceleration is in the direction of the velocity. Without thinking about the normal acceleration you know that the acceleration vector is pointed in the half plane of directions shown.

If at point B nothing is known about the rate of change of speed, you still know that the acceleration must be in the half plane shown because that is the direction of ${\accentset{\rightharpoonup}{\boldsymbol{\kappa}}}$ and $\hat{\boldsymbol{e}}_{n}$ .

Filename:sfig8-3-1a

Figure 18.11: A particle moves from 0 to D on the path shown. At A it is speeding up so

\dot{v}>0

. At B it is turning to the left so

\hat{\boldsymbol{e}}_{n}

points to the left. At C it is slowing down and turning to the right. So the acceleration must be a positive vector pointed in the quadrant shown.

If at C you know that the particle is slowing down then you know that $\dot{v}<0$ . But you also can see the curve is to the right so $\hat{\boldsymbol{e}}_{n}$ is to the right. So the acceleration must be in the quadrant shown.

One can use the information about the curvature to further restrict the possible accelerations at point A also. At point B there is nothing more to know unless you know how the speed is changing with time.

Earlier we found the curvature by assuming the particle’s path was parameterized by arc length $s$ . A second way of calculating the curvature ${\accentset{\rightharpoonup}{\boldsymbol{\kappa}}}$ (and then the unit normal $\hat{\boldsymbol{e}}_{n}$ ) is to calculate the normal part of the acceleration. First calculate the acceleration. Then subtract from the acceleration that part which is parallel to the velocity.

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}-({\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot\hat{% \boldsymbol{e}}_{t})\hat{\boldsymbol{e}}_{t}={\accentset{\rightharpoonup}{% \boldsymbol{a}}}-{({\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{v}}}){\accentset{\rightharpoonup}{% \boldsymbol{v}}}\over v^{2}}

The normal acceleration is ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}=v^{2}{\accentset{% \rightharpoonup}{\boldsymbol{\kappa}}}$ so

{\accentset{\rightharpoonup}{\boldsymbol{\kappa}}}={{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{n}\over v^{2}}={{\accentset{\rightharpoonup% }{\boldsymbol{a}}}\over v^{2}}-{({\accentset{\rightharpoonup}{\boldsymbol{a}}}% \cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}){\accentset{\rightharpoonup% }{\boldsymbol{v}}}\over v^{4}}.

Recipes for path coordinates.

Assume that you know the position as a function of time in either cartesian or polar coordinates. Then, say, at a particular time of interest when the particle is at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ , you can calculate the velocity of the particle using:

{\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{x}\hat{\boldsymbol{\imath}}% +\dot{y}\hat{\boldsymbol{\jmath}}+\dot{z}\hat{\boldsymbol{k}}\quad\mbox{or}% \quad{\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{R}\hat{\boldsymbol{e}}% _{\scriptscriptstyle\!R}+R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle% \!\theta}+\dot{z}\hat{\boldsymbol{k}}

and the acceleration using

{\accentset{\rightharpoonup}{\boldsymbol{a}}}=\ddot{x}\hat{\boldsymbol{\imath}% }+\ddot{y}\hat{\boldsymbol{\jmath}}+\ddot{z}\hat{\boldsymbol{k}}\quad\mbox{or}% \quad{\accentset{\rightharpoonup}{\boldsymbol{a}}}=(\ddot{R}-R{\dot{\theta}}^{% 2})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{% \theta})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+\ddot{z}\hat{% \boldsymbol{k}}.

From these expressions we can calculate all the quantities used in the path coordinate description. So we repeat what we have already said but in an algorithmic form. Here is one set of steps one can follow. This recipe is of little practical use, but does show that the motion explicitly determines the path base vectors as well as the osculating circle.

1.

Calculate $\hat{\boldsymbol{e}}_{t}={\accentset{\rightharpoonup}{\boldsymbol{v}}}/|{% \accentset{\rightharpoonup}{\boldsymbol{v}}}|$ .
2.

Calculate $a_{t}={\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{v}}}/v$ .
3.

Calculate ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}-{({\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{v}}}){\accentset{\rightharpoonup}{% \boldsymbol{v}}}/v^{2}}$ .
4.

Calculate $\hat{\boldsymbol{e}}_{n}={{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}% \over|{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}|}$
5.

Calculate the radius of curvature as $\rho={{{|{\accentset{\rightharpoonup}{\boldsymbol{v}}}|}^{2}}\over|{\accentset% {\rightharpoonup}{\boldsymbol{a}}}_{n}|}$ .
6.

Write a parametric equation for the osculating circle as

${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm osculating}=\underbrace{({% \accentset{\rightharpoonup}{\boldsymbol{r}}}+\rho\hat{\boldsymbol{e}}_{n})}_{% \mbox{center}}+\underbrace{\rho(-\cos\phi\hat{\boldsymbol{e}}_{n}+\sin\phi\hat% {\boldsymbol{e}}_{t})}_{\mbox{circle}}$

where $\phi$ is the parameter used to parameterize the points on the circle ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm osculating}$ of the point on the curve ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ . As $\phi$ ranges from 0 to $2\pi$ the point ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm osculating}$ goes from ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ around the circle and back. The plane of the osculating circle is determined by $\hat{\boldsymbol{e}}_{t}$ and $\hat{\boldsymbol{e}}_{n}$ . For planar curves, the osculating circle is in the plane of the curve.

Filename:sfig8-5-wiper

Figure 18.12: The path of a rock in a tire of radius

R

is shown as are the unit tangent, the unit normal and the osculating circle at the top position.

Example: Particle on the rim of a tire

A particle P on the rim of a tire whose center is moving at constant speed $v$ has position $r$ given by

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=\left(vt-R\sin(vt/R)\right)\hat{% \boldsymbol{\imath}}+R\left(1-\cos(vt/R)\right)\hat{\boldsymbol{\jmath}}

where the origin is at the ground contact time $t=0$ . When the particle is at its highest point $vt=\pi R$ and

{\accentset{\rightharpoonup}{\boldsymbol{v}}}=2v\hat{\boldsymbol{\imath}}\hbox to% 36.135pt{\hfil and\hfil}{\accentset{\rightharpoonup}{\boldsymbol{a}}}=-(v^{2}/% R)\hat{\boldsymbol{\jmath}}.

At that midpoint

	$\displaystyle\hat{\boldsymbol{e}}_{t}=\hat{\boldsymbol{\imath}},\quad\hat{% \boldsymbol{e}}_{n}=-\hat{\boldsymbol{\jmath}},{\accentset{\rightharpoonup}{% \boldsymbol{\kappa}}}=-(1/(4R))\hat{\boldsymbol{\jmath}}$
	$\displaystyle\hbox to36.135pt{\hfil and\hfil}(2v)^{2}/\rho=v^{2}/R\Rightarrow% \rho=4R$

as shown in fig. 18.12. The osculating circle has 4 times the radius of the tire. Note the intimacy of the osculating circle’s kiss with the cycloidal path.

Summary of polar(cylindrical) coordinates

See also the inside back cover, table II, row 3 for future reference.

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}=R\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!R}+z\hat{\boldsymbol{k}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}=v_{R}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}+v_{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}+v_{z}\hat{\boldsymbol{k}}=\dot{R}\hat{\boldsymbol{% e}}_{\scriptscriptstyle\!R}+R\dot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}+\dot{z}\hat{\boldsymbol{k}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}=a_{R}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}+a_{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}+a_{z}\hat{\boldsymbol{k}}=(\ddot{R}-\dot{\theta}^{% 2}R)\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+(R\ddot{\theta}+2\dot{R}\dot{% \theta})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+\ddot{z}\hat{% \boldsymbol{k}}$

	$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=({\accentset{% \rightharpoonup}{\boldsymbol{r}}}-z\hat{\boldsymbol{k}})/\|{\accentset{% \rightharpoonup}{\boldsymbol{r}}}-z\hat{\boldsymbol{k}}\|$
	$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=\hat{% \boldsymbol{k}}\times\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$

For 2-D problems just set $z=0,\dot{z}=0$ , and $\ddot{z}=0$ in these equations.

Summary of path coordinates

See the inside back cover table II, row 4 and the text under the table for future reference:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}=\mbox{no simple % expression in terms of path base vectors}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{s}\hat{% \boldsymbol{e}}_{t}=v\hat{\boldsymbol{e}}_{t}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{t}+{\accentset{\rightharpoonup}{\boldsymbol% {a}}}_{n}$
	$\displaystyle\qquad{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}=a_{t}\hat% {\boldsymbol{e}}_{t}=\dot{v}\hat{\boldsymbol{e}}_{t}=\ddot{s}\hat{\boldsymbol{% e}}_{t}=({\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot\hat{\boldsymbol{e}% }_{t})\hat{\boldsymbol{e}}_{t}$
	$\displaystyle\qquad{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}=a_{n}\hat% {\boldsymbol{e}}_{n}=(v^{2}/\rho)\hat{\boldsymbol{e}}_{n}={\accentset{% \rightharpoonup}{\boldsymbol{a}}}-({\accentset{\rightharpoonup}{\boldsymbol{a}% }}\cdot\hat{\boldsymbol{e}}_{t})\hat{\boldsymbol{e}}_{t}$

	$\displaystyle\hat{\boldsymbol{e}}_{t}=d{\accentset{\rightharpoonup}{% \boldsymbol{r}}}/ds={\accentset{\rightharpoonup}{\boldsymbol{v}}}/v$
	$\displaystyle\hat{\boldsymbol{e}}_{n}={\accentset{\rightharpoonup}{\boldsymbol% {\kappa}}}/\|{\accentset{\rightharpoonup}{\boldsymbol{\kappa}}}\|=\rho{% \accentset{\rightharpoonup}{\boldsymbol{\kappa}}}=({\accentset{\rightharpoonup% }{\boldsymbol{a}}}-({\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot\hat{% \boldsymbol{e}}_{t})\hat{\boldsymbol{e}}_{t})/\|{\accentset{\rightharpoonup}{% \boldsymbol{a}}}-({\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot\hat{% \boldsymbol{e}}_{t})\hat{\boldsymbol{e}}_{t}\|$
	$\displaystyle\hat{\boldsymbol{e}}_{b}=\hat{\boldsymbol{e}}_{t}\times\hat{% \boldsymbol{e}}_{n}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\kappa}}}=d\hat{% \boldsymbol{e}}_{t}/ds=({\accentset{\rightharpoonup}{\boldsymbol{a}}}-({% \accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot\hat{\boldsymbol{e}}_{t})\hat% {\boldsymbol{e}}_{t})/v^{2}\qquad\rho=1/\|{\accentset{\rightharpoonup}{% \boldsymbol{\kappa}}}\|$

Both polar coordinates and path coordinates define base vectors in terms of the motion of a particle of interest relative to a fixed coordinate system.

SAMPLE 18.1

Filename:sfig8-5-wiper-a

Figure 18.13:

Acceleration in polar coordinates. A bug walks along the spiral section of a natural shell. The path of the bug is described by the equation $R=R_{0}{\rm e}^{a\theta}$ where $a=0.182$ and $R_{0}=5\,{\rm mm}$ . The bug’s radial distance from the center of the spiral is seen to be increasing at a constant rate of $2\,{\rm mm}/\,\mathrm{s}$ . Find the $x$ and $y$ components of the acceleration of the bug at $\theta=\pi$ .

Solution In polar coordinates, the acceleration of a particle in planar motion is

{\accentset{\rightharpoonup}{\boldsymbol{a}}}=(\ddot{R}-R\dot{\theta}^{2})\hat% {\boldsymbol{e}}_{\scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}.

Since we know the position of the bug,

$\displaystyle R$	$\displaystyle=$	$\displaystyle R_{0}\,{\rm e}^{a\theta},$
$\displaystyle\dot{R}$	$\displaystyle=$	$\displaystyle R_{0}\,a{\rm e}^{a\theta}\dot{\theta}\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}\dot{\theta}=\frac{\dot{R}}{R_{0}a\,{\rm e}^{a\theta}},$
$\displaystyle\ddot{R}$	$\displaystyle=$	$\displaystyle R_{0}\,a{\rm e}^{a\theta}\ddot{\theta}+R_{0}\,a^{2}{\rm e}^{a% \theta}\dot{\theta}^{2}.$

Since the radial distance R of the bug is increasing at a constant rate $\dot{R}~{}=~{}2~{}\,{\rm mm}/s$ , $\ddot{R}=0$ , that is,

$\displaystyle R_{0}\,a{\rm e}^{a\theta}(\ddot{\theta}+a\dot{\theta}^{2})$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\ddot{\theta}$	$\displaystyle=$	$\displaystyle-a\dot{\theta}^{2}$
	$\displaystyle=$	$\displaystyle-\frac{\dot{R}^{2}}{R^{2}_{0}a\,{\rm e}^{2a\theta}}.$

Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle(\ddot{R}-R\dot{\theta}^{2})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!\theta}$
	$\displaystyle=$	$\displaystyle\left(0-R_{0}{\rm e}^{a\theta}\cdot\frac{\dot{R}^{2}}{R^{2}_{0}a^% {2}{\rm e}^{2a\theta}}\right)\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+% \left(\frac{2\dot{R}^{2}}{R_{0}a\,{\rm e}^{a\theta}}+R_{0}{\rm e}^{a\theta}% \cdot\frac{-\dot{R}^{2}}{R^{2}_{0}a\,{\rm e}^{2a\theta}}\right)\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
	$\displaystyle=$	$\displaystyle\frac{\dot{R}^{2}}{R_{0}a\,{\rm e}^{a\theta}}\left[-\frac{1}{a}% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+(2-1)\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}\right].$

Now substituting $R_{0}=5\,{\rm mm},\,a=0.182,\,\dot{R}=2\,{\rm mm}/s$ , and $\theta=\pi$ in the above expression, we get

{\accentset{\rightharpoonup}{\boldsymbol{a}}}=(-13.63\,{\rm mm}/{\,\mathrm{s}}% ^{2})\,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+(2.48\,{\rm mm}/{\,\mathrm% {s}}^{2})\,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}.

But, at $\theta=\pi$

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=\cos\theta\hat{\boldsymbol{\imath% }}+\sin\theta\hat{\boldsymbol{\jmath}}=-\hat{\boldsymbol{\imath}}\hbox to36.13% 5pt{\hfil and\hfil}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=-\sin% \theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{\jmath}}=-\hat{% \boldsymbol{\jmath}},

therefore,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle(13.63\,{\rm mm}/{\,\mathrm{s}}^{2})\,\hat{\boldsymbol{\imath}}-(% 2.48\,{\rm mm}/{\,\mathrm{s}}^{2})\,\hat{\boldsymbol{\jmath}},$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}a_{x}$	$\displaystyle=$	$\displaystyle 13.63\,{\rm mm}/{\,\mathrm{s}}^{2}\hbox to36.135pt{\hfil and% \hfil}a_{y}=-2.48\,{\rm mm}/{\,\mathrm{s}}^{2}.$

Answer: $a_{x}=13.63\,{\rm mm}/{\,\mathrm{s}}^{2},\;a_{y}=-2.48\,{\rm mm}/{\,\mathrm{s}% }^{2}$

SAMPLE 18.2 Going back and forth between $(x,y)$ and $(R,\theta)$ . Given the position of a particle in polar coordinates $(R,\theta)$ and its radial and angular velocity $(\dot{R},\dot{\theta})$ and radial and angular acceleration $(\ddot{R},\ddot{\theta})$ , find $(\dot{x},\dot{y})$ and $(\ddot{x},\ddot{y})$ . Also, find the inverse relationship.

Solution

•

Polar to Cartesian: In polar coordinates, we are given $R,\theta,\dot{R},\dot{\theta},\ddot{R}$ , and $\ddot{\theta}$ . We need to find $\dot{x},\dot{y},\ddot{x}$ , and $\ddot{y}$ . Let us consider the velocity first. The velocity of a point is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{x}\hat{\boldsymbol{\imath}}% +\dot{y}\hat{\boldsymbol{\jmath}}$ in cartesian coordinates and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{R}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ in polar coordinates. Thus,

\dot{x}\hat{\boldsymbol{\imath}}+\dot{y}\hat{\boldsymbol{\jmath}}=\dot{R}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}+R\dot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}.

where $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=\cos\theta\hat{\boldsymbol{\imath% }}+\sin\theta\hat{\boldsymbol{\jmath}}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=-\sin\theta\hat{\boldsymbol{% \imath}}+\cos\theta\hat{\boldsymbol{\jmath}}$ . Dotting this equation with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ , respectively, we get

$\displaystyle\dot{x}$	$\displaystyle=$	$\displaystyle\dot{R}\overbrace{(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}% \cdot\hat{\boldsymbol{\imath}})}^{\cos\theta}+R\dot{\theta}\overbrace{(\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}\cdot\hat{\boldsymbol{\imath}})}^{% -\sin\theta}$
$\displaystyle\dot{y}$	$\displaystyle=$	$\displaystyle\dot{R}\underbrace{(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}% \cdot\hat{\boldsymbol{\jmath}})}_{\sin\theta}+R\dot{\theta}\underbrace{(\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}\cdot\hat{\boldsymbol{\jmath}})}_{% \cos\theta}$
or,
$\displaystyle\dot{x}$	$\displaystyle=$	$\displaystyle\dot{R}\cos\theta+R\dot{\theta}(-\sin\theta)$
$\displaystyle\dot{y}$	$\displaystyle=$	$\displaystyle\dot{R}\sin\theta+R\dot{\theta}\cos\theta$
or,
$\displaystyle\left\{\begin{array}[]{c}\dot{x}\\ \dot{y}\end{array}\right\}$	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{rr}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{array}\right]\left[\begin{array}[]{rr}1&0\\ 0&R\end{array}\right]\left\{\begin{array}[]{c}\dot{R}\\ \dot{\theta}\end{array}\right\}.$	(18.13)

Thus given $\dot{R}$ and $\dot{\theta}$ at $(R,\theta)$ , we can find $\dot{x}$ and $\dot{y}$ . Similarly, from the acceleration formula, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=\ddot{x}\hat{\boldsymbol{\imath}% }+\ddot{y}\hat{\boldsymbol{\jmath}}=(\ddot{R}-R\dot{\theta}^{2})\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ , we derive

\displaystyle\left\{\begin{array}[]{c}\ddot{x}\\ \ddot{y}\end{array}\right\}

\displaystyle=

\displaystyle\left[\begin{array}[]{rr}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{array}\right]\left(\left[\begin{array}[]{rr}1&0\\ 0&R\end{array}\right]\left\{\begin{array}[]{c}\ddot{R}\\ \ddot{\theta}\end{array}\right\}+\left\{\begin{array}[]{c}-R\dot{\theta}^{2}\\ 2\dot{R}\dot{\theta}\end{array}\right\}\right).

(18.24)

It is not necessary to split the terms on the right hand side. We could have kept them together as $(\ddot{R}-R\dot{\theta}^{2})$ and $(2\dot{R}\dot{\theta}+R\ddot{\theta})$ but we split them to keep the radial acceleration term $\ddot{R}$ and angular acceleration $\ddot{\theta}$ in evidence.

•

Cartesian to Polar: Given $\dot{x},\dot{y},\ddot{x}$ , and $\ddot{y}$ at $(x,y)$ , we can now find $\dot{R},\dot{\theta},\ddot{R}$ , and $\ddot{\theta}$ easily by inverting eqn. (18.13) and eqn. (LABEL:eq:10.1.aptoac):

$\displaystyle\left\{\begin{array}[]{c}\dot{R}\\ \dot{\theta}\end{array}\right\}$ $\displaystyle=$ $\displaystyle\left[\begin{array}[]{rr}1&0\\ 0&\frac{1}{R}\end{array}\right]\left[\begin{array}[]{rr}\cos\theta&\sin\theta% \\ -\sin\theta&\cos\theta\end{array}\right]\left\{\begin{array}[]{c}\dot{x}\\ \dot{y}\end{array}\right\}.$ (18.34)

$\displaystyle\left\{\begin{array}[]{c}\ddot{R}\\ \ddot{\theta}\end{array}\right\}$ $\displaystyle=$ $\displaystyle\left[\begin{array}[]{rr}1&0\\ 0&\frac{1}{R}\end{array}\right]\left(\left[\begin{array}[]{rr}\cos\theta&\sin% \theta\\ -\sin\theta&\cos\theta\end{array}\right]\left\{\begin{array}[]{c}\ddot{x}\\ \ddot{y}\end{array}\right\}-\left\{\begin{array}[]{c}-R\dot{\theta}^{2}\\ 2\dot{R}\dot{\theta}\end{array}\right\}\right).$ (18.45)

Note that in eqn. (LABEL:eq:10.1.actoap) we need $\dot{R}$ and $\dot{\theta}$ in order to compute $\ddot{R}$ and $\ddot{\theta}$ . This, however, is no problem since we have $\dot{R}$ and $\dot{\theta}$ from eqn. (18.34). Of course, $R$ and $\theta$ are required too, which are easily computed as $R=\sqrt{x^{2}+y^{2}}$ and $\theta=\tan^{-1}(y/x)$ .

SAMPLE 18.3 Velocity in path coordinates. The path of a particle, stuck at the edge of a disk rolling on a level ground with constant speed, is called a cycloid. The parametric equations of a cycloid described by a particle is $x=t-\sin{t}$ , $y=1-\cos{t}$ where $t$ is a dimensionless time. Find the velocity of the particle at

1.

$t=\frac{\pi}{2}$ ,
2.

$t=\pi$ , and
3.

$t=2\pi$

and express the velocity in terms of path basis vectors $(\hat{\boldsymbol{e}}_{t},\hat{\boldsymbol{e}}_{n})$ .

Solution The position of the particle is given:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=$	$\displaystyle x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle(t-\sin{t})\hat{\boldsymbol{\imath}}+(1-\cos{t})\hat{\boldsymbol{% \jmath}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\quad{\accentset{% \rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle\equiv$	$\displaystyle\frac{d{\accentset{\rightharpoonup}{\boldsymbol{r}}}}{dt}$
	$\displaystyle=$	$\displaystyle(1-\cos{t})\hat{\boldsymbol{\imath}}+\sin{t}\hat{\boldsymbol{% \jmath}},$
$\displaystyle\mbox{and}\quad\quad\quad v$	$\displaystyle=$	$\displaystyle\|{\accentset{\rightharpoonup}{\boldsymbol{v}}}\|$
	$\displaystyle=$	$\displaystyle\sqrt{(1-\cos{t})^{2}+\sin^{2}{t}}$
	$\displaystyle=$	$\displaystyle\sqrt{2-2\cos{t}}.$	(18.49)

In terms of path basis vectors, the velocity is given by

{\accentset{\rightharpoonup}{\boldsymbol{v}}}=v\hat{\boldsymbol{e}}_{t}\quad% \mbox{where}\quad\hat{\boldsymbol{e}}_{t}={\accentset{\rightharpoonup}{% \boldsymbol{v}}}/v.

Here,

\hat{\boldsymbol{e}}_{t}=\frac{(1-\cos{t})\hat{\boldsymbol{\imath}}+\sin{t}% \hat{\boldsymbol{\jmath}}}{\sqrt{2-2\cos{t}}}.

(18.50)

Substituting the values of $t$ in equations 18.49 and 18.50 we get

1.

at $t=\frac{\pi}{2}$ :

$v=\sqrt{2},\quad\hat{\boldsymbol{e}}_{t}=\frac{1}{\sqrt{2}}(\hat{\boldsymbol{% \imath}}+\hat{\boldsymbol{\jmath}}),\quad{\accentset{\rightharpoonup}{% \boldsymbol{v}}}=\sqrt{2}\hat{\boldsymbol{e}}_{t}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\sqrt{2}\hat{\boldsymbol{e}}_{t}% ,\quad\hat{\boldsymbol{e}}_{t}=\frac{1}{\sqrt{2}}(\hat{\boldsymbol{\imath}}+% \hat{\boldsymbol{\jmath}})$
2.

at $t=\pi$ :

$v=2,\quad\hat{\boldsymbol{e}}_{t}=\hat{\boldsymbol{\imath}},\quad{\accentset{% \rightharpoonup}{\boldsymbol{v}}}=2\hat{\boldsymbol{e}}_{t}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=2\hat{\boldsymbol{e}}_{t},\quad% \hat{\boldsymbol{e}}_{t}=\hat{\boldsymbol{\imath}}$
3.

at $t=2\pi$ :

$v=0,\quad\hat{\boldsymbol{e}}_{t}={\rm undefined},\quad{\accentset{% \rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$

SAMPLE 18.4 Path coordinates in 2-D. A particle traverses a limacon $R=(1+2\cos\theta)\,\mathrm{ft},$ with constant angular speed $\dot{\theta}=3\,\mathrm{rad}/\mathrm{s}$ .

1.

Find the normal and tangential accelerations ( $a_{t}$ and $a_{n}$ ) of the particle at $\theta=\frac{\pi}{2}$ .
2.

Find the radius of the osculating circle and draw the circle at $\theta=\frac{\pi}{2}$ .

Solution

The equation of the path is

R=(1+2\cos\theta)\,\mathrm{ft}.

Filename:sfig8-5-2disks

Figure 18.14: The limacon

R=(1+2\cos\theta)\,\mathrm{ft}

The path is shown in Fig. 18.14. Since the equation of the path is given in polar coordinates, we can calculate the velocity and acceleration using the polar coordinate formulae:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle\dot{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R\dot{\theta}% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$		(18.51)
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle(\ddot{R}-R\dot{\theta}^{2})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!\theta}.$		(18.52)

So, we need to find $\dot{R},\ddot{R},\ddot{\theta}$ for computing ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ . From the given equation for $R$

$\displaystyle R$	$\displaystyle=$	$\displaystyle(1+2\cos\theta)\,\mathrm{ft}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{R}$	$\displaystyle=$	$\displaystyle-(2\,\mathrm{ft})\sin\theta\,\dot{\theta}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{R}$	$\displaystyle=$	$\displaystyle-(2\,\mathrm{ft})\sin\theta\,\ddot{\theta}\;-(2\,\mathrm{ft})\cos% \theta\,\dot{\theta}^{2}$
	$\displaystyle=$	$\displaystyle-(2\,\mathrm{ft})\,\dot{\theta}^{2}\cos\theta$

where we set $\ddot{\theta}=0$ because $\dot{\theta}=$ constant. Substituting these expressions in Eqn. (18.51) and (18.52), we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle-(2\,\mathrm{ft})\dot{\theta}\sin\theta\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+\dot{\theta}(1+\cos\theta)\,\mathrm{ft}\hat{\boldsymbol% {e}}_{\scriptscriptstyle\!\theta}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle(\ddot{R}-R\dot{\theta}^{2})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!\theta}$
	$\displaystyle=$	$\displaystyle[-2\dot{\theta}^{2}\,\cos\theta-(1+2\cos\theta)\,\dot{\theta}^{2}% ]\,\mathrm{ft}\,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+(-2\dot{\theta}^{% 2}\sin\theta)\,\mathrm{ft}\,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
	$\displaystyle=$	$\displaystyle-\dot{\theta}^{2}[(1+4\cos\theta)\,\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(2\sin\theta)\,\hat{\boldsymbol{e}}_{\scriptscriptstyle% \!\theta}]\,\mathrm{ft}$

which give velocity and acceleration at any $\theta$ . Now substituting $\theta=\pi/2$ we get the velocity and acceleration at the desired point:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}\|_{\frac{\pi}{2}}$	$\displaystyle=$	$\displaystyle\dot{\theta}[-2\sin\frac{\pi}{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(1+2\cos\frac{\pi}{2})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}]\,\mathrm{ft}$
	$\displaystyle=$	$\displaystyle 3\,{\mathrm{ft/s}}(-2\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R% }+\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta})$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}\|_{\frac{\pi}{2}}$	$\displaystyle=$	$\displaystyle-9\,{\mathrm{ft}}/{\mathrm{s}}^{2}(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+2\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}).$

Filename:sfig8-4-1

Figure 18.15: The unit vectors

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R},\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta},

and

\hat{\boldsymbol{e}}_{t},\hat{\boldsymbol{e}}_{n}

\theta=\pi/2

Thus we know the velocity and the acceleration of the particle in polar coordinates. Now we proceed to find the tangential and the normal components of acceleration (acceleration in path coordinates). In path coordinates

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{n}$
		$\displaystyle\equiv$	$\displaystyle a_{t}\hat{\boldsymbol{e}}_{t}+a_{n}\hat{\boldsymbol{e}}_{n}$

where $\hat{\boldsymbol{e}}_{t}$ and $\hat{\boldsymbol{e}}_{n}$ are unit vectors in the directions of the tangent and the principle normal of the path. We compute these unit vectors as follows.

$\displaystyle\hat{\boldsymbol{e}}_{t}$	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{v}}}}{\|{\accentset% {\rightharpoonup}{\boldsymbol{v}}}\|}$
	$\displaystyle=$	$\displaystyle(\frac{-6\,{\mathrm{ft/s}})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(3\,{\mathrm{ft/s}})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}}{\sqrt{45}\;\,{\mathrm{ft/s}}}$
	$\displaystyle=$	$\displaystyle-\frac{2}{\sqrt{5}}\,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}% +\frac{1}{\sqrt{5}}\,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}.$
So,
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}$	$\displaystyle=$	$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot\hat{% \boldsymbol{e}}_{t})\,\hat{\boldsymbol{e}}_{t}$
	$\displaystyle=$	$\displaystyle 9\,{\mathrm{ft}}/{\mathrm{s}}^{2}\left(-\frac{2}{\sqrt{5}}+\frac% {2}{\sqrt{5}}\right)\hat{\boldsymbol{e}}_{t}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}},$
and
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}-{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{t}$
	$\displaystyle=$	$\displaystyle-9\,{\mathrm{ft}}/{\mathrm{s}}^{2}(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+2\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}).$
Therefore,
$\displaystyle\hat{\boldsymbol{e}}_{n}$	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}}{\|{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}\|}$
	$\displaystyle=$	$\displaystyle-\frac{1}{\sqrt{5}}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+% 2\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}).$

Thus,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle 9\sqrt{5}\;\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{e}}_% {n}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}a_{t}$	$\displaystyle=$	$\displaystyle 0\hbox to36.135pt{\hfil and\hfil}a_{n}=20.12\,{\mathrm{ft}}/{% \mathrm{s}}^{2}.$

Answer: $a_{t}=0,\;a_{n}=20.12\,{\mathrm{ft}}/{\mathrm{s}}^{2}$

2.
Filename:sfig8-4-1a

Figure 18.16: The osculating circle of radius $\rho={5}^{1/2}\,\mathrm{ft}$ at $\theta=\pi/2$ . Note that $\hat{\boldsymbol{e}}_{n}$ points to the center of the osculating circle.

In path coordinates the acceleration is also expressed as

${\accentset{\rightharpoonup}{\boldsymbol{a}}}=\dot{v}\hat{\boldsymbol{e}}_{t}+% \frac{v^{2}}{\rho}\hat{\boldsymbol{e}}_{n}$

where $\rho$ is the radius of the osculating circle. Since we already know the speed $v$ and the normal component of acceleration $a_{n}$ we can easily compute the radius of the osculating circle.

$\displaystyle a_{n}$ $\displaystyle=$ $\displaystyle\frac{v^{2}}{\rho}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\rho$ $\displaystyle=$ $\displaystyle\frac{v^{2}}{a_{n}}=\frac{45(\,{\mathrm{ft/s}})^{2}}{9{\sqrt{5}}% \,{\mathrm{ft}}/{\mathrm{s}}^{2}}=\sqrt{5}\,\mathrm{ft}.$

Answer: $\rho=2.24\,\mathrm{ft}$

Problems for 18.1 Polar coordinates and path coordinates

18.1.1 A particle moves along the two paths (1) and (2) as shown.

(a)

In each case, determine the velocity of the particle in terms of $b$ , $\theta$ , and $\dot{\theta}$ . Answer: plot(2): $\theta(t)=br(t)$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\frac{\dot{\theta}}{b}\hat{% \boldsymbol{e}}_{r}+\frac{\theta\dot{\theta}}{b}\hat{\boldsymbol{e}}_{t}$ .
(b)

Find the $x$ and $y$ coordinates of the path as functions of $b$ and $r$ or $b$ and $\theta$ . Answer: plot(2): $x=r\cos(br)$ and $y=r\sin{br}$ or $x=\frac{\theta}{b}\cos(\theta)$ and $y=\frac{\theta}{b}\sin{\theta}$

Filename:sfig8-6-2

Figure 18.17

18.1.2 For the particle path (1) in problem 18.17, find the acceleration of the particle in terms of $b$ , $\theta$ , and $\dot{\theta}$ .

18.1.3 For the particle path (2) in problem 18.17, find the acceleration of the particle in terms of $b$ , $\theta$ , and $\dot{\theta}$ .

18.1.4 A body moves with constant velocity $V$ in a straight line parallel to and at a distance $d$ from the $x$ -axis.

(a)

Calculate $\dot{\theta}$ in terms of $V, d$ , and $\theta$ .
(b)

Calculate the $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ component of acceleration.

Carefully define, using a sketch and/or words, any variables, coordinate systems, and reference frames you use. Express your answers using any convenient coordinate system (just make sure its orientation has been clearly defined).

Filename:sfig8-6-2a

Figure 18.18

18.1.5 Picking apart the polar coordinate formula for velocity. This problem concerns a small mass $m$ that sits in a slot in a turntable. Alternatively you can think of a small bead that slides on a rod. The mass always stays in the slot (or on the rod). Assume the mass is a little bug that can walk as it pleases on the rod (or in the slot) and you control how the turntable/rod rotates. Name two situations in which one of the terms is zero but the other is not in the two term polar coordinate formula for velocity, $\dot{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ . You should thus gain some insight into the meaning of each of the two terms in that formula. Answer: One situation: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=R\dot{\theta}\hat{\boldsymbol{e}% }_{\scriptscriptstyle\!\theta}$ ;the case of a particle constrained to move in a circle.

Filename:efig1-2-28

Figure 18.19

18.1.6 Picking apart the polar coordinate formula for acceleration. Reconsider the configurations in problem 18.18. This time, name four situations in which all of the terms, but one, in the four term polar coordinate formula for acceleration, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=(\ddot{R}-R\dot{\theta}^{2})\hat% {\boldsymbol{e}}_{\scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ , are zero. Each situation should pick out a different term. You should thus gain some insight into the meaning of each of the four terms in that formula. Answer: One situation: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=\ddot{R}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}$ ; the case of no rotation ( $\dot{\theta}=\ddot{\theta}=0$ ).

18.1.7 The two differential equations:

	$\displaystyle\ddot{R}-R\dot{\theta}^{2}$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle 2\dot{R}\dot{\theta}+R\ddot{\theta}$	$\displaystyle=$	$\displaystyle 0$

have the general solution

	$\displaystyle R$	$\displaystyle=$	$\displaystyle\sqrt{d^{2}+(v(t-t_{0}))^{2}}$
	$\displaystyle\theta$	$\displaystyle=$	$\displaystyle\theta_{0}+\tan^{-1}(v(t-t_{0})/d)$

where $\theta_{0}$ , $d$ , $t_{0}$ , and $v$ are arbitrary constants. This solution could be checked by plugging back into the differential equations — you need not do this (tedious) substitution. The solution describes a curve in the plane. That is, if for a range of values of $t$ , the values of $R$ and $\theta$ were calculated and then plotted using polar coordinates, a curve would be drawn. What can you say about the shape of this curve?

[Hints:

(a)

Actually make a plot using some random values of the constants and see what the plot looks like.
(b)

Write the equation ${\bf F}=m{\bf a}$ for a particle in polar coordinates and think of a force that would be relevant to this problem.
(c)

The answer is something simple. ]

18.1.8 A car driver on a very boring highway is carefully monitoring her speed. Over a one hour period, the car travels on a curve with constant radius of curvature, $\rho=25\,{\rm mi}$ , and its speed increases uniformly from $50\,\mathrm{mph}$ to $60\,\mathrm{mph}$ . What is the acceleration of the car half-way through this one hour period, in path coordinates?

18.1.9 Find expressions for $\hat{\boldsymbol{e}}_{t}$ , $a_{t}$ , $a_{n}$ , $\hat{\boldsymbol{e}}_{n}$ , and the radius of curvature $\rho$ , at any position (or time) on the given particle paths for

(a)

problem 12.1, Answer:

$\displaystyle\hat{\boldsymbol{e}}_{t}$	$\displaystyle=$	$\displaystyle\frac{\frac{2t}{\,\mathrm{s}}\hat{\boldsymbol{\imath}}+e^{\frac{t% }{\,\mathrm{s}}}\hat{\boldsymbol{\jmath}}}{\sqrt{4\frac{t^{2}}{\,\mathrm{s}^{2% }}+e^{\frac{2t}{\,\mathrm{s}}}}}$
$\displaystyle a_{t}$	$\displaystyle=$	$\displaystyle\frac{4t\frac{\,\mathrm{m}}{\,\mathrm{s}^{3}}+e^{\frac{2t}{\,% \mathrm{s}}}\,\mathrm{m}/\mathrm{s}^{2}}{\sqrt{4\frac{t^{2}}{\,\mathrm{s}^{2}}% +e^{\frac{2t}{\,\mathrm{s}}}}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}$	$\displaystyle=$	$\displaystyle\frac{(2-2\frac{t}{\,\mathrm{s}})e^{\frac{2t}{\,\mathrm{s}}}\,% \mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{\imath}}+(4\frac{t^{2}}{\,\mathrm{s}% ^{2}}-4\frac{t}{\,\mathrm{s}})e^{\frac{t}{\,\mathrm{s}}}\,\mathrm{m}/\mathrm{s% }^{2}\hat{\boldsymbol{\jmath}}}{4\frac{t^{2}}{\,\mathrm{s}^{2}}+e^{\frac{2t}{% \,\mathrm{s}}}}$
$\displaystyle\hat{\boldsymbol{e}}_{n}$	$\displaystyle=$	$\displaystyle\frac{e^{\frac{t}{\,\mathrm{s}}}\hat{\boldsymbol{\imath}}-2\frac{% t}{\,\mathrm{s}}\hat{\boldsymbol{\jmath}}}{\sqrt{4\frac{t^{2}}{\,\mathrm{s}^{2% }}+e^{\frac{2t}{\,\mathrm{s}}}}}$
$\displaystyle\rho$	$\displaystyle=$	$\displaystyle\frac{(4\frac{t^{2}}{\,\mathrm{s}^{2}}+e^{\frac{2t}{\,\mathrm{s}}% })}{(2-2\frac{t}{\,\mathrm{s}})e^{\frac{t}{\,\mathrm{s}}}}\,\mathrm{m}$

(b)

problem 12.1,
(c)

problem 18.37,
(d)

problem 12.1,
(e)

problem 12.1, and
(f)

problem 12.1.

18.1.10 A particle travels at non-constant speed on an elliptical path given by $y^{2}=b^{2}(1-\frac{x^{2}}{a^{2}})$ . Carefully sketch the ellipse for particular values of $a$ and $b$ . For various positions of the particle on the path, sketch the position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)$ ; the polar coordinate basis vectors $\hat{\boldsymbol{e}}_{r}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ ; and the path coordinate basis vectors $\hat{\boldsymbol{e}}_{n}$ and $\hat{\boldsymbol{e}}_{t}$ . At what points on the path are $\hat{\boldsymbol{e}}_{r}$ and $\hat{\boldsymbol{e}}_{n}$ parallel(or $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ and $\hat{\boldsymbol{e}}_{t}$ parallel)?

18.2 Rotating reference frames and their time-varying base vectors

In this section you will learn about rotating reference frames, how to take the derivative of a vector ‘in’ a rotating frame, and how to use that derivative to find the derivative in a Newtonian or fixed frame. We start by showing the alternative, just using one frame with one set of fixed base vectors.

The fixed base-vector method

To motivate the sections that follow we first show the “fixed base vector” method.

Filename:pfigure-blue-118-2

Figure 18.20: A bug walks at constant speed on a straight line (dotted) marked on a tire that is rotating clockwise (

\theta

relative to a vertical line). The distance of the line from the center of the tire is

s

. The distance the bug has walked from the closest point to the tire center is

\ell

. The fixed base vectors

\hat{\boldsymbol{\imath}}

and

\hat{\boldsymbol{\jmath}}

are aligned with the fixed

x

and

y

axis. The line segment connecting the tire center with the closest point on the line is in the direction

\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{\boldsymbol{\jmath}}

. The direction along the line on which the bug walks is

\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{\jmath}}

Consider the task of determining the acceleration of a bug walking at constant speed as it walks on a straight line marked on the surface of a tire rolling at constant rate. Artificial as this problem seems, it is similar to the sort of calculation needed in the kinematics of mechanisms. For now, imagine you really care how strong the bug’s legs need to be to hold on (unreasonably neglecting air friction). So knowing the bug’s acceleration determines the net force on it by ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ . Now we try to find ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ by taking two time derivatives of position.

If we want to avoid using rotating base vectors we have to write an expression for the position of the bug in terms of $x$ and $y$ components. Choosing a suitable origin of the coordinate system we have

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0^{% \prime}/0}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0^{% \prime}}}$
	$\displaystyle=\underbrace{\left(R\theta\hat{\boldsymbol{\imath}}+R\hat{% \boldsymbol{\jmath}}\right)}_{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \!\mathrm{0^{\prime}/0}}}+\par\par\par\underbrace{\left({s}(\cos\theta\hat{% \boldsymbol{\imath}}-\sin\theta\hat{\boldsymbol{\jmath}})+\ell(\sin\theta\hat{% \boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{\jmath}})\right)}_{{\accentset% {\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0^{\prime}}}}$
	$\displaystyle=\left(R\theta+s\cos\theta+\ell\sin\theta\right)\hat{\boldsymbol{% \imath}}+\left(R-s\sin\theta+\ell\cos\theta\right)\hat{\boldsymbol{\jmath}}.$	(18.53)

To find the velocity we take the time derivative, taking account that both $\theta$ and $\ell$ are functions of $t$ . Thus, for example looking at the term $\ell\cos\theta$ both the product rule and chain rule need be applied. Proceeding we get

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{P/0}}$	$\displaystyle=$	$\displaystyle\left(R\dot{\theta}-s\dot{\theta}\sin\theta+\dot{\ell}\sin\theta+% \ell\dot{\theta}\cos\theta\right)\hat{\boldsymbol{\imath}}$		(18.54)
			$\displaystyle+\left(-s\dot{\theta}\cos\theta+\dot{\ell}\cos\theta-\ell\dot{% \theta}\sin\theta\right)\hat{\boldsymbol{\jmath}}.$		(18.54)

To get the acceleration of the bug we differentiate one more time. This time we use the product rule and chain rule again, but get to use the simplification for this problem that the rolling and bug walking are at constant rate so $\ddot{\theta}=0$ and $\ddot{\ell}=0$ :

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/0}}$	$\displaystyle=$	$\displaystyle\phantom{+}\left(-s\dot{\theta}^{2}\cos\theta+\dot{\ell}\dot{% \theta}\cos\theta+\dot{\ell}\dot{\theta}\cos\theta-\ell\dot{\theta}^{2}\sin% \theta\right)\hat{\boldsymbol{\imath}}$	(18.55)
		$\displaystyle+\left(\phantom{-}s\dot{\theta}^{2}\,\sin\theta-\dot{\ell}\dot{% \theta}\,\sin\theta-\dot{\ell}\dot{\theta}\,\sin\theta-\ell\dot{\theta}^{2}% \cos\theta\right)\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle\phantom{+}\left(-\dot{\theta}^{2}(s\cos\theta+\ell\sin\theta)+2% \dot{\ell}\dot{\theta}\cos\theta\right)\hat{\boldsymbol{\imath}}$
		$\displaystyle+\left(\phantom{-}\dot{\theta}^{2}(s\,\sin\theta-\ell\cos\theta)-% 2\dot{\ell}\dot{\theta}\,\sin\theta\right)\hat{\boldsymbol{\jmath}}$

which is a bit of a mess. We could regroup the terms, but there would still be 6 of them.

The moving-reference-frame methods that follow don’t change this answer. But they give a somewhat simpler derivation. And they also group the terms in a physically meaningful way. One would be hard pressed to make sense of all the terms in eqn. (18.55). With the time-varying base vector methods below we can interpret the terms.

Reference frames

Filename:pfigure-s95f3a

Figure 18.21: A fixed reference frame

\cal F

is defined by an origin 0 and coordinate axes

x y z

or base vectors

\hat{\boldsymbol{\imath}}\hat{\boldsymbol{\jmath}}\hat{\boldsymbol{k}}

. Once the

x y

(or

\hat{\boldsymbol{\imath}}\hat{\boldsymbol{\jmath}}

) directions are chosen the

z

(or

\hat{\boldsymbol{k}}

) direction is implicitly defined by the right hand rule.

A reference frame is a coordinate system^†^†margin: A fine point for experts: reference frame vs coordinate system. There is a semantic debate about the degree to which the phrases “coordinate system” and “ reference frame” are synonymous. For simplicity we take them to mean the same thing. An alternative definition distinguishes reference frames from coordinate systems. It turns out that coordinate systems which are rotated, but not rotating, with respect to each other both calculate the same time-derivative of a given vector. Because these coordinate systems are equivalent in this regard they are sometimes called the same reference frame. That is, some people consider one reference frame to be the set of all coordinate systems that are glued to each other, no matter what their position or orientation. In this way of thinking, a frame is made manifest by the use of one of its coordinate systems, but no particular coordinate system is unique to the frame. . It has an origin and a set of preferred mutually orthogonal directions represented by base vectors. You can think of a reference frame as a giant piece of graph paper, or in 3-D as a giant jungle gym, that permeates space. It has the look of a wire frame. Because we will use various frames, we name them. We always have one frame that we think of as fixed for the purposes of Newtonian mechanics. We call this frame $\cal F$ (or sometimes $\cal N$ ). Most often we choose a frame that is ‘glued’ to the ground with an origin at a convenient point and with at least one base vector lined up with something convenient (e.g., up, sideways, along a slope, along the edge of an important part, etc.). $\cal F$ is a frame in which the mechanics laws we use are accurate. We define it by its origin and the direction of its coordinate axes, thus we would write

${\cal F}$ is $0xyz$ or ${\cal F}$ is $O\hat{\boldsymbol{\imath}}\hat{\boldsymbol{\jmath}}\hat{\boldsymbol{k}}$ .

where we would generally have a picture showing the position of the origin and the orientation of the coordinate axes (see fig. 18.21).

When we write casually ‘position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ ’ of a point we mean ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0}}$ . When we write ‘velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ ’ we mean $\displaystyle{d\over dt}{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ as calculated in $\cal F$ . That is, if ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=x\hat{\boldsymbol{\imath}}+y\hat% {\boldsymbol{\jmath}}+z\hat{\boldsymbol{k}}$ then we define the derivative of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ with respect to $t$ in $\cal F$ as

{{}^{\cal F}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}\over dt}={}^{% \lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}}=\dot{x}\hat{\boldsymbol{\imath}}+\dot{y}\hat{\boldsymbol{% \jmath}}+\dot{z}\hat{\boldsymbol{k}}.

The script $\cal F$ shows explicitly that when we take the time derivative of the vector we take the time derivative of its components, using the components associated with $\cal F$ and holding constant the base vectors associated with $\cal F$ . That is

$\displaystyle{{}^{\cal F}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/0}}\over dt}={}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{% \accentset{\rightharpoonup}{\boldsymbol{r}}}}_{{}_{\rm P/0}}$ are just fancy ways of writing what we have been calling ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ .

The elaborate notation just makes explicit how ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ is defined. The only need for this elaborate notation is if there is ambiguity. There is only ambiguity if more than one reference frame is used in a given problem.

Filename:Danef94s1q2

Figure 18.22: A second reference frame

\cal B

is defined by an origin 0’ and coordinate axes

x^{\prime}y^{\prime}z^{\prime}

or base vectors

\hat{\boldsymbol{\imath}}^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}\hat{% \boldsymbol{k}}^{\prime}

. Once the

x^{\prime}y^{\prime}

(or

\hat{\boldsymbol{\imath}}^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}

) directions are chosen the

z^{\prime}

(or

\hat{\boldsymbol{k}}^{\prime}

) direction is implicitly defined by the right hand rule.

Using more than one reference frame

Let’s add a second reference frame called $\cal B$ glued to and oriented with the roof of the building. We will always use script capital letters ( $\cal A,B,C,D,E,F$ or $\cal N$ ) to name reference frames. We define $\cal B$ by writing

${\cal B}$ is $0^{\prime}x^{\prime}y^{\prime}z^{\prime}$ or ${\cal B}$ is $O^{\prime}\hat{\boldsymbol{\imath}}^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}% \hat{\boldsymbol{k}}^{\prime}$

and by drawing a picture (see fig. 18.22).

This new frame, as we have drawn it, is also a good Newtonian or fixed frame. So we could write all positions using the $\cal B$ coordinates and base vectors and then proceed with all of our mechanics equations. The only confusion being that gravity doesn’t point in the $-\hat{\boldsymbol{\jmath}}^{\prime}$ direction, but in some crooked direction relative to $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ (which we would have to work out from the angle of the roof).

Although one hardly notices when using just a single fixed frame, we actually use frames for three somewhat distinct purposes:

I.

To define a vector. For example if we were tracking the motion of a cannon ball at P we could define its position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ as ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0}}$ , using frame $\cal F$ to define ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ . Or we could define ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ as ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}$ using frame $\cal B$ to define ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ .
II.

To assign coordinate values to a given vector. For example, the vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}$ could be written as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}},$

$\displaystyle 4\,\mathrm{m}\hat{\boldsymbol{\imath}}+5\,\mathrm{m}\hat{% \boldsymbol{\jmath}}$ $\displaystyle=$ $\displaystyle 6\,\mathrm{m}\hat{\boldsymbol{\imath}}^{\prime}-2.24\,\mathrm{m}% \hat{\boldsymbol{\jmath}}^{\prime}.$

Alternatively, if we just want to look at the components of a given vector we use $[]_{\cal F}$ to indicate the components of the vector in the square brackets, $[\ ]$ , using the base vectors of $\cal F$ . Thus

$[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}]_{\cal F}=[4\,% \mathrm{m},\quad 5\,\mathrm{m}]^{\prime}\hbox to36.135pt{\hfil and\hfil}[{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}]_{\cal B}=[6\,% \mathrm{m},\quad-2.24\,\mathrm{m}]^{\prime}$

where we have used $[]^{\prime}$ to put the components in their standard column form (although this is a picky detail). Note that although ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}$ that

$\displaystyle[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}]_% {\cal F}$ $\displaystyle\neq$ $\displaystyle[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}]_% {\cal B}$

$\displaystyle\mbox{because}\quad\left[\begin{array}[]{c}4\,\mathrm{m}\\ 5\,\mathrm{m}\\ \end{array}\right]$ $\displaystyle\neq$ $\displaystyle\left[\begin{array}[]{c}6\,\mathrm{m}\\ -2.24\,\mathrm{m}\\ \end{array}\right]$

That is, the components of a vector are different when expressed in different frames. Two different lists of numbers represent the same vector. The magnitude, however, is independent of frame ( $4^{2}+5^{2}=6^{2}+2.24^{2}$ , to two decimal places).
III.

To find the rate of change of a given vector. The position of P relative to A changes with time. We can calculate this rate of change two different ways. First using frame $\cal F$

$\displaystyle{{}^{\cal F}d{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{P/A}}\over dt}={}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}% }\dot{{\accentset{\rightharpoonup}{\boldsymbol{r}}}}_{\rm P/A}$ $\displaystyle=$ $\displaystyle\left({d\over dt}x_{\rm P/A}\right)\hat{\boldsymbol{\imath}}+% \left({d\over dt}y_{\rm P/A}\right)\hat{\boldsymbol{\jmath}}$

$\displaystyle\hbox{or more informally as}\quad{}^{\lxWithClass{rp-bodyframe}{% \mathcal{F}}}\dot{{\accentset{\rightharpoonup}{\boldsymbol{r}}}}$ $\displaystyle=$ $\displaystyle\dot{x}\hat{\boldsymbol{\imath}}+\dot{y}\hat{\boldsymbol{\jmath}}$

if we are clear in our minds that $x$ and $y$ are the coordinates of P relative to A. But we can also calculate the rate of change of the same vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$ using frame $\cal B$ as

$\displaystyle{{}^{\cal B}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}\over dt}={}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{{% \accentset{\rightharpoonup}{\boldsymbol{r}}}}_{\rm P/A}$ $\displaystyle=$ $\displaystyle\left({d\over dt}x^{\prime}_{\rm P/A}\right)\hat{\boldsymbol{% \imath}}^{\prime}+\left({d\over dt}y^{\prime}_{\rm P/A}\right)\hat{\boldsymbol% {\jmath}}^{\prime}$

$\displaystyle\hbox{or more informally as}\quad{}^{\lxWithClass{rp-bodyframe}{% \mathcal{B}}}\dot{{\accentset{\rightharpoonup}{\boldsymbol{r}}}}$ $\displaystyle=$ $\displaystyle\dot{x}^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+\dot{y}^{% \prime}\hat{\boldsymbol{\jmath}}^{\prime}.$

For the two frames $\cal F$ and $\cal B$ ${}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}}={}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset% {\rightharpoonup}{\boldsymbol{r}}}}$ because the two frames are not rotating relative to each other. Specifically, for $\cal F$ and $\cal B$ the formula for finding $x$ and $y$ from $x^{\prime}$ and $y^{\prime}$ does not involve time. Similarly, the formulas for finding $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ from $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ do not involve time. For frames that are rotated with respect to each other but not rotating, the two time derivatives of a given vector are related the same way the vector itself is related to itself in the two frames. The vectors are the same but their coordinates are different. That is, for rotated but not relatively rotating frames

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}={% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$ and $\displaystyle{{}^{\cal F}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}\over dt}={{}^{\cal B}d{\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{\!\mathrm{P/A}}\over dt}$

$\displaystyle\hbox{but}\quad[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}]_{\cal F}\neq[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}]_{\cal B}$ and $\displaystyle[{{}^{\cal F}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}\over dt}]_{\cal F}\neq[{{}^{\cal F}d{\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{\!\mathrm{P/A}}\over dt}]_{\cal B}.$

Going back and forth between these three uses of frames with ease is one of the advanced skills of a person who can analyze the dynamics of complex systems (And being confused about the distinctions is an almost universal part of learning advanced dynamics).

Example: Two fixed frames $\cal F$ and $\cal B$

Consider $\cal F$ and $\cal B$ both to be fixed to the ground. Let’s look at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$ where P is moving up at constant rate (see fig. 18.23). First look at the position using both frames:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}% \hbox to36.135pt{\hfil and\hfil}$
$\displaystyle d\hat{\boldsymbol{\imath}}+ct\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle\left(\sqrt{2}ct/2\right)\hat{\boldsymbol{\imath}}^{\prime}+\left% (\sqrt{2}ct/2\right)\hat{\boldsymbol{\jmath}}^{\prime}\mbox{\quad but}$
$\displaystyle\ [{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}% ]_{\cal F}=[d,\quad ct]^{\prime}\quad$	$\displaystyle\neq$	$\displaystyle\quad[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/% A}}]_{\cal B}=[\sqrt{2}ct/2,\quad\sqrt{2}ct/2]^{\prime}.$

Now look at the rate of change of position using both frames. First $\cal F$ :

	$\displaystyle{{}^{\cal F}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}\over dt}={}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{% \accentset{\rightharpoonup}{\boldsymbol{r}}}}_{\rm P/A}$	$\displaystyle=$	$\displaystyle\left({d\over dt}x_{\rm P/A}\right)\hat{\boldsymbol{\imath}}+% \left({d\over dt}y_{\rm P/A}\right)\hat{\boldsymbol{\jmath}}$
		$\displaystyle=$	$\displaystyle c\hat{\boldsymbol{\jmath}}$

Filename:pfigure-blue-123-1

Figure 18.23: A particle P is moving up at speed

c

. Two fixed frames

\cal F

and

\cal B

are used to keep track of it.

Then the rate of change of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$ as calculated in $\cal B$ :

	$\displaystyle{{}^{\cal B}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}\over dt}={}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{{% \accentset{\rightharpoonup}{\boldsymbol{r}}}}_{\rm P/A}$	$\displaystyle=$	$\displaystyle\left({d\over dt}x^{\prime}_{\rm P/A}\right)\hat{\boldsymbol{% \imath}}^{\prime}+\left({d\over dt}y^{\prime}_{\rm P/A}\right)\hat{\boldsymbol% {\jmath}}^{\prime}$
		$\displaystyle=$	$\displaystyle(\sqrt{2}c/2)\hat{\boldsymbol{\imath}}^{\prime}+(\sqrt{2}c/2)\hat% {\boldsymbol{\jmath}}^{\prime}=c\hat{\boldsymbol{\jmath}}$

You can quickly verify that ${{}^{\cal F}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}% \over dt}={{}^{\cal B}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}\over dt}$ by noting that $\hat{\boldsymbol{\imath}}^{\prime}=\sqrt{2}(\hat{\boldsymbol{\imath}}+\hat{% \boldsymbol{\jmath}})/2$ and $\hat{\boldsymbol{\jmath}}^{\prime}=\sqrt{2}(-\hat{\boldsymbol{\imath}}+\hat{% \boldsymbol{\jmath}})/2$ .

So long as $\cal B$ is not rotating with respect to $\cal F$ then the rate of change of a given vector is the same in both reference frames.

Filename:pfigure-blue-119-2

Figure 18.24: A third reference frame

\cal C

is defined by an origin 0^′′ and coordinate axes

x^{\prime\prime}y^{\prime\prime}z^{\prime\prime}

or base vectors

\hat{\boldsymbol{\imath}}^{\prime\prime}\hat{\boldsymbol{\jmath}}^{\prime% \prime}\hat{\boldsymbol{k}}^{\prime\prime}

. In this case

\cal C

is moving and rotating with respect to

\cal F

and

\cal B

. Once the

x^{\prime\prime}y^{\prime\prime}

(or

\hat{\boldsymbol{\imath}}^{\prime\prime}\hat{\boldsymbol{\jmath}}^{\prime\prime}

) directions are chosen the

z^{\prime\prime}

(or

\hat{\boldsymbol{k}}^{\prime\prime}

) direction is implicitly defined by the right hand rule.

Translating and rotating reference frames

Now look at a third reference frame $\cal C$ that is glued to the roof of the car as it starts up hill (see fig. 18.24). We define $\cal C$ by the origin of its coordinate system 0^′′ and its time-varying base vectors $\hat{\boldsymbol{\imath}}^{\prime\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime\prime}$ . The issues with defining a vector with $\cal C$ and with writing components using $\cal C$ are the same as for $\cal B$ . However taking the time derivative of a given vector in $\cal C$ is different than taking the time derivative in $\cal B$ or $\cal F$ because $\cal C$ is rotating relative to them.

Rate of change of a vector relative to a rotating frame: the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula

Because dynamics involves the time derivatives of so many different vectors (e.g. ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/C}$}}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ ) it is easier to think about the derivative of some arbitrary or general vector, call it ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ , and then apply what we learn to these other vectors.

Recalling our three uses of frames:

I.

To define a vector.
II.

To express the coordinates of a given vector.
III.

To take the time derivative of a vector.

we see that items [III.] and [I.] can be combined. That is, once a vector ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ is defined clearly by some means then we can define a new vector as the derivative of that vector in, say, moving frame $\cal C$ . Once this new vector ${}^{\lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}$ is defined it can be expressed in terms of the coordinates of any convenient frame.

Example: Derivative in a moving frame of a constant vector

Consider as ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ the relative position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$ of the points A and P that do not move in the fixed frame $\cal F$ . That is, the points A and P don’t move in the ordinary sense of the words (see fig. 18.25). Now also look at the frame $\cal B$ that is rotating with respect to $\cal F$ at the rate $\dot{\theta}$ . We have

Filename:pfigure-s95q14

Figure 18.25: A fixed frame

\cal F

and a rotating frame

\cal B

both keep track of the position vector

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$
	$\displaystyle\ell\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle\ell\sin\theta\hat{\boldsymbol{\imath}}^{\prime}+\ell\cos\theta% \hat{\boldsymbol{\jmath}}^{\prime}$

So we can now calculate the derivative in each frame by holding the corresponding base vectors as constant. So

$\displaystyle{{}^{\cal F}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}\over dt}$	$\displaystyle=$	$\displaystyle\dot{\ell}\hat{\boldsymbol{\jmath}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil and\hfil}{{}^{\cal B}d{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}\over dt}$	$\displaystyle=$	$\displaystyle\ell\dot{\theta}\cos\theta\hat{\boldsymbol{\imath}}^{\prime}-\ell% \dot{\theta}\sin\theta\hat{\boldsymbol{\jmath}}^{\prime}$
	$\displaystyle=$	$\displaystyle\ell\dot{\theta}\underbrace{\left(\cos\theta\hat{\boldsymbol{% \imath}}^{\prime}-\sin\theta\hat{\boldsymbol{\jmath}}^{\prime}\right)}_{\hat{% \boldsymbol{\imath}}}$
	$\displaystyle=$	$\displaystyle\ell\dot{\theta}\hat{\boldsymbol{\imath}}$

That is, the stationary vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$ and the rotating frame $\cal B$ define a new vector, the derivative of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$ in $\cal B$ . This is also ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}-% {\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A/\cal B}}$ , the difference between the velocity of P and the velocity of A in the frame $\cal B$ . This new vector can be expressed in any coordinate system of choice for example the $\hat{\boldsymbol{\imath}}\hat{\boldsymbol{\jmath}}$ system. So we wrote above

{{}^{\cal B}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}% \over dt}=\ell\dot{\theta}\hat{\boldsymbol{\imath}}

which looks mixed up but isn’t. The frame $\cal B$ is used to help define a vector which is then expressed in the coordinates of $\cal F$ .

Using the moving-frame derivative to calculate the fixed-frame derivative

Given a new vector ${}^{\lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}$ , the derivative of ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ as calculated in a rotating frame $\cal C$ , one calculation of common use is the determination of the derivative ${}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}$ of the same vector in the fixed frame $\cal F$ .

First think of a line segment that is marked between two points that are glued to a moving frame $\cal C$ . We know (at least in 2-D and for fixed axis rotation) that

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A}.

Filename:Danef94s3q2

Figure 18.26: A fixed frame

\cal F

defined by

O x y z

and a moving frame

\cal C

defined by

O^{\prime}x^{\prime}y^{\prime}z^{\prime}

. Also shown is an arbitrary vector

{\accentset{\rightharpoonup}{\boldsymbol{Q}}}

which changes relative to both frames.

Likewise for any vector which is fixed in $\cal C$ . It is especially useful to apply this formula to unit base vectors, so

$\displaystyle{\dot{\hat{\boldsymbol{\imath}}}^{\prime\prime}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}% \times\hat{\boldsymbol{\imath}}^{\prime\prime},$
$\displaystyle{\dot{\hat{\boldsymbol{\jmath}}}^{\prime\prime}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}% \times\hat{\boldsymbol{\jmath}}^{\prime\prime},\quad\mbox{and}$	(18.57)
$\displaystyle{\dot{\hat{\boldsymbol{k}}}^{\prime\prime}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}% \times\hat{\boldsymbol{k}}^{\prime\prime}.$

In some minds, Eqns. 18.2 are the core of rigid body kinematics. Box 18.2 on 18.2 shows how these relations give ‘the Q dot’ formula: For any time dependent vector ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$

{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}={}^{\lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}+{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\mathcal{C}/F}\times{\accentset{\rightharpoonup}{\boldsymbol{Q}}}.

(18.58)

or more simply, but less explicitly,

\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}_{rel}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{Q}}}.

where $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}_{rel}$ is the time derivative of ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ relative to the moving frame of interest (in this case $\cal C$ ). The ‘Q dot’ formula says that

The derivative of a vector with respect to a Newtonian frame $\cal F$ (or ‘absolute derivative’) can be calculated as the derivative ${}^{\lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}$ of the vector with respect to a moving frame $\cal C$ , plus a term ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}$ that corrects for the rotation of frame $\cal C$ relative to frame $\cal F$ .

Note that if ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ is a constant in the frame $\cal C$ , like the relative position vector of two points glued to $\cal C$ , then ${}^{\lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}={\accentset{\rightharpoonup}{\boldsymbol{Q}}}_{rel}={% \accentset{\rightharpoonup}{\boldsymbol{0}}}$ and the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula reduces to

{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{Q}}}.

The $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula 18.58 is useful for the derivation of a variety of formulas and is also useful in the solution of problems.

While we have shown how to use this formula to calculate the rate of change of a vector with respect to a Newtonian frame, the formula can be used to calculate its rate of change with respect to a non-Newtonian frame. Letting $\cal A$ and $\cal B$ be two possibly non-Newtonian frames, the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula for the rate of change of ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ with respect to frame $\cal A$ is

{}^{\lxWithClass{rp-bodyframe}{\mathcal{A}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}={{}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}}+{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{{\cal B}/{\cal A}}\times{\accentset{\rightharpoonup}{\boldsymbol{Q}% }}.

(18.59)

Both $\cal A$ and $\cal B$ could be non-Fixed (non-Newtonian).

Summary of the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula

For a vector ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ fixed in $\cal B$ ,

\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}={\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{% Q}}}\qquad\mbox{or}\qquad{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{% \accentset{\rightharpoonup}{\boldsymbol{Q}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B/\cal F}\times{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}.

For any time dependent vector ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ ,

\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}={}^{\lxWithClass{rp-% bodyframe}{\mathcal{B}}}\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}\qquad\mbox{or}\qquad{}^{\lxWithClass{rp-% bodyframe}{\mathcal{F}}}\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}={}^{% \lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/% \cal F}\times{\accentset{\rightharpoonup}{\boldsymbol{Q}}}.

Some examples of applying the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula are:

•

Absolute velocity of a point $P$ relative to $O^{\prime}$ :

$\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{\prime}}={{}^{% \lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}}}_{P/O^{\prime}}+{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{% \prime}}$
•

Rate of change of a rotating unit vector which is fixed in ${\cal B}$ :

$\dot{\hat{\boldsymbol{\imath}}}^{\prime}=\underbrace{{\dot{\hat{\boldsymbol{% \imath}}}^{\prime}}_{rel}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times\hat{% \boldsymbol{\imath}}^{\prime}$

The varying base-vectors method of computing velocity and acceleration

One way to calculate velocity, acceleration is to express the position of a particle in terms of a combination of base vectors, some of which change in time. Velocity and acceleration are then determined by directly differentiating the expression for position, taking account that the base vectors themselves are changing. This method is sometimes convenient for bodies connected in series, one body to the next, etc.

The overall approach is as follows:

1)

Glue a coordinate system to every moving body. If needed, also create moving frames that move independently of any particular body.
2)

Call the basis vectors associated with these frames $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , $\hat{\boldsymbol{k}}$ for the fixed frame $\cal F$ ; $\hat{\boldsymbol{\imath}}^{\prime}$ , $\hat{\boldsymbol{\jmath}}^{\prime}$ , $\hat{\boldsymbol{k}}^{\prime}$ for the moving frame $\cal B$ ; and $\hat{\boldsymbol{\imath}}^{\prime\prime}$ , $\hat{\boldsymbol{\jmath}}^{\prime\prime}$ , $\hat{\boldsymbol{k}}^{\prime\prime}$ for the moving frame $\cal C$ , etc.
3)

Evaluate all of the relative angular velocities; ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B}$ , etc. in terms of the scalar angular rates $\dot{\theta}$ , $\dot{\phi}$ , etc. and the base vectors glued to the frames.
4)

Express all of the absolute angular velocities in terms of the relative angular velocities.
5)

Differentiate to get the angular accelerations using, for example,

$\displaystyle\dot{\hat{\boldsymbol{\imath}}}^{\prime}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times% \hat{\boldsymbol{\imath}}^{\prime}\mbox{ or}$

$\displaystyle\dot{\hat{\boldsymbol{\jmath}}}^{\prime\prime}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times% \hat{\boldsymbol{\jmath}}^{\prime\prime}$
6)

Write the position of all points of interest in terms of the various base vectors.
7)

Differentiate the position to get the velocities (again using $\dot{\hat{\boldsymbol{\jmath}}}^{\prime\prime}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal C}\times\hat{\boldsymbol{\jmath}}^{\prime\prime}$ , etc.)
8)

Differentiate again to get acceleration.

First, reconsider the bug crawling on the tire in fig. 18.27.

Example: Absolute velocity of a point moving relative to a moving frame: Bug crawling on a tire

We write the position of the bug in terms of the various basis vectors as

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{O^{\prime}/O}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{\prime}}$
		$\displaystyle=$	$\displaystyle\overbrace{d\hat{\boldsymbol{\imath}}+R\hat{\boldsymbol{\jmath}}}% ^{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{O^{\prime}/O}}+\overbrace{s% \hat{\boldsymbol{\imath}}^{\prime}+\ell\hat{\boldsymbol{\jmath}}^{\prime})}^{{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{\prime}}}.$

To get the absolute velocity $\frac{{}^{\cal F}d}{dt}({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O})$ of the bug at the instant shown, we differentiate the position of the bug once, using the product rule and the rates of change of the rotating basis vectors with respect to the fixed frame, to get

Filename:bikefork1-ang-accel

Figure 18.27: The

x y z

coordinate system is attached to the fixed frame with basis vectors

\hat{\boldsymbol{\imath}}

\hat{\boldsymbol{\jmath}}

\hat{\boldsymbol{k}}

and the

x^{\prime}y^{\prime}z^{\prime}

coordinate system is attached to the rolling tire with basis vectors

\hat{\boldsymbol{\imath}}^{\prime}

\hat{\boldsymbol{\jmath}}^{\prime}

\hat{\boldsymbol{k}}^{\prime}

.The absolute angular velocity of the tire is

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal B}/{\cal F}}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal B}}=-\dot{\theta}\hat% {\boldsymbol{k}}^{\prime}=-\dot{\theta}\hat{\boldsymbol{k}}

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$	$\displaystyle=$	$\displaystyle\dot{d}\hat{\boldsymbol{\imath}}+\underbrace{\dot{R}}_{0}\hat{% \boldsymbol{\jmath}}+\underbrace{\dot{s}}_{0}\hat{\boldsymbol{\imath}}^{\prime% }+s\dot{\hat{\boldsymbol{\imath}}}^{\prime}+\dot{\ell}\hat{\boldsymbol{\jmath}% }^{\prime}+\ell\dot{\hat{\boldsymbol{\jmath}}}^{\prime}$	(18.60)
	$\displaystyle=$	$\displaystyle\dot{d}\hat{\boldsymbol{\imath}}+s({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times\hat{\boldsymbol{\imath}}^{\prime})+\dot{% \ell}\hat{\boldsymbol{\jmath}}^{\prime}+\ell({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times\hat{\boldsymbol{\jmath}}^{\prime})$
	$\displaystyle=$	$\displaystyle\dot{d}\hat{\boldsymbol{\imath}}+s(-\dot{\theta}\hat{\boldsymbol{% k}}^{\prime}\times\hat{\boldsymbol{\imath}}^{\prime})+\dot{\ell}\hat{% \boldsymbol{\jmath}}^{\prime}+\ell(-\dot{\theta}\hat{\boldsymbol{k}}^{\prime}% \times\hat{\boldsymbol{\jmath}}^{\prime})$
	$\displaystyle=$	$\displaystyle\dot{d}\hat{\boldsymbol{\imath}}+\ell\dot{\theta}\hat{\boldsymbol% {\imath}}^{\prime}+(\dot{\ell}-s\dot{\theta})\hat{\boldsymbol{\jmath}}^{\prime}.$

Example: Absolute acceleration of a point moving relative to a moving frame (2-D): Bug crawling on a tire, again

Differentiating equation 18.60 from the example above again, we get the absolute acceleration $\frac{{}^{\cal F}d^{2}}{dt^{2}}({\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {P/O})=\frac{{}^{\cal F}d}{dt}({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% P})$ of the bug at the instant shown,

$\displaystyle\ddot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O}={\dot{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{P}$	$\displaystyle=$	$\displaystyle\ddot{d}\hat{\boldsymbol{\imath}}+(\ell\ddot{\theta}+\dot{\ell}% \dot{\theta})\hat{\boldsymbol{\imath}}^{\prime}+\ell\dot{\theta}(-\dot{\theta}% \hat{\boldsymbol{k}}^{\prime}\times\hat{\boldsymbol{\imath}}^{\prime})$
		$\displaystyle+(\ddot{\ell}-s\ddot{\theta}-\underbrace{\dot{s}}_{0}\dot{\theta}% )\hat{\boldsymbol{\jmath}}^{\prime}+(\dot{\ell}-s\dot{\theta})(-\dot{\theta}% \hat{\boldsymbol{k}}^{\prime}\times\hat{\boldsymbol{\jmath}}^{\prime})$
	$\displaystyle=$	$\displaystyle\ddot{d}\hat{\boldsymbol{\imath}}+(\ell\ddot{\theta}-s\dot{\theta% }^{2}+2\dot{\ell}\dot{\theta})\hat{\boldsymbol{\imath}}^{\prime}+(\ddot{\ell}-% \ell\dot{\theta}^{2}-s\ddot{\theta})\hat{\boldsymbol{\jmath}}^{\prime}.$

Summary of the varying base-vector method

In the varying base vector method, we calculate the velocity of a point by looking at the position as the sum of two position vectors, one of which is expressed in the moving base vectors. We then differentiate the position, taking account that the base vectors of the moving frame change with time. In general

Filename:bikefork-ang-accel

Figure 18.28: The position of a point P relative to the origin of a fixed frame

\cal F

is represented as the sum of two vectors: the position of the new origin relative to the old, and the position of P relative to the new origin. (Here we use “new” as an informal synonym for “moving frame”.)

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}$
	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% O^{\prime}/O}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{\prime}}\right]$
	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left[(x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{% \jmath}}+z\hat{\boldsymbol{k}})+(x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+% y^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}+z^{\prime}\hat{\boldsymbol{k}}^{% \prime})\right]$
	$\displaystyle=$	$\displaystyle(\dot{x}\hat{\boldsymbol{\imath}}+\dot{y}\hat{\boldsymbol{\jmath}% }+\dot{z}\hat{\boldsymbol{k}})+(\dot{x}^{\prime}\hat{\boldsymbol{\imath}}^{% \prime}+\dot{y}^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}+\dot{z}^{\prime}% \hat{\boldsymbol{k}}^{\prime})+$
		$\displaystyle\left[x^{\prime}({\accentset{\rightharpoonup}{\boldsymbol{\omega}% }}_{\cal B}\times\hat{\boldsymbol{\imath}}^{\prime})+y^{\prime}({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times\hat{\boldsymbol{\jmath}}% ^{\prime})+z^{\prime}({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B% }\times\hat{\boldsymbol{k}}^{\prime})\right]$

We could calculate ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ similarly using a combination of the product rule of differentiation and the facts that $\dot{\hat{\boldsymbol{\imath}}}^{\prime}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times\hat{\boldsymbol{\imath}}^{\prime}$ , $\dot{\hat{\boldsymbol{\jmath}}}^{\prime}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times\hat{\boldsymbol{\jmath}}^{\prime}$ , and $\dot{\hat{\boldsymbol{k}}}^{\prime}={\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\cal B}\times\hat{\boldsymbol{k}}^{\prime}$ ,

and would get a formula with 15 non-zero terms.

Box 18.1 The $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula

We think about some vector ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ as a quantity that could be represented by an arrow. We can write ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ using the coordinates of the fixed Newtonian frame with base vectors $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},\hat{\boldsymbol{k}}$ : ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}=Q_{x}\hat{\boldsymbol{\imath}}+Q% _{y}\hat{\boldsymbol{\jmath}}+Q_{z}\hat{\boldsymbol{k}}$ . Similarly we could write ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ in terms of the coordinates of some moving and rotating frame $\cal B$ with base vectors $\hat{\boldsymbol{\imath}}^{\prime},\hat{\boldsymbol{\jmath}}^{\prime},\hat{% \boldsymbol{k}}^{\prime}$ : ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}=Q_{x^{\prime}}\hat{\boldsymbol{% \imath}}^{\prime}+Q_{y^{\prime}}\hat{\boldsymbol{\jmath}}^{\prime}+Q_{z^{% \prime}}\hat{\boldsymbol{k}}^{\prime}$ . Now of course

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$
	$\displaystyle\hbox{so}\quad Q_{x}\hat{\boldsymbol{\imath}}+Q_{y}\hat{% \boldsymbol{\jmath}}+Q_{z}\hat{\boldsymbol{k}}\$	$\displaystyle=$	$\displaystyle\ Q_{x^{\prime}}\hat{\boldsymbol{\imath}}^{\prime}+Q_{y^{\prime}}% \hat{\boldsymbol{\jmath}}^{\prime}+Q_{z^{\prime}}\hat{\boldsymbol{k}}^{\prime}.$

Similarly, $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}$ so long as what we mean by $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ is its derivative in a fixed frame. That is, we use $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ as an informal notation for $\displaystyle{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}={{}^{\cal F}d{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}\over dt}$ . We can calculate $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ the same way we have from the start of the book, namely,

\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}=\dot{Q}_{x}\hat{\boldsymbol{% \imath}}+\dot{Q}_{y}\hat{\boldsymbol{\jmath}}+\dot{Q}_{z}\hat{\boldsymbol{k}}.

We didn’t have to use the product rule of differentiation because the unit vectors $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{k}}$ , associated with a fixed frame, are constant in time.

What if we wanted to use the coordinate information that was given to us by a person who was moving and rotating with the moving frame $\cal B$ ? Now we calculate $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ taking account that the $\cal B$ base vectors change in time.

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$	$\displaystyle=$	$\displaystyle{d\over dt}[Q_{x^{\prime}}\hat{\boldsymbol{\imath}}^{\prime}+Q_{y% ^{\prime}}\hat{\boldsymbol{\jmath}}^{\prime}+Q_{z^{\prime}}\hat{\boldsymbol{k}% }^{\prime}]$
	$\displaystyle=$	$\displaystyle\underbrace{[\dot{Q}_{x^{\prime}}\hat{\boldsymbol{\imath}}^{% \prime}+\dot{Q}_{y^{\prime}}\hat{\boldsymbol{\jmath}}^{\prime}+\dot{Q}_{z^{% \prime}}\hat{\boldsymbol{k}}^{\prime}]}_{{}^{\lxWithClass{rp-bodyframe}{% \mathcal{B}}}\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}}$
		$\displaystyle\quad+\quad\underbrace{[Q_{x^{\prime}}\dot{\hat{\boldsymbol{% \imath}}^{\prime}}+Q_{y^{\prime}}\dot{\hat{\boldsymbol{\jmath}}^{\prime}}+Q_{z% ^{\prime}}\dot{\hat{\boldsymbol{k}}^{\prime}}]}_{\hbox{\rm\Large? }}.$

The first term in the product rule is just the derivative of ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ in the moving frame ${}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}$ . That is, ${}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}$ is calculated by differentiating the components in $\cal B$ holding the base vectors in $\cal B$ fixed. The second term depends on evaluating $\dot{\hat{\boldsymbol{\imath}}^{\prime}}$ , $\dot{\hat{\boldsymbol{\jmath}}^{\prime}}$ , and $\dot{\hat{\boldsymbol{k}}^{\prime}}$ . We know (at least for 2-D and for fixed-axis rotation in 3-D) that

$\displaystyle{\dot{\hat{\boldsymbol{\imath}}}^{\prime}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}% \times\hat{\boldsymbol{\imath}}^{\prime},$
$\displaystyle{\dot{\hat{\boldsymbol{\jmath}}}^{\prime}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}% \times\hat{\boldsymbol{\jmath}}^{\prime},\quad\mbox{and}$	(18.62)
$\displaystyle{\dot{\hat{\boldsymbol{k}}}^{\prime}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}% \times\hat{\boldsymbol{k}}^{\prime}.$

Eqns. 18.2 are the core of rigid body kinematics.

Now we can go back to the second group of terms in Eqn. 18.2.

?	$\displaystyle=$	$\displaystyle[Q_{x^{\prime}}\dot{\hat{\boldsymbol{\imath}}^{\prime}}+Q_{y^{% \prime}}\dot{\hat{\boldsymbol{\jmath}}^{\prime}}+Q_{z^{\prime}}\dot{\hat{% \boldsymbol{k}}^{\prime}}]$
	$\displaystyle=$	$\displaystyle[Q_{x^{\prime}}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% _{\cal B}\times{\hat{\boldsymbol{\imath}}^{\prime}}+Q_{y^{\prime}}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\hat{\boldsymbol{\jmath}% }^{\prime}}+Q_{z^{\prime}}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \cal B}\times{\hat{\boldsymbol{k}}^{\prime}}]$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times% [Q_{x^{\prime}}{\hat{\boldsymbol{\imath}}^{\prime}}+Q_{y^{\prime}}{\hat{% \boldsymbol{\jmath}}^{\prime}}+Q_{z^{\prime}}{\hat{\boldsymbol{k}}^{\prime}}]$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times% {\accentset{\rightharpoonup}{\boldsymbol{Q}}}$

Going back to Eqn. 18.2 we get the desired result:

{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}={}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}+{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\cal B/F}\times{\accentset{\rightharpoonup}{\boldsymbol{Q}}}.

(18.63)

or more simply, but less explicitly,

\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}_{rel}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{Q}}}.

(18.64)

Geometric ‘derivation’ of the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula

Here is a geometrical ‘derivation’ of the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula in two dimensions. Referring to the figure at right, we look at a vector ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ at two successive times. We then look at how ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ seems to change in a frame that rotates slightly as ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ changes. The picture shows how to account for the difference between the change of ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ as perceived by the two different frames.

In detail the parts (a) to (e) of the picture show the following.

$\bullet$

Part (a) shows a vector ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ at time $t$ .
$\bullet$

Part (b) shows ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ at time $t+\Delta t$ and the change in ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ , $\Delta{\accentset{\rightharpoonup}{\boldsymbol{Q}}}\approx\dot{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}\cdot\Delta t$ .
$\bullet$

Part (c) is like (a) but shows a moving body or frame $\cal A$ .
$\bullet$

Part (d) shows the change in ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ , $({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}\times{\accentset{% \rightharpoonup}{\boldsymbol{Q}}})\cdot\Delta t$ , that would occur if ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ were fixed (constant) in $\cal A$ .
$\bullet$

Part (e) shows the change in ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ that would be observed in the moving frame $\cal A$ .
$\bullet$

Part (f) shows the net change in ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ , $\Delta{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ , that is the same as that in (b) above; here, it is shown as the sum of the two contributions from (d) and (e).

Thus, using $\cal A$ , $\Delta{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ for small $\Delta t$ is composed of two parts: (1) the $\Delta{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ observed in ${\cal A}(t)$ , and (2) the change in ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ which would occur if ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ were constant in ${\cal A}(t)$ and thus rotating with it. Dividing $\Delta{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ by $\Delta t$ gives the ‘ $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula’, $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}\ =\mathrel{\mathop{{% \accentset{\rightharpoonup}{\boldsymbol{Q}}}}\limits^{\mathcal{A}\mathord{% \boldsymbol{\cdot}}}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A% }\times{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ .

Filename:tfigure8-qdotnewer-mike

Two different looks at the change in the vector ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ , $\Delta{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ , over a time interval $\Delta t$ .

SAMPLE 18.5

Filename:bikefork1-alt

Figure 18.29:

Acceleration of a point moving in a rotating frame. It is given that the arm OAB rotates with counterclockwise angular acceleration $\dot{\omega}=3\,\mathrm{rad}/\mathrm{s}^{2}$ and at the instant shown the angular speed $\omega=5\,\mathrm{rad}/\mathrm{s}$ . Also, at the same instant, the particle P is sliding (roughly) down with speed $v_{\rm/tube}=4\,{\mathrm{ft/s}}$ and acceleration $a_{\rm/tube}=2\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ . Find the absolute acceleration of the particle at the given instant. Take $\ell=2\,\mathrm{ft}$ in the figure.

Solution Let us attach a body frame $\mathcal{B}$ to the rigid arm OAB. For calculations we fix a coordinate system $x^{\prime}y^{\prime}z^{\prime}$ in this frame such that the origin $O^{\prime}$ of the coordinate system coincides with O, and at the given instant, the axes are aligned with the inertial coordinate axes $x y z$ . Since $x^{\prime}y^{\prime}z^{\prime}$ is fixed in the frame $\mathcal{B}$ and $\mathcal{B}$ rotates with the rigid arm with $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=3\,% \mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=5\,\mathrm{% rad}/\mathrm{s}\hat{\boldsymbol{k}}$ , the basis vectors $\hat{\boldsymbol{\imath}}^{\prime}$ , $\hat{\boldsymbol{\jmath}}^{\prime}$ and $\hat{\boldsymbol{k}}^{\prime}$ rotate with the same $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}$ .

Filename:bikefork-alt

Figure 18.30:

In the rotating (primed) coordinate system,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}$	$\displaystyle=$	$\displaystyle x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{% \boldsymbol{\jmath}}^{\prime}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P})=% \frac{d}{dt}(x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{% \boldsymbol{\jmath}}^{\prime})$
	$\displaystyle=$	$\displaystyle\dot{x}^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+x^{\prime}\dot% {\hat{\boldsymbol{\imath}}}^{\prime}+\dot{y}^{\prime}\hat{\boldsymbol{\jmath}}% ^{\prime}+y^{\prime}\dot{\hat{\boldsymbol{\jmath}}}^{\prime}.$

Now, we use the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula to evaluate $\dot{\hat{\boldsymbol{\imath}}}^{\prime}$ and $\dot{\hat{\boldsymbol{\jmath}}}^{\prime}$ , i.e.,

	$\displaystyle\dot{\hat{\boldsymbol{\imath}}}^{\prime}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}% \times\hat{\boldsymbol{\imath}}^{\prime}=\omega\hat{\boldsymbol{k}}^{\prime}% \times\hat{\boldsymbol{\imath}}^{\prime}=\omega\hat{\boldsymbol{\jmath}}^{\prime}$
	$\displaystyle\dot{\hat{\boldsymbol{\jmath}}}^{\prime}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}% \times\hat{\boldsymbol{\jmath}}^{\prime}=\omega\hat{\boldsymbol{k}}^{\prime}% \times\hat{\boldsymbol{\jmath}}^{\prime}=-\omega\hat{\boldsymbol{\imath}}^{% \prime}.$

Also, note that $x^{\prime}$ is constant since in frame $\mathcal{B}$ , the motion of the particle is always along the tube, i.e., along the negative $y^{\prime}$ axis (see Fig. 18.30). Thus, $x^{\prime}=2\ell$ , $\dot{x}^{\prime}=0$ , $y^{\prime}=\ell$ , and $\dot{y}^{\prime}=-v_{\rm/tube}$ . Substituting these quantities in ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ , we get:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$	$\displaystyle=$	$\displaystyle x^{\prime}\omega\hat{\boldsymbol{\jmath}}^{\prime}-v_{\rm/tube}% \hat{\boldsymbol{\jmath}}^{\prime}+y^{\prime}(-\omega\hat{\boldsymbol{\imath}}% ^{\prime})$		(18.65)
		$\displaystyle=$	$\displaystyle(x^{\prime}\omega-v_{\rm/tube})\hat{\boldsymbol{\jmath}}^{\prime}% -\omega y^{\prime}\hat{\boldsymbol{\imath}}^{\prime}.$		(18.65)

Now substituting $x^{\prime}=2\ell=4\,\mathrm{ft}$ , $\omega=5\,\mathrm{rad}/\mathrm{s}$ , $y^{\prime}=\ell=2\,\mathrm{ft}$ , $v_{\rm/tube}=4\,{\mathrm{ft/s}}$ and noting that $\hat{\boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}},\;\hat{% \boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}}$ at the given instant, we get:

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}

\displaystyle=

\displaystyle[(20-4)\hat{\boldsymbol{\jmath}}-10\hat{\boldsymbol{\imath}}]\,{% \mathrm{ft/s}}=(-10\hat{\boldsymbol{\imath}}+16\hat{\boldsymbol{\jmath}})\,{% \mathrm{ft/s}}.

We can find ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ by differentiating Eq. (18.65) and noting again that $\hat{\boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}},\;\hat{% \boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}}$ at the given instant

^†^†margin: We will revisit this problem in Sample 18.48. Here we use the

\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}

formula on various base vectors. In Sample 18.48 we will use the relative velocity and acceleration formulae.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P})=% \frac{d}{dt}[(x^{\prime}\omega-v_{\rm/tube})\hat{\boldsymbol{\jmath}}^{\prime}% -\omega y^{\prime}\hat{\boldsymbol{\imath}}^{\prime}]$
	$\displaystyle=$	$\displaystyle(\underbrace{\dot{x}^{\prime}}_{0}\omega+x^{\prime}\dot{\omega}-% \overbrace{\dot{v}_{\rm/tube}}^{a_{\rm/tube}})\hat{\boldsymbol{\jmath}}^{% \prime}+(x^{\prime}\omega-v_{\rm/tube})\dot{\hat{\boldsymbol{\jmath}}}^{\prime% }-(\dot{\omega}y^{\prime}+\omega\overbrace{\dot{y}^{\prime}}^{-v_{\rm/tube}})% \hat{\boldsymbol{\imath}}^{\prime}-\omega y^{\prime}\dot{\hat{\boldsymbol{% \imath}}}^{\prime}$
	$\displaystyle=$	$\displaystyle(x^{\prime}\dot{\omega}-a_{\rm/tube})\hat{\boldsymbol{\jmath}}^{% \prime}+(x^{\prime}\omega-v_{\rm/tube})(-\omega\hat{\boldsymbol{\imath}}^{% \prime})-(\dot{\omega}y^{\prime}-\omega v_{\rm/tube})\hat{\boldsymbol{\imath}}% ^{\prime}-\omega y^{\prime}(\omega\hat{\boldsymbol{\jmath}}^{\prime})$
	$\displaystyle=$	$\displaystyle-a_{\rm/tube}\hat{\boldsymbol{\jmath}}^{\prime}+2\omega v_{\rm/% tube}\hat{\boldsymbol{\imath}}^{\prime}-\omega^{2}(x^{\prime}\hat{\boldsymbol{% \imath}}^{\prime}+y^{\prime}\hat{\boldsymbol{\jmath}}^{\prime})+\dot{\omega}(x% ^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}-y^{\prime}\hat{\boldsymbol{\imath}% }^{\prime})$
	$\displaystyle=$	$\displaystyle-2\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\jmath}}+40\,{% \mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\imath}}-25(4\hat{\boldsymbol{% \imath}}+2\hat{\boldsymbol{\jmath}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}+3(4\hat{% \boldsymbol{\jmath}}-2\hat{\boldsymbol{\imath}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}$
	$\displaystyle=$	$\displaystyle-(66\hat{\boldsymbol{\imath}}+40\hat{\boldsymbol{\jmath}})\,{% \mathrm{ft}}/{\mathrm{s}}^{2}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=-(66\hat{\boldsymbol{\imath}% }+40\hat{\boldsymbol{\jmath}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}$

SAMPLE 18.6

Filename:pfigure-blue-128-1

Figure 18.31:

Rate of change of unit vectors. A circular disk $\mathcal{D}$ is welded to a rigid rod AB. The rod rotates about point A with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ . A frame $\mathcal{B}$ is attached to the disk and therefore rotates with the same ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . Two coordinate systems, $(\hat{\boldsymbol{\imath}}^{\prime},\hat{\boldsymbol{\jmath}}^{\prime})$ and $(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R},\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta})$ are fixed in frame $\mathcal{B}$ as shown in the figure.

1.

Find the rate of change of unit vectors $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R},\;\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta},\;\hat{\boldsymbol{\imath}}^{\prime}\;{\rm and}\;% \hat{\boldsymbol{\jmath}}^{\prime}$ using the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula.
2.

Express the $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ vectors in terms of $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ and verify the results obtained above for $\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}$ and $\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!\theta}$ by direct differentiation.

Solution Since the disk is welded to the rod and frame $\mathcal{B}$ is fixed in the disk, the frame rotates with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\omega\hat{% \boldsymbol{k}}$ .

To find the rate of change of the unit vectors using the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula, we substitute the desired unit vector in place of ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ in the formula (eqn. (18.58)). For example,

\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}=\;^{\mathcal{B}}\dot{\hat{% \boldsymbol{e}}}_{\scriptscriptstyle\!R}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{B}}\times\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}.

It should be clear that ${}^{\mathcal{B}}\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}=0$ , since $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ does not change with respect to an observer sitting in frame $\mathcal{B}$ . Therefore,

\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}={\accentset{\rightharpoonup% }{\boldsymbol{\omega}}}_{\mathcal{B}}\times\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}=\omega\hat{\boldsymbol{k}}\times\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}=\omega\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}.

Similarly,

$\displaystyle\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle\;\underbrace{{}^{\mathcal{B}}\dot{\hat{\boldsymbol{e}}}_{% \scriptscriptstyle\!\theta}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}=\omega\overbrace{\hat{\boldsymbol% {k}}\times\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}}^{-\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}}=-\omega\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}.$
$\displaystyle{\dot{\hat{\boldsymbol{\imath}}}^{\prime}}$	$\displaystyle=$	$\displaystyle\;\underbrace{{}^{\mathcal{B}}{\dot{\hat{\boldsymbol{\imath}}}^{% \prime}}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times\hat{\boldsymbol{% \imath}}^{\prime}=\omega\hat{\boldsymbol{k}}\times\hat{\boldsymbol{\imath}}^{% \prime}=\omega\hat{\boldsymbol{\jmath}}^{\prime}.$
$\displaystyle{\dot{\hat{\boldsymbol{\jmath}}}^{\prime}}$	$\displaystyle=$	$\displaystyle\;\underbrace{{}^{\mathcal{B}}{\dot{\hat{\boldsymbol{\jmath}}}^{% \prime}}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times\hat{\boldsymbol{% \jmath}}^{\prime}=\omega\hat{\boldsymbol{k}}\times\hat{\boldsymbol{\jmath}}^{% \prime}=-\omega\hat{\boldsymbol{\imath}}^{\prime}.$

Answer: $\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}=\omega\hat{\boldsymbol{e}}_% {\scriptscriptstyle\!\theta},\quad\dot{\hat{\boldsymbol{e}}}_{% \scriptscriptstyle\!\theta}=-\omega\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R% },\quad\dot{\hat{\boldsymbol{\imath}}}^{\prime}=\omega\hat{\boldsymbol{\jmath}% }^{\prime},\mbox{ and }\dot{\hat{\boldsymbol{\jmath}}}^{\prime}=-\omega\hat{% \boldsymbol{\imath}}^{\prime}$

Filename:tfigure8-syst-bods

Figure 18.32:

Since $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=\cos\theta\hat{\boldsymbol{\imath% }}^{\prime}+\sin\theta\hat{\boldsymbol{\jmath}}^{\prime}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=-\sin\theta\hat{\boldsymbol{% \imath}}^{\prime}+\cos\theta\hat{\boldsymbol{\jmath}}^{\prime}$ , we get their rates of change by direct differentiation as

$\displaystyle\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}$	$\displaystyle=$	$\displaystyle\cos\theta\dot{\hat{\boldsymbol{\imath}}}^{\prime}+\sin\theta\dot% {\hat{\boldsymbol{\jmath}}}^{\prime}$
	$\displaystyle=$	$\displaystyle\cos\theta(\omega\hat{\boldsymbol{\jmath}}^{\prime})+\sin\theta(-% \omega\hat{\boldsymbol{\imath}}^{\prime})$
	$\displaystyle=$	$\displaystyle\omega(-\sin\theta\hat{\boldsymbol{\imath}}^{\prime}+\cos\theta% \hat{\boldsymbol{\jmath}}^{\prime})=\omega\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta},$
$\displaystyle\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle-\sin\theta\dot{\hat{\boldsymbol{\imath}}}^{\prime}+\cos\theta% \dot{\hat{\boldsymbol{\jmath}}}^{\prime}$
	$\displaystyle=$	$\displaystyle-\sin\theta(\omega\hat{\boldsymbol{\jmath}}^{\prime})+\cos\theta(% -\omega\hat{\boldsymbol{\imath}}^{\prime})$
	$\displaystyle=$	$\displaystyle-\omega(\cos\theta\hat{\boldsymbol{\imath}}^{\prime}+\sin\theta% \hat{\boldsymbol{\jmath}}^{\prime})=-\omega\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}.$

Here we have used the fact that $\theta$ , the angle between the unit vectors $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{\imath}}^{\prime}$ , remains constant during the motion. The results obtained are the same as in part (a).

SAMPLE 18.7

Filename:sfig8-7-2

Figure 18.33:

Rate of change of a position vector. A rigid rod OAB rotates counterclockwise about point O with constant angular speed $\omega=5\,\mathrm{rad}/\mathrm{s}$ . A collar C slides out on the bent arm AB with constant speed $v=0.5\,\mathrm{m}/\mathrm{s}\;$ with respect to the arm. Find the velocity of the collar using the ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula.

^†^†margin:

Solution Let ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$ be the position vector of the collar. Then the velocity of the collar is $\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$ . Let the rod OAB be the rotating frame $\mathcal{B}$ . Now we can find $\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$ using the ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula:

\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}=\;^{\mathcal{B}}\dot{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{C}

To compute $\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$ , let us first find ${}^{\mathcal{B}}\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$ , the rate of change of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$ as seen in frame $\mathcal{B}$ (this term represents the velocity of the collar you see if you sit on the rod and watch the collar; also called ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{rel}$ ).

Filename:sfig8-7-2a

Figure 18.34:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{C/A}$
	${}^{\mathcal{B}}\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$	$\displaystyle=$	${}^{\mathcal{B}}\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}+^{% \mathcal{B}}\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/A}$

Note that the vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}=\ell\hat{\boldsymbol{\lambda}}$ does not change in frame $\mathcal{B}$ since both its magnitude, $\ell$ , and direction, $\hat{\boldsymbol{\lambda}}$ , remain fixed in $\mathcal{B}$ . Therefore,

{}^{\mathcal{B}}\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}=0

\rm{Now}\quad{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/A}=r\hat{% \boldsymbol{\imath}}^{\prime}.\;\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\;^{% \mathcal{B}}\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/A}=\;\dot{r}% \hat{\boldsymbol{\imath}}^{\prime}=0.5\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{% \imath}}^{\prime}

because $\hat{\boldsymbol{\imath}}^{\prime}$ does not change in $\mathcal{B}$ and $\dot{r}$ = speed of the collar with respect to the arm (see Figure 18.34). Thus,

{}^{\mathcal{B}}\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}=0.5\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}^{\prime}.

Hence,

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$	$\displaystyle=$	${}^{\mathcal{B}}\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}+\omega% \hat{\boldsymbol{k}}\times(\ell\hat{\boldsymbol{\lambda}}+r\hat{\boldsymbol{% \imath}}^{\prime})$
	$\displaystyle=$	${}^{\mathcal{B}}\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}+\omega% \ell(\hat{\boldsymbol{k}}\times\hat{\boldsymbol{\lambda}})+\omega r(\hat{% \boldsymbol{k}}\times\hat{\boldsymbol{\imath}}^{\prime})$
	$\displaystyle=$	$\displaystyle\dot{r}\hat{\boldsymbol{\imath}}^{\prime}+\omega\ell(\sin\theta% \hat{\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{\jmath}})+\omega r\hat{% \boldsymbol{\jmath}}^{\prime}$
	$\displaystyle=$	$\displaystyle 0.5\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}^{\prime}+5\,% \mathrm{m}/\mathrm{s}(\frac{\sqrt{3}}{2}\hat{\boldsymbol{\imath}}+\frac{1}{2}% \hat{\boldsymbol{\jmath}})+1\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}^{\prime}$
	$\displaystyle=$	$\displaystyle 4.83\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}+3.5\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}$

where we have used the fact that at the given instant, $\hat{\boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}=4.83\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\imath}}+3.5\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}$

Problems for 18.2 Rotating reference frames

18.2.1 Express the basis vectors ( $\hat{\boldsymbol{\imath}}^{\prime},\hat{\boldsymbol{\jmath}}^{\prime}$ ) associated with axes $x^{\prime}$ and $y^{\prime}$ in terms of the standard basis ( $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}}$ ) for $\theta=30^{{}^{\circ}}$ .

Filename:sfig8-7-2again

Figure 18.35

18.2.2 Body frames are frames of reference attached to a body in motion. The orientation of a coordinate system attached to a body frame $\cal B$ is shown in the figure at some instant of interest. For $\theta=60^{{}^{\circ}}$ , express the basis vectors $(\hat{\boldsymbol{b}}_{1},\hat{\boldsymbol{b}}_{2})$ in terms of the standard basis ( $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}}$ ).

Filename:sfig8-7-2disks

Figure 18.36

18.2.3 Find the components of (a) ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=3\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}}+2\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}+4\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{k}}$ in the rotated basis ( $\hat{\boldsymbol{\imath}}^{\prime},\hat{\boldsymbol{\jmath}}^{\prime},\hat{% \boldsymbol{k}}^{\prime}$ ) and (b) ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-0.5\,\mathrm{m}/\mathrm{s}^{2}% \hat{\boldsymbol{\jmath}}^{\prime}+3.0\,\mathrm{m}/\mathrm{s}^{2}\hat{% \boldsymbol{k}}^{\prime}$ in the standard basis ( $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},\hat{\boldsymbol{k}}$ ). ( $y$ and $y^{\prime}$ are in the same direction and $x^{\prime}$ and $z^{\prime}$ are in the $x z$ plane.)

Filename:sfig8-4-4

Figure 18.37

18.2.4 A particle travels in a straight line in the $x y$ -plane parallel to the $x$ -axis at a distance $y=\ell$ in the positive $x$ direction. The position of the particle is denoted by ${\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)$ . The angle of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ measured positive counter-clockwise from the $x$ axis is decreasing at a constant rate with magnitude $\omega$ . If the particle starts on the $y$ axis at $\theta=\pi/2$ , what is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)$ in cartesian coordinates?

18.2.5 Given that $\,{\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)=ct^{2}\hat{\boldsymbol{% \imath}}^{\prime}\,$ and that $\,\theta(t)=d\sin(\lambda t)\,$ , find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}(t)$

(a)

in terms of $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ ,
(b)

in terms of $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ .

Filename:sfig8-4-4a

Figure 18.38

18.2.6 A bug walks on a turntable. In polar coordinates, the position of the bug is given by ${\accentset{\rightharpoonup}{\boldsymbol{R}}}=R\hat{\boldsymbol{e}}_{r}$ , where the origin of this coordinate system is at the center of the turntable. The $(\hat{\boldsymbol{e}}_{r},\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta})$ coordinate system is attached to the turntable and, hence, rotates with the turntable. The kinematical quantities describing the bug’s motion are $\dot{R}=-\dot{R}_{0}$ , $\ddot{R}=0$ , $\dot{\theta}=\omega_{0}$ , and $\ddot{\theta}=0$ . A fixed coordinate system $O x y$ has origin $O$ at the center of the turntable. As the bug walks through the center of the turntable:

(a)

What is its speed?
(b)

What is its acceleration?
(c)

What is the radius of the osculating circle (i.e., what is the radius of curvature of the bug’s path?).

18.3 General expressions for velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ of a moving point

Now that we have some comfort with moving frames we can develop formulas that are not so strongly attached to base vectors. That is, we take account that the base vectors rotate with the frame, but develop formulas that don’t use the base vectors explicitly. Thus the formulas we develop here work equally for any frame that is glued to the rotating frame of choice, independent of its orientation.

Absolute velocity of a point moving relative to a moving frame

Imagine that you know the absolute velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{0^{\prime}/% \mathcal{F}}}$ of some point $O^{\prime}$ on an object $\cal B$ , say the center of a car tire and the angular velocity of the tire, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal B}/{\cal F}}$ . Finally, imagine you also know the relative velocity of point P, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}$ , say of a bug crawling on the tire.

If the frame $\cal B$ is translating or rotating, the velocity of particle P relative to the frame ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}$ is not the absolute velocity (the velocity relative to a Newtonian frame). The absolute velocity in this case is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal F}}$ , or more simply ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}$ , or more simply still, just ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ . We want to know the relationship between the relative velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}$ and the absolute velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal F}}$ (otherwise known as just ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ ).

Filename:sfig8-4-4b

Figure 18.39: The position of point P relative to the origin

O^{\prime}

of moving frame

{\cal B}

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}

. The position of the origin of the frame

{\cal B}

relative to the origin

O

of the fixed frame

{\cal F}

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O^{\prime}/O}}

. The position of point P relative to

O

is the sum of

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}

and

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O^{\prime}/O}}

. The motion of point P relative to the fixed frame may be complicated.

Let’s start by looking at the position. The position of a point P that is moving is:

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O^{\prime}/O}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}

where $O^{\prime}$ is the origin of a coordinate system which is glued to the rigid object, as shown in fig. 18.39.

To find the absolute velocity of point P we will use the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula, equation 18.58, for computing the rate of change of a vector. The velocity of P is the rate of change of its position. Here, we use ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{P/\mathcal{F}}}$	$\displaystyle=$	$\displaystyle{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}}_{\rm P/O}$
	$\displaystyle=$	$\displaystyle{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}}_{O^{\prime}/O}+{}^{\lxWithClass{rp-% bodyframe}{\mathcal{F}}}\dot{{\accentset{\rightharpoonup}{\boldsymbol{r}}}}_{% \rm P/O^{\prime}}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{O^{\prime}/\cal F}}+\overbrace{\underbrace{{}^{\lxWithClass{rp-% bodyframe}{\mathcal{B}}}\dot{{\accentset{\rightharpoonup}{\boldsymbol{r}}}}_{% \rm P/O^{\prime}}}_{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5% mu\mathrm{P/\cal B}}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B% /F}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P^{\prime}/O% ^{\prime}}}}^{{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}}_{\rm P/O^{\prime}}}$

The $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula 18.58 was used in the calculation to compute ${}^{\cal F}{\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{\prime}}}$

{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}}_{P/O^{\prime}}={}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}% \dot{{\accentset{\rightharpoonup}{\boldsymbol{r}}}}_{P/O^{\prime}}+{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/F}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P^{\prime}/O^{\prime}}}.

(18.66)

Thus, the ‘three term velocity formula’.

\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{% P}}}^{\text{Absolute velocity of point P}}=\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{O^{\prime}}}}_{\text{1.% Velocity of reference point $O^{\prime}$ on object \mathcal{B}}}\par+\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{P/\cal B}}}^{\text{2. Velocity of point P relative to moving frame \mathcal{B}}}+\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P^{\prime}/O^{\prime}}}}_{\text{3% . Velocity relative to point $O^{\prime}$ of the point $P^{\prime}$ glued to object \mathcal{B}\ % which is instantaneously coincident with point P}}

(18.67)

Another way to write the formula for absolute velocity is as

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P^{\prime}}}% +{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\mathcal% {B}}}

where P’ is a point glued to $\mathcal{B}$ which is instantaneously coincident with P, so the absolute velocity of $P^{\prime}$ is

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P^{\prime}}% }={\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{O^{\prime% }}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P^{\prime}/O^{\prime}}}.

(18.68)

Reconsider the bug crawling on the tire, object $\cal B$ , in fig. 18.40. To find the absolute velocity of the bug, we need to be concerned with how the bug moves relative to the tire and how the tire moves relative to the ground.

Example: Absolute velocity of a point moving relative to a moving frame (2-D): Bug crawling on a tire, again

Filename:sfig8-4-4c

Figure 18.40:

Referring to equation 18.3 on page 18.3, the absolute velocity of the bug is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{P/\mathcal{F}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{O^{\prime}/\cal F}}+\underbrace{{}^{\cal B}{\dot{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\rm P/O^{\prime}}}}_{{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\mathcal{B}}}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/F}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}$
	$\displaystyle=$	$\displaystyle R\dot{\theta}\hat{\boldsymbol{\imath}}\quad+\quad\dot{\ell}\hat{% \boldsymbol{\jmath}}^{\prime}\quad+\quad(-\dot{\theta}\hat{\boldsymbol{k}}^{% \prime})\times(s\hat{\boldsymbol{\imath}}^{\prime}+\ell\hat{\boldsymbol{\jmath% }}^{\prime})$
	$\displaystyle=$	$\displaystyle R\dot{\theta}\hat{\boldsymbol{\imath}}\quad+\quad\ell\dot{\theta% }\hat{\boldsymbol{\imath}}^{\prime}\quad+\quad(\dot{\ell}-s\dot{\theta})\hat{% \boldsymbol{\jmath}}^{\prime}.\par$

At the instant of interest, the direction of the bug’s absolute velocity depends upon the relative magnitudes of $\dot{\ell}$ and $\dot{\theta}$ as well as the orientation of $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ .

As we noted earlier, another way to write the formula for absolute velocity is

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P^{\prime}}}% +{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\mathcal% {B}}}

where, in the example above, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P^{\prime}}% }=R\dot{\theta}\hat{\boldsymbol{\imath}}+\dot{\theta}(\ell\hat{\boldsymbol{% \imath}}^{\prime}-s\hat{\boldsymbol{\jmath}}^{\prime})$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\mathcal{% B}}}=\dot{\ell}\hat{\boldsymbol{\jmath}}^{\prime}$ . At the instant of concern, we can think of the absolute velocity of the bug as the velocity of the mark labeled $P^{\prime}$ under the bug plus the velocity of the bug relative to the tire.

Acceleration

We would like to find acceleration of a point using information about its motion relative to a moving frame. The result, the ‘five term acceleration formula’ is the most complicated formula in this book. (For reference, it is in Table II, 5c).

Acceleration relative to an object or frame

Filename:pfigure-s94h13p2

Figure 18.41: An object or reference frame

\mathcal{B}

and a point P move around. We are interested in the acceleration of point P.

The acceleration of a point relative to an object or frame is the acceleration you would calculate if you were looking at the particle while you translated and rotated with the frame and took no account of the outside world. That is, if the position of a particle P relative to the origin $O^{\prime}$ of a coordinate system in a moving frame $\cal B$ is given by:

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}=% \underbrace{r_{{\rm P}x^{\prime}/O^{\prime}}}_{x^{\prime}}\hat{\boldsymbol{% \imath}}^{\prime}+\underbrace{r_{{\rm P}y^{\prime}/O^{\prime}}}_{y^{\prime}}% \hat{\boldsymbol{\jmath}}^{\prime}+\underbrace{r_{{\rm P}z^{\prime}/O^{\prime}% }}_{z^{\prime}}\hat{\boldsymbol{k}}^{\prime},

then the acceleration of the particle P relative to the frame is:

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{\rm P/\cal B}}=% \underbrace{\ddot{r}_{{\rm P}x^{\prime}/O^{\prime}}}_{\ddot{x}^{\prime}}\hat{% \boldsymbol{\imath}}^{\prime}+\underbrace{\ddot{r}_{{\rm P}y^{\prime}/O^{% \prime}}}_{\ddot{y}^{\prime}}\hat{\boldsymbol{\jmath}}^{\prime}+\underbrace{% \ddot{r}_{{\rm P}z^{\prime}/O^{\prime}}}_{\ddot{z}^{\prime}}\hat{\boldsymbol{k% }}^{\prime}.

That is, the acceleration relative to the frame takes no account of (a) the motion of the frame or of (b) the rotation of the base vectors with the frame to which they are fixed.

Filename:pfigure-f93f5

Figure 18.42: A bug crawling on a tire at point P. Point P’ is on the tire and instantaneously corresponds to point P.

Reconsider the bug labeled point P crawling on the tire, object B, in fig. 18.42.

Example: Acceleration relative to a frame (2-D): Bug crawling on a tire, again

If we are sitting on the tire, all that we see is the bug crawling in a straight line at non-constant rate relative to us. Thus, its acceleration relative to the tire is

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}=\ddot{\ell}% \hat{\boldsymbol{\jmath}}^{\prime}.

So, at the instant of interest, the bug has an acceleration relative to the tire frame parallel to the $y^{\prime}$ -axis.

Absolute acceleration of a point P^′ glued to a moving frame

Imagine that you know the absolute acceleration of some point $O^{\prime}$ at the center of a frame $\cal B$ , say the center of a car tire. Imagine you also know the angular velocity of the tire, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/{\cal F}}$ , and the angular acceleration, ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}/{\cal F}}$ . Then, you can find the absolute acceleration of a piece of gum labeled point $D$ stuck to the sidewall (see fig. 18.44).

If we start with the equation 18.68 for the absolute velocity of a point glued to a moving frame on page 18.68 and differentiate with respect to time, we get the absolute acceleration of a point $D$ fixed in a moving frame $\cal B$ as follows:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{D}}$	$\displaystyle=$	$\displaystyle{d\over dt}[{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0% .5mu\mathrm{O^{\prime}}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \mathcal{B}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/O% ^{\prime}}}]$	(18.69)
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{O^{\prime}% /O}}+[\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/O^{\prime}}}+% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/O^{\prime}}})]$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{O^{\prime}% /O}}+{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/O^{\prime}}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/O^{\prime}}})\par$

Example: Absolute acceleration of a point glued to a moving frame (2-D): Bug crawling on a tire, again

Filename:pfigure-s94h13p3

Figure 18.43: Tracking a moving particle from two different reference frames.

Filename:p-s96-p3-3

Figure 18.44: Gum is glued to a rolling tire at D. The point P^′ glued to the tire is just where the point P happens to be passing at the instant of interest.

Here, the acceleration of point P^′ glued to the tire, relative to the tire is zero, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P^{\prime}/\cal B}}=0$ (see fig. 18.44). The angular velocity of the wheel with respect to the ground is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/{\cal F}}=-% \dot{\theta}\hat{\boldsymbol{k}}=-\dot{\theta}\hat{\boldsymbol{k}}^{\prime}$ . The angular speed is increasing at a rate $\ddot{\theta}$ . Thus, ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}/{\cal F}}=-% \ddot{\theta}\hat{\boldsymbol{k}}=-\ddot{\theta}\hat{\boldsymbol{k}}^{\prime}$ . The position of $P^{\prime}$ relative to $O^{\prime}$ is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P^{\prime}/O^{\prime}% }}=s\hat{\boldsymbol{\imath}}^{\prime}+\ell\hat{\boldsymbol{\jmath}}^{\prime}$ .

Using equation 18.69 on page 18.69, we get the absolute acceleration of point $P^{\prime}$ to be

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P^{\prime}% /\cal F}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{O^{\prime}% /\cal F}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/F}\times(% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/F}\times{\accentset% {\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P^{\prime}/O^{\prime}}})+{% \accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{{\cal B}/{\cal F}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P^{\prime}/O^{\prime}}}$
		$\displaystyle=$	$\displaystyle\underbrace{R\ddot{\theta}\hat{\boldsymbol{\imath}}}_{\text{{% acceleration of origin of moving frame}}}-\overbrace{\dot{\theta}^{2}(s\hat{\boldsymbol{% \imath}}^{\prime}+\ell\hat{\boldsymbol{\jmath}}^{\prime})}^{\text{{centripetal% term}}}+\overbrace{R\ddot{\theta}(s\hat{\boldsymbol{\jmath}}^{\prime}-\ell% \hat{\boldsymbol{\imath}}^{\prime})}^{\text{tangential term}}$

In this example, the absolute acceleration $P^{\prime}$ is due to:

1.

the increase in the translational speed of the tire relative to the ground (acceleration of origin of moving frame),
2.

its going in circles at a non-constant rate about point $O^{\prime}$ relative to the ground (‘tangential term’), and
3.

‘centripetal term’ towards the origin of the moving frame. (In three-dimensional problems, this term is directed towards an axis through ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ that goes through O’).

Absolute acceleration of a point moving relative to a moving frame

Filename:bikefork1-ang-mom

Figure 18.45: Keeping track of P which instantaneously coincides with point P’ which is glued to frame

\mathcal{B}

If we start with the equation for absolute velocity 18.3 on page 18.3 and differentiate with respect to time we get the absolute acceleration of a point P using a moving frame $\cal B$ . To do this calculation we need to use the product rule of differentiation. Refer to the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula, eqn. (18.63) on page 18.63. Here is the calculation:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}$	$\displaystyle=$	$\displaystyle{d\over dt}[{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0% .5mu\mathrm{O^{\prime}/O}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{P/\cal B}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}_{\mathcal{B}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/O^{\prime}}}]$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{O^{\prime}% /O}}+({\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}})$
		$\displaystyle+\quad[\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \mathcal{B}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O% ^{\prime}}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}% \times{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/% \cal B}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}% \times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}})]$
	$\displaystyle=$	$\displaystyle\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \mathrm{O^{\prime}/O}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \mathcal{B}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \mathcal{B}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O% ^{\prime}}})+{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}}% _{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P^{\prime}}}}+{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}+2{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}.$

The collection of terms ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P^{\prime}}}$ is the acceleration of a point $P^{\prime}$ which is glued to body $\mathcal{B}$ and is instantaneously coincident with P. It is the same as ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{D}}$ using $D=P^{\prime}$ in equation 18.69. To repeat, the result is

The ‘five term’ acceleration formula

	$\displaystyle\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \mathrm{P}}}_{\framebox{The absolute acceleration}}$	$\displaystyle\hskip-36.135pt=\quad\overbrace{{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{\mathrm{O^{\prime}/O}}}^{\text{1. acceleration of reference % point $O^{\prime}$ on object \mathcal{B}}}+\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}})}_{\text{2. % centripetal acceleration: acceleration relative to $O^{\prime}$ of point glued to object, coincident with P, and going in circles around $O^{\prime}$ at constant rate}}+\par\overbrace{{\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}}^{\text{3. % tangential acceleration: acceleration relative to $O^{\prime}$ of point glued to object, coincident with P, due to non-constant ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$}}+\par\underbrace{{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}}_{\text{4. % acceleration of P relative to moving frame $\mathcal{B}$}}+\!\!\!\!\!\!\!% \underbrace{2\ {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}% }\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/% \cal B}}}_{\framebox{5. Coriolis accel.}}\par$		(18.70)
		$\displaystyle=\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \mathrm{P^{\prime}}}}_{\framebox{The first 3 terms above}}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}+2{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}$		(18.71)

This ‘five-term-acceleration’ formula is both famous and infamous. It’s famous because it is given a lot of emphasis by some instructors

^†^†margin: True story. Once a teacher wrote eqn. (18.70) in big letters on a fresh clean black board, including the explanations in boxes. Then he talked it through, telling the class that this was what they had been building up to. That after all the vector notation, the more involved kinematic development and the messy notation that this was finally the climax. There was still silence in the room as, so the teacher was hoping, the students could feel satisfied with the accomplishment. After time for a slow breath a student asked “Was it as good for you as it was for us?”

and infamous because it takes some getting used to. Eqn. 18.71 is the ‘three term acceleration formula’. It combines the first three terms in the 5-term formula and interprets them as the acceleration of the point P^′ on B that instantaneously coincides with P. The best way to get used to the five term acceleration formula is to find situations where some of the terms drop out.

Reconsider the bug labeled point P crawling on the tire, object $\cal B$ , in fig. 18.44. To find the absolute acceleration of the bug we need to think about how the bug moves relative to the tire and how the tire moves relative to the ground.

Example: Absolute acceleration of a point moving relative to a moving frame (2-D): Bug crawling on a tire, again

Filename:bikefork-ang-mom

Figure 18.46:

From the previous bug examples on page 18.3 and page 18.3 we know that

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}=% \dot{\ell}\hat{\boldsymbol{\jmath}}^{\prime},\ \mbox{ and }\par{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}=\ddot{\ell}\hat{% \boldsymbol{\jmath}}^{\prime}.

Referring to the five term acceleration formula, equation 18.70 on page 18.70, the absolute acceleration of the bug is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal F}}$	$\displaystyle=$	$\displaystyle\phantom{+}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm% {O^{\prime}/O}}$
		$\displaystyle+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}% \times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}})$
		$\displaystyle+{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}$
		$\displaystyle+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}$
		$\displaystyle+2{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}% }\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/% \cal B}}$
	$\displaystyle=$	$\displaystyle\underbrace{R\ddot{\theta}\hat{\boldsymbol{\imath}}-\dot{\theta}^% {2}(s\hat{\boldsymbol{\imath}}^{\prime}+\ell\hat{\boldsymbol{\jmath}}^{\prime}% )+\ddot{\theta}(\ell\hat{\boldsymbol{\imath}}^{\prime}-s\hat{\boldsymbol{% \jmath}}^{\prime})}_{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P^% {\prime}/\cal F}}}+\ddot{\ell}\hat{\boldsymbol{\jmath}}^{\prime}+2\dot{\theta}% \dot{\ell}\hat{\boldsymbol{\imath}}^{\prime}$
	$\displaystyle=$	$\displaystyle R\ddot{\theta}\hat{\boldsymbol{\imath}}+(\ell\ddot{\theta}-s\dot% {\theta}^{2}+2\dot{\theta}\dot{\ell})\hat{\boldsymbol{\imath}}^{\prime}+(-s% \ddot{\theta}+\ddot{\ell}-\ell\dot{\theta}^{2})\hat{\boldsymbol{\jmath}}^{% \prime}.$

So, at the instant of interest, the bug’s absolute acceleration is due to:

1.

the translational acceleration of the tire, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{O^{\prime}/O}}=R\ddot{% \theta}\hat{\boldsymbol{\imath}}$
2.

the centripetal acceleration of going in circles of radius $\sqrt{s^{2}+\ell^{2}}$ about the center of the tire as it rolls, $-\dot{\theta}^{2}(s\hat{\boldsymbol{\imath}}^{\prime}+\ell\hat{\boldsymbol{% \jmath}}^{\prime})$ , pointing at the center of the tire,
3.

the tangential acceleration of going in circles about the center of the tire as the tire rolls at non-constant rate, $\ddot{\theta}(\ell\hat{\boldsymbol{\imath}}^{\prime}-s\hat{\boldsymbol{\jmath}% }^{\prime})$ ,
4.

the acceleration of the bug relative to the tire as it crawls on the line, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}=\ddot{\ell}% \hat{\boldsymbol{\jmath}}^{\prime}$ , and
5.

the Coriolis acceleration caused, in part, by the change in direction, relative to the ground, of the velocity of the bug relative to the tire, $2\dot{\theta}\dot{\ell}\hat{\boldsymbol{\imath}}^{\prime}$ .

Items 1, 2 and 3 sum to be the acceleration of point $P^{\prime}$ on the tire but instantaneously coinciding with moving point P.

Motion relative to a point versus motion relative to a frame

We can now give a different interpretation of the expressions we have been using ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B/A}}$ .

Rather than thinking of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}$ as the difference between ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A}}$ we can think of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}$ as the ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/\cal A}}$ where $\cal A$ is a frame with origin that moves with point A and which has no rotation rate relative to $\cal F$ . That is

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}\quad% \mbox{means}\quad{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B/\cal A}}

Similarly,

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B/A}}\quad\mbox{means}% \quad{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B/\cal A}}.

Box 18.2 Relation between moving frame formulae and polar coordinate formulae

A similarity exists between the polar coordinate velocity formula

{\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{R}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}

and the second two terms in the ‘three-term’ velocity formula

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{O^{\prime}/O% }}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}+{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}.

In fact, we have tried to build your understanding of moving frames by means of that connection.

Similarly, the polar coordinate formula for acceleration

{\accentset{\rightharpoonup}{\boldsymbol{a}}}=(\ddot{R}-R\dot{\theta}^{2})\hat% {\boldsymbol{e}}_{\scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}

is somehow closely linked to the last four terms of the ‘5-term’ acceleration formula

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{O^{\prime}% /O}}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}})$
			$\displaystyle+\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}+2{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}.$

Let’s make these connections explicit. Imagine a particle P moving around on the $x y$ -plane.

Filename:tfigure8-polarcomp1

Let’s create a moving frame $\cal B$ with rotating coordinate system $x^{\prime}y^{\prime}$ attached to it whose origin $O^{\prime}$ is coincident with origin $O$ of a coordinate system $x y$ attached to a fixed frame $\cal F$ . Let this frame rotate in exactly such a way so that the particle is always on the $x^{\prime}$ -axis. So, in this frame, ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}=R\hat{% \boldsymbol{\imath}}^{\prime}$ , ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}=% \dot{R}\hat{\boldsymbol{\imath}}^{\prime}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}=\ddot{R}\hat% {\boldsymbol{\imath}}^{\prime}$ . Also, the frame motion is characterized by ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{O^{\prime}/% O}}={\accentset{\rightharpoonup}{\boldsymbol{0}}},{\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{\mathrm{O^{\prime}/O}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\dot{\theta}% \hat{\boldsymbol{k}}$ , and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\ddot{% \theta}\hat{\boldsymbol{k}}$ . So, if we plug in the three-term velocity formula, we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{P}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{O^{\prime}/O}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \mathcal{B}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O% ^{\prime}}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm% {P/\cal B}}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}+(\dot{\theta}\hat{% \boldsymbol{k}})\times(R\hat{\boldsymbol{\imath}}^{\prime})+\dot{R}\hat{% \boldsymbol{\imath}}^{\prime}$
	$\displaystyle=$	$\displaystyle R\dot{\theta}\hat{\boldsymbol{\jmath}}^{\prime}+\dot{R}\hat{% \boldsymbol{\imath}}^{\prime}$
	$\displaystyle=$	$\displaystyle R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+% \dot{R}\hat{\boldsymbol{e}}_{r}\qquad\mbox{ (because $\hat{\boldsymbol{\imath}% }^{\prime}=\hat{\boldsymbol{e}}_{r}$ and $\hat{\boldsymbol{\jmath}}^{\prime}=% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$)}$

which is the polar coordinate velocity formula.

Similarly, if we plug into the five-term acceleration formula, we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{O^{\prime}% /O}}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\cal B}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}})$
		$\displaystyle+\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}+2{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\cal B}}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}+\ddot{R}\hat{% \boldsymbol{\imath}}^{\prime}+\dot{\theta}\hat{\boldsymbol{k}}\times(\dot{% \theta}\hat{\boldsymbol{k}}\times R\hat{\boldsymbol{\imath}}^{\prime})$
		$\displaystyle+(\ddot{\theta}\hat{\boldsymbol{k}})\times(R\hat{\boldsymbol{% \imath}}^{\prime})+2(\dot{\theta}\hat{\boldsymbol{k}})\times\dot{R}\hat{% \boldsymbol{\imath}}^{\prime}$
	$\displaystyle=$	$\displaystyle(\ddot{R}-R\dot{\theta}^{2})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!\theta}$

Again, we recover the appropriate polar coordinate formula.

We have just shown how the polar coordinate formulae are special cases of the relative motion formulae.

Warning!

In problems where we want the rotating frame to be a rotating object on which a particle moves, the polar coordinate formulae only correspond term by term with the relative motion formulae if the particle path is a straight radial line fixed on a $2D$ object, as in the example of a bug walking on a straight line scribed on the surface of a rotating CD or a bead sliding in a tube rotating about an axis perpendicular to the tube.

Filename:tfigure8-polarcomp2

Because we are sometimes interested in more general relative motions, the polar coordinate formulae do not always apply and we must make use of the more general relative motion formulae.

SAMPLE 18.8

Filename:summer95f-5-a

Figure 18.47:

A ‘T’ shaped tube is welded to a massless rigid arm OAB which rotates about O at a constant rate $\omega=5\,\mathrm{rad}/\mathrm{s}$ . See also Sample 18.29. At the instant shown a particle P is falling down in the vertical section of the tube with speed $v_{\rm/tube}=4\,{\mathrm{ft/s}}$ . Find the absolute velocity of the particle. Take $\ell=2\,\mathrm{ft}$ in the figure.

^†^†margin:

Solution Let us attach a frame $\cal B$ to arm OAB. Thus $\cal B$ rotates with OAB with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\omega\hat{% \boldsymbol{k}}$ where $\omega=5\,\mathrm{rad}/\mathrm{s}$ . To do calculations in $\cal B$ we attach a coordinate system $x^{\prime}y^{\prime}z^{\prime}$ to $\cal B$ at point O. At the instant of interest the rotating coordinate system $x^{\prime}y^{\prime}z^{\prime}$ coincides with the fixed coordinate system $x y z$ . (Since the entire motion is in the $x y$ -plane, the $z$ -axis is not shown in the figure). Let $P^{\prime}$ be a point coincident with $P$ but fixed in $\cal B$ . Now,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{P^{\prime}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \rm rel}\mbox{\hskip 36.135pt}

Filename:pfigure4-2-rp10

Figure 18.48:

where

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}$	$\displaystyle=$	$\displaystyle\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{O^{% \prime}}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}\;\;+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}/O^{\prime}}$
	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times(2\ell\hat{\boldsymbol{\imath}}+% \ell\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle 2\omega\ell\hat{\boldsymbol{\jmath}}-\omega\ell\hat{\boldsymbol{% \imath}},\mbox{\hskip 72.26999pt}$

and

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}$	$\displaystyle=$	Velocity relative to the frame $\cal B$
		$\displaystyle=$	$\displaystyle-v_{\rm/tube}\hat{\boldsymbol{\jmath}}.$

Thus,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}+{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{rel}$
	$\displaystyle=$	$\displaystyle-\omega\ell\hat{\boldsymbol{\imath}}+(2\omega\ell-v_{\rm/tube})% \hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-5\,\mathrm{rad}/\mathrm{s}\cdot 2\,\mathrm{ft}\hat{\boldsymbol{% \imath}}+(2\cdot 5\,\mathrm{rad}/\mathrm{s}\cdot 2\,\mathrm{ft}-4\,{\mathrm{ft% /s}})\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-10\,{\mathrm{ft/s}}\hat{\boldsymbol{\imath}}+16\,{\mathrm{ft/s}}% \hat{\boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}=-10\,{\mathrm{ft/s}}\hat{% \boldsymbol{\imath}}+16\,{\mathrm{ft/s}}\hat{\boldsymbol{\jmath}}$

Comments: The kinematics calculation is equivalent to the vector addition shown in Figure 18.48. The velocity of P is the sum of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}={\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{P/\mathcal{B}}$ .

SAMPLE 18.9

Filename:pfigure-blue-125-2

Figure 18.49:

Acceleration of a point in a rotating frame. Consider the rotating tube of Sample 18.3 again. The arm OAB rotates with counterclockwise angular acceleration $\dot{\omega}=3\,\mathrm{rad}/\mathrm{s}^{2}$ and, at the instant shown, its angular speed $\omega=5\,\mathrm{rad}/\mathrm{s}$ . Also, at the same instant, the particle P falls down with speed $v_{\rm/tube}=4\,{\mathrm{ft/s}}$ and acceleration $a_{\rm/tube}=2\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ . Find the absolute acceleration of the particle at the given instant. Take $\ell=2\,\mathrm{ft}$ in the figure.

Solution

Filename:pfigure-blue-68-1

Figure 18.50: Acceleration of point P^′. P^′ is fixed to the frame

\mathcal{B}

and at the moment coincides with point P. Therefore,

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}={\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}/O^{\prime}}-\omega^{2}_{\mathcal% {B}}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}/O^{\prime}}

We consider a frame $\mathcal{B}$ , with coordinate axes $x^{\prime}y^{\prime}z^{\prime}$ , fixed to the arm OAB and thus rotating with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\omega\hat{% \boldsymbol{k}}=5\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}}=\dot{\omega}% \hat{\boldsymbol{k}}=3\,\mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{k}}$ . The acceleration of point P is given by

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{P^{\prime}}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \rm cor}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm rel}

where

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}$	$\displaystyle=$	acceleration of a point P^′ that is fixed in $\mathcal{B}$
		and at the moment coincides with P,
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cor}}$	$\displaystyle=$	Coriolis acceleration, and
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{rel}}$	$\displaystyle=$	acceleration of P relative to frame $\mathcal{B}$ .

Now we calculate each of these terms separately. For calculating ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}$ , imagine a rigid rod from point O to point P, rotating with the frame $\mathcal{B}$ . Mark the far end of the rod as P^′ (same as point P). The acceleration of this end of the rod is ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}$ . To find the relative terms ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{rel}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{rel}}$ , freeze the motion of the frame $\mathcal{B}$ at the given moment and watch the motion of point P. The non-intuitive term ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cor}}$ has no such simple physical interpretation but has a simple formula. Thus,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}$	$\displaystyle=$	$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{O^{% \prime}}}^{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+{\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}/O^{\prime}}+\overbrace{{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}/O^{\prime}})}^{-% \omega^{2}_{\mathcal{B}}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P^{% \prime}/O^{\prime}}}$
	$\displaystyle=$	$\displaystyle\dot{\omega}\hat{\boldsymbol{k}}\times(2\ell\hat{\boldsymbol{% \imath}}+\ell\hat{\boldsymbol{\jmath}})-\omega^{2}(2\ell\hat{\boldsymbol{% \imath}}+\ell\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle 2\dot{\omega}\ell\hat{\boldsymbol{\jmath}}-\dot{\omega}\ell\hat{% \boldsymbol{\imath}}-2\omega^{2}\ell\hat{\boldsymbol{\imath}}-\omega^{2}\ell% \hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-\ell(\dot{\omega}+2\omega^{2})\hat{\boldsymbol{\imath}}+\ell(2% \dot{\omega}-\omega^{2})\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-(106\hat{\boldsymbol{\imath}}+38\hat{\boldsymbol{\jmath}})\,{% \mathrm{ft}}/{\mathrm{s}}^{2},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm cor}$	$\displaystyle=$	$\displaystyle 2{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}% }\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}$
	$\displaystyle=$	$\displaystyle 2\omega\hat{\boldsymbol{k}}\times v_{\rm/tube}(-\hat{\boldsymbol% {\jmath}})$
	$\displaystyle=$	$\displaystyle 2\omega v_{\rm/tube}\hat{\boldsymbol{\imath}}=2\cdot 5\,\mathrm{% rad}/\mathrm{s}\cdot 4\,{\mathrm{ft/s}}$
	$\displaystyle=$	$\displaystyle 40\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\imath}},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm rel}$	$\displaystyle=$	$\displaystyle a_{\rm/tube}(-\hat{\boldsymbol{\jmath}})=-2\,{\mathrm{ft}}/{% \mathrm{s}}^{2}\hat{\boldsymbol{\jmath}}.$

Adding the three terms together, we get

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$	$\displaystyle=$	$\displaystyle-106\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\imath}}-38% \,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\jmath}}+40\,{\mathrm{ft}}/{% \mathrm{s}}^{2}\hat{\boldsymbol{\imath}}-2\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat% {\boldsymbol{\jmath}}$
		$\displaystyle=$	$\displaystyle-66\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\imath}}-40\,% {\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=-(66\hat{\boldsymbol{\imath}% }+40\hat{\boldsymbol{\jmath}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}$

Note that the single term ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}$ encompasses three terms of the five term acceleration formula.

SAMPLE 18.10

Filename:pfigure-blue-58-1

Figure 18.51:

A small collar P is pinned to a rigid rod AB at length $\ell=1\,\mathrm{m}$ along the rod. The collar is free to slide in a straight track on a disk of radius $r=400\,{\rm mm}$ . The disk rotates about its center O at a constant $\omega=2\,\mathrm{rad}/\mathrm{s}$ . At the instant shown, when $\theta=45^{{}^{\circ}}$ and the collar is at a distance $\frac{3}{4}r$ in the track from the center O, find

1.

the angular velocity of the rod AB and
2.

the velocity of point P relative to the disk.

Solution

Filename:pfigure-blue-157-1

Figure 18.52:

We will think of P in two ways: one as attached to the rod and the other as sliding in the slot. First, let us attach a frame $\cal B$ to the disk. Thus $\cal B$ rotates with the disk with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\omega\hat{% \boldsymbol{k}}=2\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}$ . We attach a coordinate system $x^{\prime}y^{\prime}z^{\prime}$ to $\cal B$ at point O. At the instant of interest, the rotating coordinate system $x^{\prime}y^{\prime}z^{\prime}$ coincides with the fixed coordinate system $x y z$ . Now let us consider point $P^{\prime}$ which is fixed on the disk (and hence in $\cal B$ ) and coincides with point P at the instant of interest. We can write the velocity of P as:

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{P^{\prime}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \rm rel}\mbox{\hskip 36.135pt}

where

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}$	$\displaystyle=$	$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{O^{% \prime}}}^{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}/O^{\prime}}=\omega\hat{% \boldsymbol{k}}\times\left(\frac{3}{4}r\hat{\boldsymbol{\imath}}^{\prime}% \right)=\frac{3}{4}\omega r\hat{\boldsymbol{\jmath}}^{\prime}=\frac{3}{4}% \omega r\hat{\boldsymbol{\jmath}},$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}$	$\displaystyle\equiv$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P/{\cal B}}=v_{\rm rel% }\hat{\boldsymbol{\imath}}^{\prime}=v_{\rm rel}\hat{\boldsymbol{\imath}}.$

In the last expression, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}=v_{\rm rel}\hat{% \boldsymbol{\imath}}$ , we do not know the magnitude of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}$ and hence have left it as an unknown $v_{\rm rel}$ , but its direction is known because ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}$ has to be along the track and the track at the given instant is along the $x$ -axis. Thus,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}=\frac{3}{4}\omega r\hat{% \boldsymbol{\jmath}}+v_{\rm rel}\hat{\boldsymbol{\imath}}.

(18.72)

Now let us consider the motion of rod AB. Let ${\accentset{\rightharpoonup}{\boldsymbol{\Omega}}}=\Omega\hat{\boldsymbol{k}}$ be the angular velocity of AB at the instant of interest where $\Omega$ is unknown. Since P is pinned to the rod, it executes circular motion about A with radius $AP=\ell$ . Therefore,

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\Omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/A}=\Omega\hat{\boldsymbol{k}}% \times\ell(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{% \jmath}})=\Omega\ell(\cos\theta\hat{\boldsymbol{\jmath}}-\sin\theta\hat{% \boldsymbol{\imath}}).

(18.73)

But, and this trivial formula is the key, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{P}$ . Therefore, from Eqn. (18.72) and (18.73),

\frac{3}{4}\omega r\hat{\boldsymbol{\jmath}}+v_{\rm rel}\hat{\boldsymbol{% \imath}}=\Omega\ell(\cos\theta\hat{\boldsymbol{\jmath}}-\sin\theta\hat{% \boldsymbol{\imath}}).

(18.74)

Taking dot product of both sides of the above equation with $\hat{\boldsymbol{\jmath}}$ we get

	$\displaystyle\frac{3}{4}\omega r$	$\displaystyle=$	$\displaystyle\Omega\ell\cos\theta$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\Omega$	$\displaystyle=$	$\displaystyle\frac{3\omega r}{4\ell\cos\theta}=\frac{3\cdot 2\,\mathrm{rad}/% \mathrm{s}\cdot 0.4\,\mathrm{m}}{4\cdot 1\,\mathrm{m}\cdot\frac{1}{\sqrt{2}}}=% 0.85\,\mathrm{rad}/\mathrm{s}.$

Again taking the dot product of both sides of Eqn. (18.74) with $\hat{\boldsymbol{\imath}}$ we get

v_{\rm rel}=-\Omega\ell\sin\theta=-0.85\,\mathrm{rad}/\mathrm{s}\cdot 1\,% \mathrm{m}\cdot\frac{1}{\sqrt{2}}=-0.6\,\mathrm{m}/\mathrm{s}.

Answer: $(i)\;{\accentset{\rightharpoonup}{\boldsymbol{\Omega}}}=0.85\,\mathrm{rad}/% \mathrm{s}\hat{\boldsymbol{k}},\quad(ii)\;{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{\rm rel}=-0.6\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}$

SAMPLE 18.11

Filename:summer95f-5

Figure 18.53:

Spinning wheel on a rotating rod in 2-D . A rigid body OA is attached to a wheel that is massless except for three point masses P, Q, and R, placed symmetrically on the wheel. Each of the three masses is $m=0.5\,\mathrm{kg}$ . The rod OA rotates about point O in the counterclockwise direction at a constant rate $\omega_{1}=3\,\mathrm{rad}/\mathrm{s}$ . The wheel rotates with respect to the arm about point A with angular acceleration $\dot{\omega}_{2}=1\,\mathrm{rad}/\mathrm{s}^{2}$ and at the instant shown it has angular speed $\omega_{2}=5\,\mathrm{rad}/\mathrm{s}$ . Note that both $\omega_{2}$ and $\dot{\omega}_{2}$ are given with respect to the arm.

Using a rotating frame $\mathcal{B}$ attached to the rod and a coordinate system attached to the frame with origin at O, find

1.

the velocity of the mass P and
2.

the acceleration of the mass P.

Solution

Filename:pfigure-blue-127-2

Figure 18.54: P^′ goes in circles around point O with radius OP^′. Therefore,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}

is tangential to the circular path at P^′.

Frame $\mathcal{B}$ is attached to the rod. We choose a coordinate system $x^{\prime}y^{\prime}z^{\prime}$ in frame $\mathcal{B}$ with its origin at O and, at the instant, aligned with the fixed coordinate system $x y z$ . We consider a point P^′ momentarily coincident with point P but fixed in frame $\mathcal{B}$ . Since P^′ is fixed in $\mathcal{B}$ , it rotates with $\mathcal{B}$ with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\omega_{1}% \hat{\boldsymbol{k}}$ . To visualize the motion of P^′ imagine a rigid rod from O to P^′ (see Fig. 18.54). Now we can calculate the velocity and acceleration of point P^′ as follows.

Velocity of point P:

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{P^{\prime}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mathrm{rel}}.

Now we calculate the two terms separately:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}$	$\displaystyle=$	$\displaystyle\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{O^{% \prime}}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}/O^{\prime}}$
	$\displaystyle=$	$\displaystyle\omega_{1}\hat{\boldsymbol{k}}\times({\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{A/O^{\prime}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{P^{\prime}/A})$
	$\displaystyle=$	$\displaystyle\omega_{1}\hat{\boldsymbol{k}}\times[\underbrace{\ell(\cos\theta% \hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})}_{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{A/O^{\prime}}}+\underbrace{r(\cos\theta\hat% {\boldsymbol{\imath}}-\sin\theta\hat{\boldsymbol{\jmath}})}_{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}/A}}]$
	$\displaystyle=$	$\displaystyle\omega_{1}(\ell+r)\cos\theta\hat{\boldsymbol{\jmath}}-\omega_{1}(% \ell-r)\sin\theta\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle 3\,\mathrm{rad}/\mathrm{s}\cdot 2.5\,\mathrm{m}\cdot\cos 30^{{}^% {\circ}}\hat{\boldsymbol{\jmath}}-3\,\mathrm{rad}/\mathrm{s}\cdot 1.5\,\mathrm% {m}\cdot\sin 30^{{}^{\circ}}\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle(6.50\hat{\boldsymbol{\jmath}}-2.25\hat{\boldsymbol{\imath}})\,% \mathrm{m}/\mathrm{s}.$

Filename:pfigure-s94h13p4

Figure 18.55: Velocity of point P with respect to the frame

\mathcal{B}

Since the wheel rotates with angular speed $\omega_{2}$ with respect to the rod, an observer sitting in frame $\mathcal{B}$ would see a circular motion of point P about point A. Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{rel}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\rm wheel}/% \mathcal{B}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/A}$
	$\displaystyle=$	$\displaystyle-\omega_{2}\hat{\boldsymbol{k}}^{\prime}\times r(\cos\theta\hat{% \boldsymbol{\imath}}^{\prime}-\sin\theta\hat{\boldsymbol{\jmath}}^{\prime})$
	$\displaystyle=$	$\displaystyle-\omega_{2}r(\cos\theta\hat{\boldsymbol{\jmath}}^{\prime}+\sin% \theta\hat{\boldsymbol{\imath}}^{\prime})$
	$\displaystyle=$	$\displaystyle-(2.16\hat{\boldsymbol{\jmath}}^{\prime}+1.25\hat{\boldsymbol{% \imath}}^{\prime})\,\mathrm{m}/\mathrm{s}.$

But at the instant of interest, $\hat{\boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}},\,\hat{% \boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}},\mbox{and }\hat{% \boldsymbol{k}}^{\prime}=\hat{\boldsymbol{k}}.$ So,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{rel}}=-(2.16\hat{% \boldsymbol{\jmath}}+1.25\hat{\boldsymbol{\imath}})\,\,\mathrm{m}/\mathrm{s}\qquad

Therefore,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{P^{\prime}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mathrm{rel}}=4.33\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}-3.50\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}=(-3.50\hat{\boldsymbol{% \imath}}+4.33\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}$

Acceleration of point P: We can similarly find the acceleration of point P:

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{P^{\prime}}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \mathrm{cor}}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{rel}}

Filename:pfigure-blue-90-2

Figure 18.56:

where

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}$	$\displaystyle=$	acceleration of point P^′
	$\displaystyle=$	$\displaystyle\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{O^{% \prime}}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+\underbrace{{% \accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}}}_{{\accentset{% \rightharpoonup}{\boldsymbol{0}}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{P^{\prime}/O^{\prime}}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{B}}\times({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{P^{\prime}/O^{\prime}})$
	$\displaystyle=$	$\displaystyle-\omega^{2}_{\mathcal{B}}{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{P^{\prime}/O^{\prime}}$
	$\displaystyle=$	$\displaystyle-\omega_{1}^{2}[(\ell+r)\cos\theta\hat{\boldsymbol{\imath}}+(\ell% -r)\sin\theta\hat{\boldsymbol{\jmath}}]$
	$\displaystyle=$	$\displaystyle-9(\,\mathrm{rad}/\mathrm{s})^{2}[2.5\,\mathrm{m}\cdot\cos 30^{{}% ^{\circ}}\hat{\boldsymbol{\imath}}+1.5\,\mathrm{m}\cdot\sin 30^{{}^{\circ}}% \hat{\boldsymbol{\jmath}}]$
	$\displaystyle=$	$\displaystyle-(19.48\hat{\boldsymbol{\imath}}+6.75\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}^{2},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cor}}$	$\displaystyle=$	Coriolis acceleration
	$\displaystyle=$	$\displaystyle 2{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}% }\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{rel}}$
	$\displaystyle=$	$\displaystyle 2\omega_{1}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup% }{\boldsymbol{v}}}_{\mathrm{rel}}\quad\qquad\mbox{(see part (a) above for ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{rel}}$).}$
	$\displaystyle=$	$\displaystyle(6\,\mathrm{rad}/\mathrm{s})\hat{\boldsymbol{k}}\times(-2.16\hat{% \boldsymbol{\jmath}}-1.25\hat{\boldsymbol{\imath}})\,\mathrm{m}/\mathrm{s}$
	$\displaystyle=$	$\displaystyle(12.99\hat{\boldsymbol{\imath}}-7.50\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}^{2},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{rel}}$	$\displaystyle=$	acceleration of P relative to frame $\mathcal{B}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P/\mathcal{B}}=% \dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{2}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P/A}-\omega_{2}^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P/A}$
	$\displaystyle=$	$\displaystyle-\dot{\omega}_{2}\hat{\boldsymbol{k}}^{\prime}\times r(\cos\theta% \hat{\boldsymbol{\imath}}^{\prime}-\sin\theta\hat{\boldsymbol{\jmath}}^{\prime% })-\omega_{2}^{2}r(\cos\theta\hat{\boldsymbol{\imath}}^{\prime}-\sin\theta\hat% {\boldsymbol{\jmath}}^{\prime})$
	$\displaystyle=$	$\displaystyle-r[(\dot{\omega}_{2}\sin\theta+\omega_{2}^{2}\cos\theta)\hat{% \boldsymbol{\imath}}^{\prime}+(\dot{\omega}_{2}\cos\theta-\omega_{2}^{2}\sin% \theta)\hat{\boldsymbol{\jmath}}^{\prime}]$
	$\displaystyle=$	$\displaystyle-0.5\,\mathrm{m}[(1\,\mathrm{rad}/\mathrm{s}^{2}\cdot\sin 30^{{}^% {\circ}}+25(\,\mathrm{rad}/\mathrm{s})^{2}\cdot\cos 30^{{}^{\circ}})\hat{% \boldsymbol{\imath}}^{\prime}$
		$\displaystyle\qquad+(1\,\mathrm{rad}/\mathrm{s}^{2}\cdot\cos 30^{{}^{\circ}}-2% 5(\,\mathrm{rad}/\mathrm{s})^{2}\cdot\sin 30^{{}^{\circ}})\hat{\boldsymbol{% \jmath}}^{\prime}]$
	$\displaystyle=$	$\displaystyle(-11.08\hat{\boldsymbol{\imath}}^{\prime}+5.82\hat{\boldsymbol{% \jmath}}^{\prime})\,\mathrm{m}/\mathrm{s}^{2}$
	$\displaystyle=$	$\displaystyle(-11.08\hat{\boldsymbol{\imath}}+5.82\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}^{2}.$

The term ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}$ encompasses three terms of the five term acceleration formula. The last line in the calculation of ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{rel}}$ follows from the fact that at the instant of interest $\hat{\boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}}$ .

Now adding the three parts of ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ we get

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}+{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cor}}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{rel}}$
		$\displaystyle=$	$\displaystyle-17.57\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{\imath}}-8.43\,% \mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=-(17.57\hat{\boldsymbol{% \imath}}+8.43\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}^{2}$

Problems for 18.3 General expressions for velocity and acceleration

18.3.1 A bug walks on a straight line engraved on a rotating turntable (the bug’s path in the room is not a straight line). The line passes through the center of the turntable. The bug’s speed on this line is 1 $\,{\mathrm{in/s}}$ (the bug’s absolute speed is not 1 $\,{\mathrm{in/s}}$ ). The turntable rotates at a constant rate of 2 revolutions every $\pi\,{\rm s}$ in a positive sense about the $z$ -axis. Its surface is always in the $x y$ -plane. At time $t=0$ , the engraved line is aligned with the $x$ -axis, the bug is at the origin and headed towards the positive $x$ direction.

(a)

What is the $x$ component of the bug’s velocity at $t=\pi\,{\rm s}$ ?
(b)

What is the $y$ component of the bug’s velocity at $t=\pi$ seconds?
(c)

What is the $y$ component of the bug’s acceleration at $t=\pi$ ?
(d)

What is the radius of curvature of the bug’s path at $t=0$ ?

18.3.2 Actual path of bug trying to walk a straight line. A straight line is inscribed on a horizontal turntable. The line goes through the center. Let $\phi$ be the angle of rotation of the turntable which spins at constant rate $\dot{\phi}_{0}$ . A bug starts on the outside edge of the turntable of radius $R$ and walks towards the center, passes through it, and continues to the opposite edge of the turntable. The bug walks at a constant speed $v_{A}$ , as measured by how far her feet move per step, on the line inscribed on the table. Ignore gravity.

(a)

Picture. Make an accurate drawing of the bug’s path as seen in the room (which is not rotating with the turntable). In order to make this plot, you will need to assume values of $v_{A}$ and $\dot{\phi}_{0}$ and initial values of $R$ and $\phi$ . You will need to write a parametric equation for the path in terms of variables that you can plot (probably $x$ and $y$ coordinates). You will also need to pick a range of times. Your plot should include the instant at which the bug walks through the origin. Make sure your $x$ and $y$ - axes are drawn to the same scale. A computer plot would be nice.
(b)

Calculate the radius of curvature of the bug’s path as it goes through the origin.
(c)

Accurately draw (say, on the computer) the osculating circle when the bug is at the origin on the picture you drew for (a) above.
(d)

Force. What is the force on the bug’s feet from the turntable when she starts her trip? Draw this force as an arrow on your picture of the bug’s path.
(e)

Force. What is the force on the bug’s feet when she is in the middle of the turntable? Draw this force as an arrow on your picture of the bug’s path.

18.3.3 A small bug is crawling on a straight line scratched on an old record. The scratch is a distance $\ell=6\,\mathrm{cm}$ from the center of the turntable. The turntable is turning clockwise at a constant angular rate $\omega=2\,\mathrm{rad}/\mathrm{s}$ . The bug is walking, relative to the turntable, at a constant rate $v_{B/T}=12\,\mathrm{cm}/{\rm s}$ , straight along the scratch in the $y$ -direction. At the instant of interest, everything is aligned as shown in the figure. The bug has a mass $m_{B}=1$ gram.

(a)

What is the bug’s velocity?
(b)

What is the bug’s acceleration?
(c)

What is the sum of all forces acting on the bug?
(d)

Sketch the path of the bug in the neighborhood of its location at the time of interest (indicate the direction the bug is moving on this path).

Filename:s92f1p7

Figure 18.57

18.3.4 Arm OC rotates with constant rate $\omega_{1}$ . Disc D of radius $r$ rotates about point C at constant rate $\omega_{2}$ measured with respect to the arm OC. What are the absolute velocity and acceleration of point P on the disc, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ ? (To do this problem will require defining a moving frame of reference. More than one choice is possible.)

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}=\omega_{1}L(\cos\theta\hat{% \boldsymbol{\jmath}}-\sin\theta\hat{\boldsymbol{\imath}})+(\omega_{1}+\omega_{% 2})r\hat{\boldsymbol{\imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=-\omega_{1}^{2}L(\cos\theta% \hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})+(\omega_{1}+% \omega_{2})^{2}\hat{\boldsymbol{\jmath}}$ .

(a)

Pick a suitable moving frame and do the problem.
(b)

Pick another suitable moving frame and redo the problem. Make sure the answers are the same.

Filename:twodisks-ang-mom

Figure 18.58

18.3.5 For the configuration in problem 18.3 what is the absolute angular velocity of the disk, $D$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{D}=(\omega_{1}+\omega_{2})% \hat{\boldsymbol{k}}$ .

18.3.6 For the configuration in problem 18.3, taking $\omega_{2}$ to be the angular velocity of disk $\mathcal{D}$ relative to the rod, what is the absolute angular acceleration of the disk, $D$ ? What is the absolute angular acceleration of the disk if $\omega_{1}$ and $\omega_{2}$ are not constant?

18.3.7 A turntable oscillates with displacement ${\accentset{\rightharpoonup}{\boldsymbol{x}}}_{C}(t)=A\sin(st)\hat{\boldsymbol% {\imath}}$ . The disc of the turntable rotates with angular speed and acceleration $\omega_{D}$ and ${\dot{\omega}}_{D}$ . A small bug walks along line $D E$ with velocity $v_{b}$ relative to the turntable. At the instant shown, the turntable is at its maximum amplitude $x=A$ , the line $D E$ is currently aligned with the $z$ -axis, and the bug is passing through point $B$ on line $D E$ . Point $B$ is a distance $a$ from the center of the turntable, point $C$ . Find the absolute acceleration of the bug, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{b}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{b}=-(s^{2}A+a\omega_{D}^{2}+2% \omega_{D}v_{b})\hat{\boldsymbol{\imath}}-a\dot{\omega}_{D}\hat{\boldsymbol{k}}$ .

Filename:pfigure-blue-90-1

Figure 18.59

18.3.8 A small $0.1\,$ kg toy train engine is going clockwise at a constant rate (relative to the track) of $2\,\mathrm{m}/\mathrm{s}$ on a circular track of radius $1\,\mathrm{m}$ . The track itself is on a turntable $\cal B$ that is rotating counter- clockwise at a constant rate of $1\,\mathrm{rad}/\mathrm{s}$ . The dimensions are as shown. At the instant of interest the train is pointing due south ( $-\bf j$ ) and is at the center $O$ of the turntable.

(a)

What is the velocity of the train relative to the turntable ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{/\cal B}$ ?
(b)

What is the absolute velocity of the train ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ ?
(c)

What is the acceleration of the train relative to the turntable ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{/\cal B}$ ?
(d)

What is the absolute acceleration of the train ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ ?
(e)

What is the total force acting on the train?
(f)

Sketch the path of the train for one revolution of the turntable (surprise)?

Filename:pfigure-blue-110-1

Figure 18.60

18.3.9 A giant bug walks on a horizontal disk. An $x y z$ frame is attached to the disk. The disk is rotating about the $z$ axis (out of the paper) and simultaneously translating with respect to an inertial frame $X Y Z$ plane. In each of the cases shown in the figure, determine the total force acting on the bug. In each case, the dotted line is scratched on the disk and is the path the bug follows walking in the direction of the arrow. The location of the bug is marked with a dot. At the instants shown, the $x y z$ coordinate system shown is aligned with the inertial $X Y Z$ frame (which is not shown in the pictures). In this problem:

${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ = the position of the center of the disk,
${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ = the velocity of the center of the disk,
${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ = the acceleration of the center of the disk,
$\Omega$ the angular velocity of the disk (i.e., ${\accentset{\rightharpoonup}{\boldsymbol{\Omega}}}=\Omega\boldsymbol{k}$ ),
${\dot{\Omega}}$ = the angular acceleration of the disk,
$u$ = the speed of the bug traversing the dotted line (arc length on the disk per unit time),
${\dot{u}}$ = $du/dt$ , and
$m=1\,\mathrm{kg}$ = mass of the bug.

Filename:pfigure-blue-107-1

Figure 18.61

18.3.10 Repeat questions (a)-(f) for the toy train in problem 18.3 going counter-clockwise at constant rate (relative to the track) of $2\,\mathrm{m}/\mathrm{s}$ on the circular track.

18.3.11 A nostalgic honeybee alights on a phonograph turntable that is being carried by a carnival goer who is riding on a carousel. The situation is sketched below. The carousel has angular velocity of $5\,{\rm rpm}$ , which is increasing (accelerating) at $10\,{{\rm rev}/{\rm min}^{2}}$ ; the phonograph rotates at a constant 33 1/3 $\,{\rm rpm}$ . The honeybee is at the outer edge of the phonograph record in the position shown in the figure; the radius of the record is 7 inches. Calculate the magnitude of the acceleration of the honeybee.

Filename:pfigure-s94h14p4

Figure 18.62

18.3.12 Consider a turntable on the back of a pick-up truck. A bug walks on a line on the turntable. The line may or may not be drawn through the center of the turntable. The truck may or may not be going at constant rate. The turntable may or may not be spinning, and, if it spins, it may or may not go at a constant rate. The bug may be anywhere on the line and may or may not be walking at a constant rate.

(a)

Draw a picture of the situation. Clearly define all variables you are going to use. What is the moving frame, and what is the reference point on that frame?
(b)

One term at a time. For each term in the five term acceleration formula, find a situation where all but one term is zero. Use the turntable as the moving frame, the bug as the particle of interest. Answer: One case: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{O^{\prime}}$ , where $P$ refers to the bug; the bug stands still on the turntable, the truck accelerates, and the turntable does not rotate.
(c)

Two terms at a time. (harder) How many situations can you find where a pair of terms is not zero but all other terms are zero? (Don’t try to do all 10 cases unless you really think this infamous formula is fun. Try at least one or two.) Answer: One case: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{O^{\prime}}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% P/\cal D}$ , where $\cal D$ is the turntable; the bug accelerates along a straight line on the turntable, the truck accelerates, and the turntable does not rotate.

Filename:pfigure-blue-112-1

Figure 18.63

Picking apart the five term acceleration formula.

18.4 Kinematics of 2-D mechanisms

An ideal mechanism or linkage is a collection of rigid objects constrained to move relative to the ground and each other by hinges but which still has some possible motion(s). People also use the word mechanism or linkage more loosely to include any collection of machine parts connected by any means.

The analysis of the kinematics of mechanisms is an important part of machine design. Mechanisms synthesis, coming up with a mechanism design which has desired motions, is obviously key in creative design, and nowadays in computer aided design.

Finally, the determination of the dynamics of a mechanism, how it will move and with what forces, is completely dependent on understanding the kinematics of the mechanism. The whole subject of mechanism kinematic analysis, although in some sense a subset of dynamics, is actually a huge and infinitely complex subject in itself

^†^†margin: Very advanced aside. The general kinematics of planar linkages is as complicated as the classification of closed surfaces in any number of dimensions: spheres, donuts, spheres with two holes, etc. For every surface there is a corresponding 2D mechanism. In impenetrable math-speak: Any orientable manifold is a connected component of the configuration space of some planar linkage. This idea was proposed by Bill Thurston in the 1970’s and later proved by Kapovich and Millson. We report this just so you don’t feel bad that you don’t know now, and never will know, all there is to know about planar mechanisms.

and also a useful subject in itself. Often kinematics is the central interest in machine design, and mechanics (force and acceleration) analysis is only carried out if something that shouldn’t do so, breaks or shakes. This section presents some of the basic ideas in kinematic analysis. The overall question in mechanism kinematic analysis is this:

Given a collection of parts and a description of how they are connected, in what ways can they move?

Without getting into the details of the motions yet, the first question to answer is simpler than finding the motions, but just counting them: In how many ways can the mechanism move?

Degrees of freedom (DOF)

The number of degrees of freedom (DOF) $n_{\rm DOF}$ of a mechanism is the number of different ways it can move. More precisely

The number of degrees of freedom $n_{\rm DOF}$ of a mechanism is the minimum number of configuration variables needed to describe all possible configurations of the mechanism.

The minimum number of configuration variables $n_{\rm DOF}$ is a property of the mechanism. The choice of what these variables are, however is not unique to a given mechanism.

Example: A particle in a plane has 2 degrees of freedom.

The set of ‘configurations’ of a particle in a plane is the set of positions of the particle. This is fully described by its $x$ and $y$ coordinates. Thus $n_{\rm DOF}=2$ . But the configuration is also determined by the particle’s polar coordinates $R$ and $\theta$ . And there are an infinite number of other pairs of numbers that could be used to describe the configurations (e.g., the $x^{\prime}$ and $y^{\prime}$ coordinates, the $w$ and $z$ coordinates with $w=e^{x}$ and $z=e^{y}$ , etc). The minimum number of configuration variables, 2, is unique, but the choice of variables is not.

Filename:tfigure8-alt-app2c

Figure 18.64: The number of degrees of freedom that are eliminated by various constraints. See text for explanation.

Filename:sfig4-6-4a

Figure 18.65: Some simple mechanisms and their number of degrees of freedom,

n

(called

n_{\rm DOF}

in the text).

For planar mechanisms one can often determine the number of degrees of freedom by the following formula

3\cdot(\underbrace{\hbox{\# of rigid objects}}_{n_{\rm bod}})+2\cdot(% \underbrace{\hbox{\# of particles}}_{n_{\rm part}})-(\underbrace{\hbox{\# of % constraints}}_{n_{\rm con}})=n_{\rm DOF}

(18.75)

The formula starts with the number of ways one rigid object can move (2 translations and a rotation makes 3) and one particle can move (just 2 translations) and then subtracts the restrictions to the motion. In eqn. (18.75) the number $n_{\rm con}$ of constraints is counted as the number of degrees of freedom restricted by the connections.

Examples of connections and their effect on $n_{\rm DOF}$

See fig. 18.64 for some standard idealized connections and their number of constraints (assuming they are already constrained to a plane).

a)

2 for a pin joint: a pin joint restricts relative motion in 2 directions but still allows relative rotation. If three objects are connected at one pin then it counts as two pin joints and thus $2\times 2=4$ reductions in the number of degrees of freedom. There are 6 reductions for 4 objects connected at one pin, etc.
b)

3 for a welded connection: a weld restricts relative translation in two directions as well as relative rotation ( $2+1=3$ ). So two parts that are welded together have $2\cdot 3-3=3$ degrees of freedom. That is, any collection of rigid objects welded together is the same as one rigid object. The word ‘weld’ is meant to include any collection of bolts, glue, string, rivets or bailing wire that prevents any relative motion.
c)

1 for a sliding contact: the sliding contact restricts relative translation normal to the contact surfaces and allows translation tangent to the surfaces. Relative rotation is also allowed.
d)

2 for a keyed sliding contact: allows relative translation in one direction but disallows translation in one direction as well as rotation.
e)

1 for a massless link hinged at its ends to two objects: this keeps the distance between two points fixed which is one restriction (alternatively the bar adds 3 degrees of freedom and each hinge subtracts 2 ( $+3-2\times 2=-1$ degree of freedom).
f)

2 for a rolling contact: relative slip is not allowed nor is interpenetration.

Be warned

If some of the constraints are redundant then a system can have more degrees of freedom than eqn. (18.75) indicates. But a mechanism never has fewer degrees of freedom than eqn. (18.75) indicates.

Some simple mechanisms

Figure 18.65 shows some examples of simple mechanisms and the number of degrees of freedom. In each case we look at eqn. (18.75): $3\cdot n_{\rm bod}+2\cdot n_{\rm part}-n_{\rm con}=n_{\rm DOF}$ .

a)

An object connected to ground by a hinge has 1 degree of freedom; the set of all possible configurations can be described by the angle of the object: $n_{\rm bod}=1,n_{\rm con}=2\quad\Rightarrow\quad n_{\rm DOF}=1$ .
b)

An unconstrained object has 3 degrees of freedom; the set of all configurations can be described by the $x$ and $y$ coordinates of a reference point and by the rotation $\theta$ : $n_{\rm bod}=1,n_{\rm con}=0\quad\Rightarrow\quad n_{\rm DOF}=3$ .
c)

A bead on a wire has 1 degree of freedom; its configuration is fully determined by the distance the bead has advanced along the wire relative to a reference mark: $n_{\rm part}=1,n_{\rm con}=1\quad\Rightarrow\quad n_{\rm DOF}=1$ .
d)

A statically determinate truss has 0 degrees of freedom; a statically determinate truss has no way to deform: $n_{\rm bod}=n_{\rm bars},n_{\rm part}=0,n_{\rm con}=2\cdot n_{\rm pins}+n_{\rm ground% \ const}\quad\Rightarrow n_{\rm DOF}=0$ . Note that the number of pin restrictions in our count here is more than twice the number of joints in the study of trusses in statics. In statics, we focussed on joints as restricted by bars. Here we look at bars as restricted by joints. A given joint counts 2, 4 or 6 pin restrictions depending on whether it connects 2, 3 or 4 bars. Thus the 11 bar truss shown has 7 joints (by truss-analysis counting) but, by the counting here it has $2\times 2+2\times 4+3\times 6=30$ pin restrictions plus 3 ground restrictions for 33 restrictions in all. Without restrictions the 11 bars had 33 degrees of freedom. So, because 33-33 = 0, and none of the restrictions is redundant, the truss has zero degrees of freedom.
e)

A rolling wheel has 1 degree of freedom; its configuration is fully determined either by the net angle $\theta$ it has rolled or by the $x$ coordinate of its center: $n_{\rm bod}=1$ , $n_{\rm con}=2\quad\Rightarrow\quad n_{\rm DOF}=1$ .
f)

A double pendulum or two-link robot arm has 2 degrees of freedom; its configuration is determined by the net rotations of its two links (or by the rotation of the first link and the relative rotation of the second link): $n_{\rm bod}=2,n_{\rm con}=4\quad\Rightarrow\quad n_{\rm DOF}=2$ .
g)

A cart with two rolling wheels has 1 degree of freedom; given the position of the cart, say $x$ , the rotation of the wheels is determined. Similarly for a bicycle. For this model of a bicycle we are neglecting the steering degree of freedom and also the freedoms of the crank and pedals. For a vehicle with 2 wheels in 2D we have: $n_{\rm bod}=3,n_{\rm con}=4\times 2=8$ (2 hinges and 2 rolling contacts) $\quad\Rightarrow\quad n_{\rm DOF}=9-8=1$ .
h)

A “four” bar linkage has 1 degree of freedom; the angle of any one of the bars determines the angles of the others: $n_{\rm bod}=4,n_{\rm con}=2\times 4+2+1=11$ (there are 4 pin joints between the bars, one pin joint to ground and one roller connection to the ground) $\quad\Rightarrow\quad n_{\rm DOF}=1$ .
i)

A slider crank has 1 degree of freedom; the rotation of the crank determines the configuration of the system $n_{\rm bod}=3,n_{\rm con}=3\cdot 2+2$ (there are 3 pins and one keyed connection) $\quad\Rightarrow\quad n_{\rm DOF}=1$ .
j)

An ideal gear train (with all gears pinned to ground) has 1 degree of freedom; the amount of rotation of any one gear determines the rotation of all of the gears: In this case the counting formula is wrong. Say there are 2 gears, then $n_{\rm bod}=2,n_{\rm con}=3\cdot 2=6$ (two pins and one rolling contact) $\Rightarrow\quad n_{\rm DOF}=0\neq 1$ . The rolling constraint prevents interpenetration, but this was already prevented by the hinges at the center of the gears. The constraints are redundant and the system has more degrees of freedom than eqn. (18.75) indicates.
k)

A redundant swing with one horizontal bar suspended by 3 parallel struts has 1 degree of freedom; the angle of one upright link determines the full configuration of the mechanism. The counting formula is again wrong: $n_{\rm bod}=4,n_{\rm con}=6\cdot 2=12\quad\Rightarrow\quad n_{\rm DOF}=0$ underestimates the number of degrees of freedom because the constraints are redundant.
l)

A 2-D 10-link model of a person with one foot on the ground has 10 degrees of freedom; the angles of the 10 links (4 arm links, 4 leg links, a body and a head) determine the full configuration of the mechanism. There are no redundant constraints so the counting formula also works using $n_{\rm bod}=10,n_{\rm hinges}=10$ :

$\displaystyle n_{\rm DOF}$ $\displaystyle=3*n_{\rm bod}-n_{\rm con}$ (18.76)

$\displaystyle=3*n_{\rm bod}-2\cdot n_{\rm hinges}=30-2\cdot 10=10$ (18.77)

What are the 10 hinges? One at the ground, 1 at each knee and elbow, 2 at the hip and 3 at the shoulders (one hinge for each object linked to the trunk).

Configuration variables

Once we know the number of degrees of freedom $n_{\rm DOF}$ of a system it is often useful to settle on one set of $n_{\rm DOF}$ configuration variables. In this book $n_{\rm DOF}$ will be 1, 2 or at most 3. Thus we pick 1,2 or 3 variables.

Example: Straight line motion

Chapter 14 on straight line motion was mostly about one-degree-of-freedom systems ( $n_{\rm DOF}=1$ ). These systems could all be characterized by the single configuration variable $x$ , the displacement along the line of a reference point on the object relative to a reference point on the ground. All the positions,velocities and accelerations of all points in the system could be found in terms of $x$ , $\dot{x}$ and $\ddot{x}$ (in fact all points had ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{x}\hat{\boldsymbol{\imath}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=\ddot{x}\hat{\boldsymbol{\imath}}$ ).

Example: Circular motion about a fixed axis

In chapters 15 and 16 we were almost entirely focussed on systems with one degree of freedom well characterized by the one configuration variable, the rotation angle $\theta$ . For such motions positions, velocities, and accelerations of all points were determined by the initial positions of the points $\theta,\dot{\theta}$ and $\ddot{\theta}$ by equations which you know well by now.

For more general motions we almost always take inspiration from the two examples above. We use the translation of a conspicuous point, or we use the rotation of a conspicuous object for a configuration variable. And more of the same if the system has more than 1 degree of freedom. The natural choice of configuration variables for some simple mechanisms is given in the text discussing fig. 18.65.

Often our main kinematic task is to express the full configuration of the system as well as all the velocities and accelerations of all its parts in terms of the positions of the parts, the configuration variables, and their first and second time derivatives.

Adding relative angular velocities

One last simple kinematic fact is needed before we can plug and chug with the theory we have so far and apply it to general kinematic mechanisms. It concerns the addition of rotations and rotation rates. The following example basically tells the whole story

Example: Double pendulum and the addition of rotation rates

The commonly used configuration variables for the double pendulum shown in fig. 18.66 are $\theta_{1}$ and $\theta_{2}$ . To actually know the configuration of the system obviously we need to know $\phi$ which is given by

\phi=\theta_{1}+\theta_{2}\quad\mbox{so}\quad\dot{\phi}=\dot{\theta}_{1}+\dot{% \theta}_{2}\quad\mbox{and so}\quad\ddot{\phi}=\ddot{\theta}_{1}+\ddot{\theta}_% {2}.

[Aside: One reason for choosing $\theta_{2}$ instead of $\phi$ as a configuration variable is that if one was measuring or controlling the second link, say as a robotic arm, the angle $\theta_{2}$ can be measured more easily than $\phi$ . Also, it turns out (in hindsight) that the differential equations of motion are slightly simpler using $\theta_{2}$ instead of $\phi$ .]

Filename:sfig4-6-3

Figure 18.66: A double pendulum’s configuration variables are most often taken to be

\theta_{1}

and

\theta_{2}

. But one then needs to know that

\phi=\theta_{1}+\theta_{2}

and that

\dot{\phi}=\dot{\theta}_{1}+\dot{\theta}_{2}

and

\ddot{\phi}=\ddot{\theta}_{1}+\ddot{\theta}_{2}

Looking at the bars as being glued to reference frames ( $\cal F$ for the fixed frame, $\cal B$ for bar AB, and $\cal C$ for bar BC), the above example shows that

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/F}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/F}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/B}$	(18.78)
which is often written with the simple notation
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{1}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{2}$

Which can only be given strict meaning by the more elaborate eqn. (18.78) above it.

Example: Double pendulum (see previous example)

Take ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/F}=\dot{\theta}_{1}% \hat{\boldsymbol{k}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/B}=\dot{\theta}_{2}% \hat{\boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/F}=\dot{\phi}\hat{% \boldsymbol{k}}$ and eqn. (18.78) is self evident from the addition of angles.

Similarly we have

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C/F}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B/F}+{% \accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C/B}$	(18.79)
which is often written with the simple notation
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{1}+{% \accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{2}.$

For those going on to study 3-D mechanics, one should note that, unlike eqn. (18.78), eqn. (18.79) does not hold in 3-D.

Kinematics of mechanisms

One approach to mechanisms is to do what one can with high-school geometry and trigonometry, the laws of sines (see page 1.5), and so on.

Example: Rod on step using geometry and trigonometry

One end of a rod slides on the ground. The other end slides on a corner at A (see fig. 18.67). Given that ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}=v_{B}% \hat{\boldsymbol{\imath}}$ we can find $\dot{\phi}$ as follows:

	$\displaystyle{h\over\ell_{OB}}=\tan\phi\Rightarrow\left\{{\ell_{OB}\over h}={% \cos\phi\over\sin\phi}\right\}$
	$\displaystyle{d\over dt}\left\{\phantom{d\over dt}\right\}\Rightarrow{\dot{% \ell}_{OB}\over h}={-\dot{\phi}\over\sin^{2}\phi}\Rightarrow\dot{\phi}=-{v_{B}% \sin^{2}\phi\over h}$

Filename:sfig4-6-3a

Figure 18.67: A rigid rod slides so that it always touches the ground and the corner at A. One would like to know the relation between

\dot{\phi}

and

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}

As the above calculation shows, this problem doesn’t need the heavy machinery of our moving-frame vector methods. But it provides an instructive example.

Example: Rod on step using moving-frame methods (see previous example)

We look at point A and note that we can think of it as a fixed point in the fixed frame $\cal F$ and also as a point that is moving relative to the translating and rotating frame $\cal B$ . We evaluate its velocity both ways.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{A}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{A}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{{\rm A}/\cal B}$	(eqn. (18.67))
$\displaystyle\{{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle v_{B}\hat{\boldsymbol{\imath}}-\dot{\phi}\hat{\boldsymbol{k}}% \times\left(\ell_{BA}\hat{\boldsymbol{\lambda}}_{BA}\right)+v_{A/\cal B}\hat{% \boldsymbol{\lambda}}_{BA}\}$
$\displaystyle\Rightarrow 0$	$\displaystyle=$	$\displaystyle v_{B}\underbrace{\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{% n}}_{BA}}_{\sin\phi}+\dot{\phi}\ell_{BA}$	( $\{\}\cdot\hat{\boldsymbol{n}}_{BA}$ )
$\displaystyle\Rightarrow\dot{\phi}$	$\displaystyle=$	$\displaystyle-{v_{B}\sin\phi\over\ell_{BA}}=-{v_{B}\sin^{2}\phi\over h}$	( $\displaystyle\sin\phi={h\over\ell_{AB}}$ )

as we had before. The key equation was the ‘three term velocity formula’ eqn. (18.67) on page 18.67 and the observation that relative to frame $\cal B$ point A slides along the rod. Note that we never had to explicitly use the rotating coordinates associated with frame $\cal B$ to do this calculation.

You should understand the examples above, and the needed background material, before going on to the following examples.

Filename:pfigure4-4-rp12

Figure 18.68: The slider crank mechanism.

Example: Slider Crank using geometry and trigonometry

The slider crank mechanism (fig. 18.68) was briefly introduced in the context of statics where its forces could be analyzed assuming inertial terms were negligible (see 7.3). But it is a commonly used mechanism (e.g., in every car) and its motions are of central interest. The angle $\theta$ is the most natural configuration variable for this $n_{\rm DOF}=1$ system. One would like to know the position, velocity and acceleration of the slider ( $x_{C},\dot{x}_{C}$ and $\ddot{x}_{C}$ ) in terms of $\theta,\dot{\theta}$ and $\ddot{\theta}$ .

$\displaystyle x_{C}$	$\displaystyle=$	$\displaystyle x_{D}+\ell_{DC}$	(18.80)
	$\displaystyle=$	$\displaystyle d\cos\theta+\sqrt{\ell^{2}-h^{2}}$
	$\displaystyle=$	$\displaystyle d\cos\theta+\sqrt{\ell^{2}-(d\sin\theta)^{2}}$
	$\displaystyle=$	$\displaystyle d\left(\cos\theta+\sqrt{(\ell/d)^{2}-\sin^{2}\theta}\right)$
	$\displaystyle=$	$\displaystyle d\left(\cos\theta+\sqrt{(\ell/d)^{2}+(\cos 2\theta-1)/2}\right)$

The positive $\sqrt{\ }$ corresponds to C being to the right of 0. The negative $\sqrt{\ }$ corresponds to point C being to the left of 0. The mechanism just doesn’t work for a full revolution of the link 0A if $\ell<d$ as you can see from the picture or from that the $\sqrt{\ }$ above giving imaginary values for $\cos 2\theta$ near -1, $\sin\theta$ near 1, and $\theta$ near $\pi/2$ .

To get the velocity of point C we just take the derivative of eqn. (18.80) above.

$\displaystyle v_{C}$	$\displaystyle=$	$\displaystyle\dot{x}_{C}$
	$\displaystyle=$	$\displaystyle{d\over dt}\left\{d\left(\cos\theta+\sqrt{(\ell/d)^{2}+(\cos 2% \theta-1)/2}\right)\right\}$
	$\displaystyle=$	$\displaystyle\dot{\theta}d\left\{-\sin(\theta)-\sin(2\,\theta){\frac{1}{\sqrt{% 4\,{\frac{{l}^{2}}{{d}^{2}}}+2\,\cos(2\,\theta)-2}}}\right\}$

Filename:pfigure-blue-38-2

Figure 18.69: Position

x_{C}

of the slider, its velocity

v_{C}

and its acceleration

a_{c}

(Strictly, e.g.,

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}=a_{C}\hat{% \boldsymbol{\imath}}

). Two sets of curves are shown. (a) a long connecting rod, and (b) a connecting rod a hair longer than the crank. In case (a) the connecting rod is nearly horizontal at all times and the displacement of point C is almost entirely due to the horizontal displacement of the end of the crank arm. Thus point C moves with almost exactly a cosine wave with amplitude equal to the length of the crank. The velocity and acceleration curves are thus also sine and cosine waves. In case (b) the motion is close to a cosine curve approximately when

-60^{{}^{\circ}}<\theta<60^{{}^{\circ}}

. That is, the displacement of C is about twice the horizontal displacement of the end of the crank arm when the end is to the right of its base bearing. When the end of the crank arm is to the left of the bearing point C is nearly stationary and is just to the right of the crank base bearing. The transition between these two cases involves a sudden large acceleration.

To get the acceleration we differentiate once again. For simplicity let’s assume the crank rotates at constant rate, so $\dot{\theta}$ is a constant and $\ddot{\theta}=0$ . Cranking out the derivative of eqn. (18.4), so to speak, we get

$\displaystyle a_{c}=\dot{v}_{C}$	$\displaystyle=$	$\displaystyle{d\over dt}\left\{\mbox{the mess on the right of eqn.~{}(\ref{eq:% crankvel})}\right\}$
	$\displaystyle=$	$\displaystyle-d\dot{\theta}^{2}\left\{\vphantom{+2\,\cos(2\,\theta){\frac{1}{% \sqrt{4\,{\frac{{l}^{2}}{{d}^{2}}}+2\,\cos(2\,\theta)-2}}}}\cos(\theta)\right.$
		$\displaystyle\qquad\quad+2\,\sin^{2}(2\,\theta)\left(4\,{\frac{{l}^{2}}{{d}^{2% }}}+2\,\cos(2\,\theta)-2\right)^{-3/2}$
		$\displaystyle\qquad\quad\left.+2\,\cos(2\,\theta){\frac{1}{\sqrt{4\,{\frac{{l}% ^{2}}{{d}^{2}}}+2\,\cos(2\,\theta)-2}}}\qquad\right\}$

So we now know the position, velocity and acceleration of point C in terms of $\theta,\dot{\theta}$ and $\ddot{\theta}$ . You should commit the solution eqn. (18.4) to memory. Just kidding.

Plots of $x_{C},v_{C}$ , and $a_{C}$ from these equations are shown in fig. 18.69ab for two different extremes of slider crank design: one with a very long connecting rod that gives sinusoidal motion, and one with a connecting rod just barely long enough to prevent locking that gives intermittent motion.

Unlike some more complex mechanisms, the slider crank is solvable in that one can write a formula for the position of any point of interest in terms of the single configuration variable $\theta$ . For more complex mechanisms this may not be possible. Further, even if possible the above example shows that the differentiation required to find velocity and acceleration can lead to a bit of a mess.

A different approach is to assume that at some value of the configuration variable ( $\theta$ for the slider crank) that the full configuration of the system is known. That is, that the locations of all points are known. Then we can use our vector methods to find velocities and accelerations of all points of interest.

Example: Slider crank using vector methods (see previous example)

Take the slider crank of fig. 18.68 to be in some known configuration. We now try to find the velocity and acceleration of point C in terms of the positions of the points 0, B, and C as well as $\theta$ and $\dot{\theta}$ .

The basic approach is to write true things, and then solve for unknowns. First work on velocities. The basic idea is to look at the closure condition. That is, the velocity of point C as calculated by working down the linkage from 0 to A to C has to be consistent with the velocity of C as calculated in the fixed frame.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}$
$\displaystyle v_{C}\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{A/0}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C/A}}$
$\displaystyle v_{C}\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle(\dot{\theta}\hat{\boldsymbol{k}})\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/0}}+(-\dot{\phi}\hat{% \boldsymbol{k}})\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm% {C/A}}$	(18.83)

eqn. (18.83) is a 2-D vector equation in the 2 unknown scalars $v_{C}$ and $\dot{\phi}$ . It could be solved as a pair of equations, or solved directly by first dotting both sides with $\hat{\boldsymbol{\jmath}}$ to find $\dot{\phi}$ and dotting both sides with ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}$ to find $v_{C}$ . These yield

	$\displaystyle\dot{\phi}$	$\displaystyle=$	$\displaystyle{\left((\dot{\theta}\hat{\boldsymbol{k}})\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/0}}\right)\cdot\hat{\boldsymbol% {\jmath}}\over\left(\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{C/A}}\right)\cdot\hat{\boldsymbol{\jmath}}}\quad% \mbox{and}$
	$\displaystyle v_{C}$	$\displaystyle=$	$\displaystyle{\dot{\theta}\left(\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/0}}\right)\cdot{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}\over\hat{\boldsymbol{\imath% }}\cdot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}}$

where everything on the right of these equations is assumed known. Without grinding out the vector products in terms, say, of components, we can just use that we know $\dot{\phi}$ and $v_{C}$ .

We proceed to find the accelerations by similar means, assuming $\dot{\theta}$ is a constant so ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal A}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ :

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}$
$\displaystyle a_{C}\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{A/0}}+{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C/A}}$
$\displaystyle a_{C}\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle-\dot{\theta}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \!\mathrm{A/0}}+(-\ddot{\phi}\hat{\boldsymbol{k}})\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}-\dot{\phi}^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}$	(18.84)

eqn. (18.84) is a 2-D vector equation in the two scalar unknowns $a_{c}$ and $\ddot{\phi}$ . We can set this up as two equations in two unknowns. Or we can solve for $\ddot{\phi}$ directly by dotting both sides with $\hat{\boldsymbol{\jmath}}$ and we can solve for $a_{C}$ directly by crossing both sides with ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}$ or by dotting with a vector perpendicular to ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}$ (e.g., ${\accentset{\rightharpoonup}{\boldsymbol{r}}}^{\perp}_{{}_{C/A}}=\hat{% \boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{% C/A}}$ )

Although we have presented an algorithm rather than a formula, we have found the velocity and acceleration of C without writing any large equations of the type needed in the previous example. The shortcoming is that this method depends on knowing the full configuration at the time of interest.

Filename:pfigure-blue-49-2

Figure 18.70: A four bar linkage.

Example: Four bar linkage using geometry and trigonometry

fig. 18.70 shows a “four-bar linkage”. Please see 7.3 for an introduction to 4-bar linkages in the context of statics. Four bar linkages are solvable in the sense that one can write equations for the positions of any point of interest in terms of the single configuration variable $\theta$ marked in fig. 18.70. But the formulas are really a mess. And the first and second time derivatives are an unbelievable mess.

The four-bar linkage is about as complex a system as can be solved in this sense, and it is probably too-complex for this solution to be useful in the kinematic analysis of accelerations.

For complex mechanisms one is often stuck using vector methods, like we are for practical purposes stuck with the 4-bar linkage. But the vector methods based on the current configuration are not crippled by complexity.

Example: Four-bar linkage using relative velocities and accelerations

Assuming the configuration is known (i.e., that $\theta$ , $\phi_{1}$ , and $\phi_{2}$ are known), we can proceed with the 4-bar linkage just as we did for the slider crank. We enforce closure by picking a point and thinking about its velocity two different ways

^†^†margin: Relative vs absolute angles. In this example, and the examples above, we use the absolute rotations of the links. However, we could also have done the calculations using the relative rotations and then used the formulas for velocity and acceleration relative to a rotating frame. For these examples this would make the calculations slightly, but not intractably, longer.

We could pick any point, say C. From the fixed frame we know that the velocity of C is zero. Working around the linkage, link by link, we know it is the sum of relative velocities as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{A/0}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B/A}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C/B}}$
	$\displaystyle=$	$\displaystyle\left(\dot{\theta}\hat{\boldsymbol{k}}\right)\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/0}}+\left(\dot{\phi}_{1}\hat{% \boldsymbol{k}}\right)\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{B/A}}+\left(\dot{\phi}_{2}\hat{\boldsymbol{k}}\right)\times{\accentset% {\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}$

which is equivalent to two scalar equations in the two unknowns $\dot{\phi}_{1}$ and $\dot{\phi}_{2}$ . This equation can be solved directly for $\dot{\phi}_{2}$ by taking the dot product of both sides with a vector perpendicular to ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}$ (such as $\hat{\boldsymbol{\jmath}}^{\prime\prime}$ or $\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{B/A}}$ ) and for $\dot{\phi}_{1}$ by taking the dot product of both sides with for a vector perpendicular to ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}$ (such as $\hat{\boldsymbol{\jmath}}^{\prime\prime\prime}$ or $\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{C/B}}$ ) to get

	$\displaystyle\dot{\phi}_{1}$	$\displaystyle=$	$\displaystyle-\dot{\theta}\frac{\left(\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/0}}\right)\cdot\hat{\boldsymbol% {\jmath}}^{\prime\prime\prime}}{\left(\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}\right)\cdot\hat{\boldsymbol% {\jmath}}^{\prime\prime\prime}}\hbox to36.135pt{\hfil and\hfil}$
	$\displaystyle\dot{\phi}_{2}$	$\displaystyle=$	$\displaystyle-\dot{\theta}\frac{\left(\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/0}}\right)\cdot\hat{\boldsymbol% {\jmath}}^{\prime\prime}}{\left(\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}\right)\cdot\hat{\boldsymbol% {\jmath}}^{\prime\prime}}.$

The dot product with $\hat{\boldsymbol{k}}$ is used to get a scalar on the top and bottom of the fraction, both vectors are already only in the $\hat{\boldsymbol{k}}$ direction. Now that $\dot{\phi}_{1}$ and $\dot{\phi}_{2}$ are known the velocity of any point on the mechanism is known. For example

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}=\left(% \dot{\phi}_{2}\hat{\boldsymbol{k}}\right)\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{B/C}}.

The angular accelerations of the two links are found by the same method. For simplicity let’s assume that the driving crank 0A spins at constant rate so $\ddot{\theta}=0$ . Looking at the acceleration of point C two ways we have

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{A/0}}+{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B/A}}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C/B}}$
	$\displaystyle=$	$\displaystyle\phantom{+}\underbrace{\left(\ddot{\theta}\hat{\boldsymbol{k}}% \right)}_{0}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/0% }}-\dot{\theta}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/% 0}}$
		$\displaystyle+\left(\ddot{\phi}_{1}\hat{\boldsymbol{k}}\right)\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}-\dot{\phi}_{1}^{% 2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}$
		$\displaystyle+\left(\ddot{\phi}_{2}\hat{\boldsymbol{k}}\right)\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}-\dot{\phi}_{2}^{% 2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}$

Because $\dot{\phi}_{1}$ and $\dot{\phi}_{2}$ are already known, this is one equation in the two unknowns $\ddot{\phi}_{1}$ and $\ddot{\phi}_{2}$ . They can be solved for $\ddot{\phi}_{1}$ by taking the dot product of both sides with $\hat{\boldsymbol{\jmath}}^{\prime\prime\prime}$ and for $\ddot{\phi}_{2}$ by taking the dot product of both sides with $\hat{\boldsymbol{\jmath}}^{\prime\prime}$ .

At this point you know $\theta$ , $\dot{\theta}$ , $\phi_{1}$ , $\dot{\phi}_{1}$ , $\ddot{\phi}_{1}$ , $\phi_{2}$ , $\dot{\phi}_{2}$ , and $\ddot{\phi}_{2}$ and can thus calculate the position, velocity and acceleration of any point in the mechanism.

SAMPLE 18.12

Filename:pfigure4-4-rp13

Figure 18.71:

Two rods connected in a one-DOF mechanism. A mechanism consists of two rods AB and CD connected together at P with a collar pinned to AB but free to slide on CD. Rod AB is driven with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=10\,\mathrm{rad}/% \mathrm{s}\hat{\boldsymbol{k}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=4\,\mathrm{rad}/% \mathrm{s}^{2}\hat{\boldsymbol{k}}$ . At the instant shown, $\theta=45^{{}^{\circ}}$ and $\beta=30^{{}^{\circ}}$ . The length of rod AB is $R=0.4\,\mathrm{m}$ . At the instant shown,

1.

Find the angular velocity and angular acceleration of rod CD.
2.

Find the velocity and acceleration of the collar with respect to rod CD.

Solution Here, we are interested in instantaneous kinematics of this mechanism. Since point P is on rod AB as well as on rod CD, its velocity and acceleration can be expressed in terms of the angular motion of rod AB or that of rod CD. Let us consider rod AB first. Let $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ be basis vectors attached to rod AB that rotate with the rod. Since P is fixed on rod AB, it executes simple circular motion about A with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=\dot{\theta}\hat{% \boldsymbol{k}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=\ddot{\theta}\hat{% \boldsymbol{k}}$ where $\dot{\theta}=10\,\mathrm{rad}/\mathrm{s}$ and $\ddot{\theta}=4\,\mathrm{rad}/\mathrm{s}^{2}$ , respectively. Then

Filename:f92h7p1

Figure 18.72: The position of point P can be written as

R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}

r\hat{\boldsymbol{\lambda}}

where

r=R\sin\theta/\sin\beta

(since

r\sin\beta=R\sin\theta

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{P}}$	$\displaystyle=$	$\displaystyle R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$		(18.85)
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}$	$\displaystyle=$	$\displaystyle-R\dot{\theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R% \ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}.$		(18.86)

Now let us consider rod CD and express the velocity and acceleration of point P in terms of motion of rod CD. Let the angular velocity and angular acceleration of rod CD be ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{CD}=\dot{\beta}\hat{% \boldsymbol{k}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{CD}=\ddot{\beta}\hat{% \boldsymbol{k}}$ , respectively. Let $\hat{\boldsymbol{\lambda}}$ and $\hat{\boldsymbol{n}}$ be basis vectors attached to rod CD. Let the instantaneous position of point P on CD be ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}=r\hat{\boldsymbol% {\lambda}}$ . Since the collar can slide along CD, we can write the velocity and acceleration of point P as

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{P}}$	$\displaystyle=$	$\displaystyle\dot{r}\hat{\boldsymbol{\lambda}}+r\dot{\beta}\hat{\boldsymbol{n}}$		(18.87)
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}$	$\displaystyle=$	$\displaystyle(\ddot{r}-r{\dot{\beta}}^{2})\hat{\boldsymbol{\lambda}}+(2\dot{r}% \dot{\beta}+r\ddot{\beta})\hat{\boldsymbol{n}}.$		(18.88)

Filename:pfigure-blue-52-2

Figure 18.73: Geometry of the unit vectors.

From eqn. (18.85) and 18.87 we get,

$\displaystyle\dot{r}\hat{\boldsymbol{\lambda}}+r\dot{\beta}\hat{\boldsymbol{n}}$	$\displaystyle=$	$\displaystyle R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{r}$	$\displaystyle=$	$\displaystyle R\dot{\theta}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}% \cdot\hat{\boldsymbol{\lambda}})$
$\displaystyle\dot{\beta}$	$\displaystyle=$	$\displaystyle(1/r)R\dot{\theta}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!% \theta}\cdot\hat{\boldsymbol{n}})$
Similarly, from eqn. (18.86) and 18.88 we get,
$\displaystyle\ddot{r}-r{\dot{\beta}}^{2}$	$\displaystyle=$	$\displaystyle-R\dot{\theta}^{2}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}% \cdot\hat{\boldsymbol{\lambda}})+R\ddot{\theta}(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}\cdot\hat{\boldsymbol{\lambda}})$
$\displaystyle r\ddot{\beta}+2\dot{r}\dot{\beta}$	$\displaystyle=$	$\displaystyle-R\dot{\theta}^{2}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}% \cdot\hat{\boldsymbol{n}})+R\ddot{\theta}(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}\cdot\hat{\boldsymbol{n}}).$

Thus, to find all kinematic quantities of interest, all we need now is to figure out a few dot products between the two sets of basis vectors. This is easily done by writing out $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R},\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta},\hat{\boldsymbol{\lambda}}$ , and $\hat{\boldsymbol{n}}$ . ^†^†margin: We have,
$\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=\cos\theta\hat{\boldsymbol{\imath% }}+\sin\theta\hat{\boldsymbol{\jmath}}$ ,
$\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=-\sin\theta\hat{\boldsymbol{% \imath}}+\cos\theta\hat{\boldsymbol{\jmath}}$ ,
$\hat{\boldsymbol{\lambda}}=\cos\beta\hat{\boldsymbol{\imath}}+\sin\beta\hat{% \boldsymbol{\jmath}}$ ,
$\hat{\boldsymbol{n}}=-\sin\beta\hat{\boldsymbol{\imath}}+\cos\beta\hat{% \boldsymbol{\jmath}}$ .
Therefore,
$\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\cdot\hat{\boldsymbol{\lambda}}=% \cos(\beta-\theta),$
$\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\cdot\hat{\boldsymbol{n}}=\sin(% \theta-\beta),$
$\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}\cdot\hat{\boldsymbol{\lambda% }}=\sin(\beta-\theta),$
$\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}\cdot\hat{\boldsymbol{n}}=% \cos(\theta-\beta).$ Substituting the dot products in the expressions for $\dot{r},\dot{\beta},\ddot{r}$ , and $\ddot{\beta}$ we get

$\displaystyle\dot{r}$	$\displaystyle=$	$\displaystyle R\dot{\theta}\sin(\beta-\theta),$
$\displaystyle\dot{\beta}$	$\displaystyle=$	$\displaystyle r^{-1}R\dot{\theta}\cos(\theta-\beta)$
$\displaystyle\ddot{r}$	$\displaystyle=$	$\displaystyle r\dot{\beta}^{2}-R\dot{\theta}^{2}\cos(\beta-\theta)+R\ddot{% \theta}\sin(\beta-\theta),$
$\displaystyle\ddot{\beta}$	$\displaystyle=$	$\displaystyle{r^{-1}}[-R\dot{\theta}^{2}\sin(\theta-\beta)+R\ddot{\theta}\cos(% \theta-\beta)-2\dot{r}\dot{\beta}].$

Substituting the given values of $\dot{R},\dot{\theta},\ddot{\theta},R,\theta,\beta$ , and $r=R\sin\theta/\sin\beta$ , we get

\dot{r}=-1.04\,\mathrm{m}/\mathrm{s},\;\dot{\beta}=6.83\,\mathrm{rad}/\mathrm{% s},\quad\ddot{r}=-12.66\,\mathrm{m}/\mathrm{s}^{2},\;\ddot{\beta}=9.43\,% \mathrm{rad}/\mathrm{s}^{2}.

Answer: $\begin{array}[]{llcl}\mbox{(a) }&{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{CD}=6.83\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}},&&\dot{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{CD}=9.43\,\mathrm{rad}/% \mathrm{s}^{2}\hat{\boldsymbol{k}}\\ \mbox{(b) }&{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm/CD}=-1.04\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\lambda}},&&{\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{\rm/CD}=-12.66\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{% \lambda}}\end{array}$

SAMPLE 18.13

Filename:pfigure4-4-rp16

Figure 18.74:

Kinematics of a Link rod in a one-DOF mechanism. In machines we often encounter mechanisms and links in which the ends of a link or a rod are constrained to move on a specified geometric path. A simplified typical link AB is shown in Fig. 18.74.

Link AB is a uniform rigid rod of length $\ell=2\,\mathrm{m}$ . End A of the rod is attached to a collar which slides on a horizontal track. End B of the rod is attached to a uniform disk of radius $R=0.5\,\mathrm{m}$ which rotates about its center O. At the instant shown, when $\theta=30^{{}^{\circ}}$ and $\phi=\theta$ , end A is observed to move at $2\,\mathrm{m}/\mathrm{s}$ to the left.

1.

Find the angular velocity of the rod.
2.

Find the angular velocity of the disk.
3.

Find the velocity of the center-of-mass of the rod.

Solution Let the angular velocities of the rod and the disk be ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=\dot{\theta}\hat{% \boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{D}=\dot{\phi}\hat{% \boldsymbol{k}}$ respectively, where $\dot{\theta}$ and $\dot{\phi}$ are unknowns. We are given ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}=-v_{A}\hat{\boldsymbol{% \imath}}$ where $v_{A}=2\,\mathrm{m}/\mathrm{s}$ .

Point B is on the rod as well as the disk. Hence, the velocity of point B can be found by considering either the motion of the rod or the disk. Considering the motion of the rod we write,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B/A}$	(18.89)
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{AB}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/A}$
	$\displaystyle=$	$\displaystyle-v_{A}\hat{\boldsymbol{\imath}}+\dot{\theta}\hat{\boldsymbol{k}}% \times\ell(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{% \jmath}})$
	$\displaystyle=$	$\displaystyle-(v_{A}+\dot{\theta}\ell\sin\theta)\hat{\boldsymbol{\imath}}+\dot% {\theta}\ell\cos\theta\hat{\boldsymbol{\jmath}}.$

Now considering the motion of the disk (and noting that $\phi=\theta$ ), we write,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{D}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/O}$	(18.90)
	$\displaystyle=$	$\displaystyle\dot{\phi}\hat{\boldsymbol{k}}\times R(-\sin\theta\hat{% \boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-\dot{\phi}R\sin\theta\hat{\boldsymbol{\jmath}}-\dot{\phi}R\cos% \theta\hat{\boldsymbol{\imath}}.$

But ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{B}$ , therefore, from equations (18.89) and (18.90) we get

-(v_{A}+\dot{\theta}\ell\sin\theta)\hat{\boldsymbol{\imath}}+\dot{\theta}\ell% \cos\theta\hat{\boldsymbol{\jmath}}=-\dot{\phi}R\sin\theta\hat{\boldsymbol{% \jmath}}-\dot{\phi}R\cos\theta\hat{\boldsymbol{\imath}}

By equating the $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ components of the above equation we get

$\displaystyle-(v_{A}+\dot{\theta}\ell\sin\theta)$	$\displaystyle=$	$\displaystyle-\dot{\phi}R\cos\theta,$	(18.91)
$\displaystyle\hbox to36.135pt{and\hfil}\dot{\theta}\ell\cos\theta$	$\displaystyle=$	$\displaystyle-\dot{\phi}R\sin\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\theta}$	$\displaystyle=$	$\displaystyle-\dot{\phi}\frac{R}{\ell}\tan\theta.$	(18.92)

Dividing Eqn. (18.91) by (18.92) we get

$\displaystyle-\left(\frac{v_{A}}{\dot{\theta}}+\ell\sin\theta\right)$	$\displaystyle=$	$\displaystyle\frac{\ell\cos\theta}{\tan\theta}=\ell\frac{\cos^{2}\theta}{\sin\theta}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-\frac{v_{A}}{\dot{% \theta}}$	$\displaystyle=$	$\displaystyle\ell\left(\frac{\cos^{2}\theta}{\sin\theta}+\sin\theta\right)=% \ell\left(\frac{\cos^{2}\theta+\sin^{2}\theta}{\sin\theta}\right)=\frac{\ell}{% \sin\theta}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\theta}$	$\displaystyle=$	$\displaystyle-\frac{v_{A}}{\ell}\sin\theta=-\frac{v_{A}}{\ell}\sin 30^{{}^{% \circ}}$
	$\displaystyle=$	$\displaystyle-\frac{2\,\mathrm{m}/\mathrm{s}}{2\,\mathrm{m}}\cdot\frac{1}{2}=-% 0.5\,\mathrm{rad}/\mathrm{s}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=-0.5\,\mathrm{rad}/% \mathrm{s}\hat{\boldsymbol{k}}$

2.

From eqn. (18.92)

$\displaystyle\dot{\phi}$ $\displaystyle=$ $\displaystyle-\frac{\dot{\theta}\ell}{R\tan\theta}=-\frac{\overbrace{-\frac{v_% {A}}{\ell}\sin\theta}^{\dot{\theta}}\ell}{R\frac{\sin\theta}{\cos\theta}}=% \frac{v_{A}}{R}\cos\theta$

$\displaystyle=$ $\displaystyle-\frac{2\,\mathrm{m}/\mathrm{s}}{0.5\,\mathrm{m}}\frac{\sqrt{3}}{% 2}=2\sqrt{3}\,\mathrm{rad}/\mathrm{s}.$

Thus ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{D}=\dot{\phi}\hat{% \boldsymbol{k}}=3.46\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{D}=3.46\,\mathrm{rad}/% \mathrm{s}\hat{\boldsymbol{k}}$

[At this point, it is a good idea to check our algebra by substituting the values of $\dot{\theta}$ and $\dot{\phi}$ in equations (18.89) and (18.90) to calculate ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$ .] ^†^†margin: Substituting $\dot{\theta}=-0.5\,\mathrm{rad}/\mathrm{s}$ in Eqn. (18.89) and plugging in the given values of other variables we get ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=-(\frac{3}{2}\hat{% \boldsymbol{\imath}}+\frac{\sqrt{3}}{2}\hat{\boldsymbol{\jmath}})\,\mathrm{m}/% \mathrm{s}.$ Similarly, substituting $\dot{\phi}=2\sqrt{3}\,\mathrm{rad}/\mathrm{s}$ in Eqn. (18.89) and plugging in the other given values we get ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=-(\frac{3}{2}\hat{% \boldsymbol{\imath}}+\frac{\sqrt{3}}{2}\hat{\boldsymbol{\jmath}})\,\mathrm{m}/% \mathrm{s},$ which checks with the ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$ found above.

Now we can calculate the velocity of the center-of-mass of the rod by considering either point A or point B as a reference:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{G/A}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{AB}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{G/A}$
	$\displaystyle=$	$\displaystyle-v_{A}\hat{\boldsymbol{\imath}}+\dot{\theta}\hat{\boldsymbol{k}}% \times\frac{\ell}{2}(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-(v_{A}+\dot{\theta}\frac{\ell}{2}\sin\theta)\hat{\boldsymbol{% \imath}}+\dot{\theta}\frac{\ell}{2}\cos\theta\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-(2\,\mathrm{m}/\mathrm{s}-0.5\,\mathrm{rad}/\mathrm{s}\cdot\frac% {2\,\mathrm{m}}{2}\cdot\frac{1}{2})\hat{\boldsymbol{\imath}}+(-0.5\,\mathrm{% rad}/\mathrm{s}\cdot\frac{2\,\mathrm{m}}{2}\cdot\frac{\sqrt{3}}{2})\hat{% \boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-\frac{7}{4}\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}-% \frac{\sqrt{3}}{4}\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}=-(1.75\hat{\boldsymbol{% \imath}}+0.43\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}$

We could easily check our calculation by taking point B as a reference and writing

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{G/B}$
		$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{AB}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{G/B}$

By plugging in appropriate values we get, of course, the same value as above.

Comment: We used the standard basis vectors $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ for all our vector calculations in this sample. We can shorten these calculations by choosing other appropriate basis vectors as we show in the following samples.

SAMPLE 18.14

Filename:pfigure4-4-rp14

Figure 18.75:

A two-DOF mechanism. A two degree-of-freedom mechanism made of three rods and two sliders is shown in the figure. At the instant shown, the crank AB is rotating with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=12\,\mathrm{rad}/% \mathrm{s}\hat{\boldsymbol{k}}$ and angular acceleration $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=10\,\mathrm{rad}/% \mathrm{s}^{2}\hat{\boldsymbol{k}}$ . At the same instant, the collar at end C of the link rod CD is sliding on the vertical rod with velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}=0.5\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}$ and acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}=10\,\mathrm{m}/% \mathrm{s}^{2}\hat{\boldsymbol{\jmath}}$ . Find the angular velocity and angular acceleration of the link rod CD.

Solution Once again, we are interested in instantaneous kinematics — we wish to find the angular velocity and acceleration of rod CD at the given instant. This problem is just like the previous sample problem except that end C of the link rod CD is not fixed but free to slide on the vertical bar. But the velocity and acceleration of point C is given; so it is exactly like the previous sample (there, the velocity and acceleration of point C was identically zero). So, we adopt the same line of attack. We figure out the velocity and acceleration of point B using the kinematics of rod AB. We then write the velocity and acceleration of the same point using the kinematics of rod CD (this will involve the unknown angular velocity and acceleration of CD that we are interested in). Equate the two and solve for the unknowns we are interested in.

Filename:pfigure4-4-rp15

Figure 18.76:

Let the angular velocity and acceleration of rod CD be ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{CD}=\dot{\beta}\hat{% \boldsymbol{k}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{CD}=\ddot{\beta}\hat{% \boldsymbol{k}}$ , respectively. Let $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ be base vectors rotating with rod AB, and $\hat{\boldsymbol{\lambda}}$ and $\hat{\boldsymbol{n}}$ be the base vectors rotating with rod CD (see fig. 18.76). Considering rod AB, we have

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B}}$	$\displaystyle=$	$\displaystyle R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$		(18.93)
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B}}$	$\displaystyle=$	$\displaystyle-R\dot{\theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R% \ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}.$		(18.94)

Considering rod CD, we have

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B}}={\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm% {C}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/C}}$	$\displaystyle=$	$\displaystyle v_{C}\hat{\boldsymbol{\jmath}}+\dot{r}\hat{\boldsymbol{\lambda}}% +r\dot{\beta}\hat{\boldsymbol{n}}$		(18.95)
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B}}={% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B/C}}$	$\displaystyle=$	$\displaystyle a_{C}\hat{\boldsymbol{\jmath}}+(\ddot{r}-r{\dot{\beta}}^{2})\hat% {\boldsymbol{\lambda}}+(2\dot{r}\dot{\beta}+r\ddot{\beta})\hat{\boldsymbol{n}}.$		(18.96)

Now equating eqn. (18.93) and (18.95), and dotting both sides with $\hat{\boldsymbol{\lambda}}$ and $\hat{\boldsymbol{n}}$ , we get

	$\displaystyle\dot{r}$	$\displaystyle=$	$\displaystyle R\dot{\theta}\overbrace{(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}\cdot\hat{\boldsymbol{\lambda}})}^{-\sin(\theta-% \beta)}-v_{C}\overbrace{(\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{% \lambda}})}^{\sin\beta}=-R\dot{\theta}\sin(\theta-\beta)-v_{C}\sin\beta$		(18.97)
	$\displaystyle r\dot{\beta}$	$\displaystyle=$	$\displaystyle R\dot{\theta}\underbrace{(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}\cdot\hat{\boldsymbol{n}})}_{\cos(\theta-\beta)}-v_% {C}\underbrace{(\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{n}})}_{\cos% \beta}=R\dot{\theta}\cos(\theta-\beta)-v_{C}\cos\beta$		(18.98)

where the dot products among the basis vectors are easily found from either their geometry (see fig. 18.77) or from their component representation (see previous sample). Following exactly the same procedure, we get, from eqn. (18.94) and 18.96,

Filename:tfigure4-spherical-rotaxis

Figure 18.77:

	$\displaystyle\ddot{r}$	$\displaystyle=$	$\displaystyle-R\dot{\theta}^{2}\cos(\theta-\beta)+R\ddot{\theta}(-\sin(\theta-% \beta))-a_{C}\sin\beta+r\dot{\beta}^{2}$		(18.99)
	$\displaystyle r\ddot{\beta}$	$\displaystyle=$	$\displaystyle-R\dot{\theta}^{2}\sin(\theta-\beta)+R\ddot{\theta}\cos(\theta-% \beta)-a_{C}\cos\beta-2\dot{r}\dot{\beta}.$		(18.100)

Now, note that although we are only interested in finding $\dot{\beta}$ and $\ddot{\beta}$ . So, we only need eqn. (18.98) and eqn. (18.100). But, eqn. (18.100) requires $\dot{r}$ on the right hand side and, therefore, we do need eqn. (18.97). We can, however, happily ignore eqn. (18.99).

Now, to find the numerical values of $\dot{\beta}$ and $\ddot{\beta}$ , we need to find $r$ and $\theta$ in addition to all other given values. Consider triangle ABC in fig. 18.76. Using the law of sines ( $\frac{R}{\sin\beta}=\frac{r}{\sin\theta}=\frac{d}{\sin(\theta-\beta)}$ ), we get $r=0.7\,\mathrm{m}$ and $\theta=79.45^{{}^{\circ}}$ . Now, substituting all known numerical values in eqns. (18.97), (18.98), and (18.100), we get

\dot{r}=-3.75\,\mathrm{m}/\mathrm{s},\quad\dot{\beta}=6.61\,\mathrm{rad}/% \mathrm{s},\quad\ddot{\beta}=8.43\,\mathrm{rad}/\mathrm{s}^{2}.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{CD}=(6.61\,\mathrm{rad}/% \mathrm{s})\hat{\boldsymbol{k}},\quad\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{CD}=(8.43\,\mathrm{rad}/\mathrm{s}^{2})\hat{\boldsymbol% {k}}$

Problems for 18.4 Kinematics of 2-D mechanisms

18.4.1 Slider crank kinematics (No FBD required!). 2-D . Assume $R,\ell,\theta,\dot{\theta},\ddot{\theta}$ are given. The crank mechanism parts move on the $x y$ plane with the $x$ direction being along the piston. Vectors should be expressed in terms of $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},$ and $\hat{\boldsymbol{k}}$ components.

(a)

What is the angular velocity of the crank OA? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{OA}=\dot{\theta}\hat{% \boldsymbol{k}}$ .
(b)

What is the angular acceleration of the crank OA? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{OA}=\ddot{\theta}\hat{% \boldsymbol{k}}$ .
(c)

What is the velocity of point A? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}=-R\dot{\theta}\sin\theta\hat% {\boldsymbol{\imath}}+R\dot{\theta}\cos\theta\hat{\boldsymbol{\jmath}}$ .
(d)

What is the acceleration of point A?

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=-(R\dot{\theta}^{2}\cos% \theta+R\ddot{\theta}\sin\theta)\hat{\boldsymbol{\imath}}+(R\ddot{\theta}\cos% \theta-R\dot{\theta}^{2}\sin\theta)\hat{\boldsymbol{\jmath}}$ .
(e)

What is the angular velocity of the connecting rod AB? [Geometry fact: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{AB}=\sqrt{\ell^{2}-R^{2}\sin^{2% }\theta}\ \hat{\boldsymbol{\imath}}-R\sin\theta\hat{\boldsymbol{\jmath}}$ ] Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=\frac{-R\dot{\theta}% \cos\theta}{\sqrt{\ell^{2}-R^{2}\sin^{2}\theta}}$ .
(f)

For what values of $\theta$ is the angular velocity of the connecting rod AB equal to zero (assume $\dot{\theta}\neq 0$ )? (you need not answer part (e) correctly to answer this question correctly.) Answer: $\theta=(2n+1)\frac{\pi}{2}$ , $n=0,\pm 1,\pm 2,\ldots$ .

Filename:tfigure5-7

Figure 18.78

18.4.2 Slider-Crank. Consider a slider-crank mechanism. Given $\theta$ , $\dot{\theta}$ , $\ddot{\theta}$ , $L$ , and $R$ , can you find the velocity and acceleration of B? There are many ways to do this problem.

Filename:tfigure5-gen-rigid-body

Figure 18.79

18.4.3 The crank AB with length $L_{{\rm AB}}=1\,{\mathrm{inch}}$ in the crank mechanism shown rotates at a constant rate $\dot{\theta}_{AB}=\omega_{AB}=2\pi\,\mathrm{rad}/\mathrm{s}$ counter-clockwise. The initial angle of rotation is $\theta=0$ at $t=0$ . The connecting rod BC has a length of $L_{{\rm BC}}=2\in$ .

(a)

What is the velocity of point B at the end of the crank when $\theta=\pi/2\,{\rm rad}$ ?
(b)

What is the velocity of point C at the end of the connecting rod when $\theta=\pi/2\,{\rm rad}$ ?
(c)

What is the angular velocity of the connecting rod BC when $\theta=\pi/2\,{\rm rad}$ ?
(d)

What is the angular velocity of the connecting rod BC as a function of time?

Filename:tfigure5-term1-a

Figure 18.80

18.4.4 The two rods AB and DE, connected together through a collar C, rotate in the vertical plane. The collar C is pinned to the rod AB but is free to slide on the frictionless rod DE. At the instant shown, rod AB is rotating clockwise with angular speed $\omega=3\,\mathrm{rad}/\mathrm{s}$ and angular acceleration $\alpha=2\,\mathrm{rad}/\mathrm{s}^{2}$ . Find the angular velocity of rod DE. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{DE}=-\frac{\omega}{2}\hat{% \boldsymbol{k}}=-1.5\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}$

Filename:tfigure5-term1-b

Figure 18.81

18.4.5 Reconsider problem 18.81. The two rods AB and DE, connected together through a collar C, rotate in the vertical plane. The collar C is pinned to the rod AB but is free to slide on the frictionless rod DE. At the instant shown, rod AB is rotating clockwise with angular speed $\omega=3\,\mathrm{rad}/\mathrm{s}$ and angular acceleration $\alpha=2\,\mathrm{rad}/\mathrm{s}^{2}$ . Find the angular acceleration of rod DE.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{DE}=3.5\,\mathrm{rad}/% \mathrm{s}^{2}\hat{\boldsymbol{k}}$

Filename:sfig1-2-12

Figure 18.82

18.4.6 Collar $A$ is constrained to slide on a horizontal rod to the right at constant speed $v_{A}$ . It is connected by a pin joint to one end of a rigid bar $A B$ with length $\ell$ which makes an angle $\theta$ with the horizontal at the instant of interest. A second collar $B$ connected by a pin joint to the other end of the rigid bar $A B$ slides on a vertical rod.

(a)

Find the velocity of point $B$ . Answer in terms of $v_{A}$ , $\theta$ , $\ell$ , $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ .
(b)

Find the angular speed $\dot{\theta}$ of the rod? Answer in terms of $v_{A}$ , $\theta$ , $\ell$ , $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ .

Filename:sfig4-6-5

Figure 18.83

18.4.7 A bar of length $\ell=5\,\mathrm{ft}$ , body $\mathcal{D}$ , connects sliders A and B on an L-shaped frame, body $\mathcal{L}$ , which itself is rotating at constant speed about an axle perpendicular to the plane of the figure through the point $O$ and relative to a fixed frame $\mathcal{F}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{L}/\mathcal{F}}=0% .5\,\mathrm{rad}/\mathrm{s}\boldsymbol{k}$ . At the instant shown, body $\mathcal{L}$ is aligned with the $x-y$ axes, slider A is $x_{A}=4\,\mathrm{ft}$ from point $O$ , slider B is $y_{B}=3\,\mathrm{ft}$ from $O$ .

The speed and acceleration of slider B relative to the frame $\mathcal{L}$ are ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/\mathcal{L}}=-2\,{\mathrm{ft/% s}}\hat{\boldsymbol{\jmath}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B/\mathcal{L}}=-2\,{\mathrm{ft}% }/{\mathrm{s}}^{2}\hat{\boldsymbol{\jmath}}$ , respectively.

Determine:

(a)

The absolute velocity of the slider A, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A/\mathcal{F}}$ , and
(b)

The absolute angular velocity of bar AB, body $\mathcal{D}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/\mathcal{F}}$ .

Filename:sfig4-6-5a

Figure 18.84

18.4.8 The link $A B$ is supported by a wheel at D and its end A is constrained so that it only has horizontal velocity. No slipping occurs between the wheel and the link. The wheel has an angular velocity $\omega$ and radius $r$ . The distance $OA=\ell$ .

Given: $\omega$ , $r$ , and $\ell$ . $(\beta=\sin^{-1}\frac{r}{\ell})$ .

Determine the angular velocity of the link AB and velocity of the point A.

Filename:sfig4-6-5b

Figure 18.85

18.4.9 A solid cylinder of radius $R$ and mass $M$ rolls without slip along the ground. A thin rod of mass $m$ and length $L$ is attached by a frictionless pin P to the cylinder’s rim and its right end is dragged at a constant speed ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm A}$ along the (frictionless) horizontal ground.

(a)

For the position shown (where P is directly above the contact point), find P’s velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm P}$ and the rod’s angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ .
(b)

Find P’s acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm P}$ and the angular acceleration ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$ of the rod.

Filename:sfig4-6-5c

Figure 18.86

18.4.10 The slotted link CB is driven in an oscillatory motion by the link ED which rotates about D with constant angular velocity $\dot{\theta}=\omega_{D}$ . The pin P is attached to ED at fixed radius $d$ and engages the slot on CB as shown. Find the angular velocity and acceleration $\dot{\phi}$ and $\ddot{\phi}$ of CB when $\theta=\pi/2$ .

Filename:sfig4-6-5d

Figure 18.87

18.4.11 The rod of radius $r=50\,{\rm mm}$ shown has a constant angular velocity of $\omega=30\,\mathrm{rad}/\mathrm{s}$ counterclockwise. Knowing that rod AD is $250\,{\rm mm}$ long and distance $d=150\,{\rm mm}$ , determine the acceleration of collar D when $\theta=90^{\circ}$ .

Filename:sfig4-5-6

Figure 18.88

18.5 Advanced kinematics of planar motion

In this section we consider three types of problems where the kinematics involves solution of differential equations. In most cases this means computer solution is involved for this type of problem. Here are the three problem types:

•

I. Closed kinematic chains. The main simple example is a 4-bar linkage with one bar grounded. This system has one degree of freedom, but it is difficult to directly calculate the positions, velocities and accelerations of all points in terms of one variable. Instead, the constraint that the linkage is closed is sometimes most-easily expressed as a differential equation.
•

II. Rolling contact with not-round objects. For non-round rollers and cams solving for configuration, velocity and acceleration can sometimes be best done with integration. A side benefit from studying this topic is the observation that all non-translational motions are equivalent to rolling of some kind.
•

III. Contact with ideal wheels and skates, looking down. Cars, tricycles, trailers, grocery carts and sleighs have wheels and have dynamics that is sometimes well characterized by planar analysis, where the plane is the horizontal plane. In this view the simple model of a wheel is as something that prevents sideways motion but allows motion in the direction of travel (like for some of the trike and car problems in 1-D constrained motion). Such problems are called non-holonomic (see box 18.5 on page 18.5).

Closed kinematic chains

When a series of mechanical links is open you can not go from one link to the next successively and get back to your starting point. Such chains include a pendulum (1 link), a double pendulum (2 links), a 100 link pendulum, and a model of the human body (so long as only one foot is on the ground). A closed chain has at least one loop in it. You can go from link to the next and get back to where you started. A slider-crank, a 4-bar linkage, and a person with two feet on the ground are closed chains.

Closed chains are kinematically difficult because they have fewer degrees of freedom than do they have joints. So some of the joint angles depend on the others. The values of any minimal set of configuration variables, say some of the joint angles, determines all of the joint angles, but by geometry that is difficult or impossible to express with formulas.

Example: Four bar linkage: configuration variables

It is impractically difficult to write the positions velocities and accelerations of a 4-bar linkage in terms of $\theta$ , $\dot{\theta}$ and $\ddot{\theta}$ of any one of its joints.

However, given a configuration, the constraint on the rates and accelerations is relatively easy to express, always yielding linear equations.

Box 18.3 Skates, wheels and non-holonomic constraints

Of the words in this book “non-holonomic” is probably the most obscure. This is because the subject of mechanics was mostly stolen from engineers by physicists about 100 years ago. And physicists, the authors of most introductions to mechanics, had no use for non-holonomic mechanics as it wasn’t useful for the development of quantum mechanics. So many people are unaware of the word, the subject or its utility.

In two dimensions the word non-holonomic in effect means the mechanics of objects constrained by ideal skates or massless ideal wheels. Often these non-holonomic constraints are described as “non integrable”. Literally, the word non-holonomic means “not whole”. But in what sense is a rolling ideal wheel “non-integrable” or less “whole” than anything else?

A constrained rigid body. Consider a rigid body that is free to slide on a plane. It has three degrees of freedom described by $x_{O^{\prime}},y_{O^{\prime}}$ and $\theta$ , all measured relative to a fixed reference frame $O\hat{\boldsymbol{\imath}}\hat{\boldsymbol{\jmath}}$ . Point C on the body has relative position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/O^{\prime}}}=x^{% \prime}_{C}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}_{C}\hat{\boldsymbol{% \jmath}}^{\prime}$ where $x^{\prime}_{C}$ and $y^{\prime}_{C}$ are constants. Now let’s constrain the body at point C one of these two different ways:

•

a) Pin the body to the ground with an ideal hinge at point C. This keeps point C from moving but allows the body to rotate (holonomic).
•

b) Put an ideal wheel or skate under the body at C that prevents sliding sideways to the skate but allows point C to move parallel to the skate and also allows rotation about the skate (non-holonomic).

Pin Constraint. In the first case, for the pin, we could describe the constraint with the phrase ‘point C on the body can have no velocity’ and then write and calculate:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{O^{\prime}}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/O^{\prime}}}$	(18.101)
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle\dot{x}_{O^{\prime}}\hat{\boldsymbol{\imath}}+\dot{y}_{O^{\prime}% }\hat{\boldsymbol{\jmath}}+\dot{\theta}\hat{\boldsymbol{k}}\times\left(x^{% \prime}_{C}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}_{C}\hat{\boldsymbol{% \jmath}}^{\prime}\right)$
$\displaystyle\bigm{\{}{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle\dot{x}_{O^{\prime}}\hat{\boldsymbol{\imath}}+\dot{y}_{O^{\prime}% }\hat{\boldsymbol{\jmath}}+\left(\dot{\theta}x^{\prime}_{C}\hat{\boldsymbol{% \jmath}}^{\prime}-\dot{\theta}y^{\prime}_{C}\hat{\boldsymbol{\imath}}^{\prime}% \right)\bigm{\}}$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{\imath}}\ \hbox to36.135% pt{\hfil$\Rightarrow$\hfil}$ $\displaystyle\dot{x}_{O^{\prime}}-\dot{\theta}\sin\theta x^{\prime}_{C}-\dot{% \theta}\cos\theta y^{\prime}_{C}=0$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{\jmath}}\ \hbox to36.135% pt{\hfil$\Rightarrow$\hfil}$ $\displaystyle\dot{y}_{O^{\prime}}+\dot{\theta}\cos\theta x^{\prime}_{C}-\dot{% \theta}\sin\theta y^{\prime}_{C}=0.$
$\Rightarrow$ $\displaystyle{d\over dt}\left(x_{O^{\prime}}+\cos\theta x^{\prime}_{C}-\sin% \theta y^{\prime}_{C}\right)=0$
$\Rightarrow$ $\displaystyle{d\over dt}\left(y_{O^{\prime}}+\sin\theta x^{\prime}_{C}+\cos% \theta y^{\prime}_{C}\right)=0$

The last two equations are two differential equations in the three variables $x_{O^{\prime}},y_{O^{\prime}}$ and $\theta$ . They are “integrable” in the sense that they are equivalent to

	$\displaystyle x_{O^{\prime}}+\cos\theta x^{\prime}_{C}-\sin\theta y^{\prime}_{C}$	$\displaystyle=$	$\displaystyle C_{1}$
	$\displaystyle\hbox to36.135pt{\hfil and\hfil}y_{O^{\prime}}+\sin\theta x^{% \prime}_{C}+\cos\theta y^{\prime}_{C}$	$\displaystyle=$	$\displaystyle C_{2}$		(18.102)

where $C_{1}$ and $C_{2}$ are integration constants that need to be set by the starting configuration. Solving for $x_{O^{\prime}}$ and $y_{O^{\prime}}$ in terms of $\theta$ :

	$\displaystyle x_{O^{\prime}}$	$\displaystyle=$	$\displaystyle C_{1}-\cos\theta x^{\prime}_{C}+\sin\theta y^{\prime}_{C}$
	$\displaystyle y_{O^{\prime}}$	$\displaystyle=$	$\displaystyle C_{2}-\sin\theta x^{\prime}_{C}-\cos\theta y^{\prime}_{C}.$

Here we have derived the obvious, that a pinned body has one independent configuration variable $\theta$ , but we did so starting with a vector expression of constraint in terms of velocities (eqn. (18.101)). Then we wrote the constraint as two scalar constraints on the derivatives of configuration variables and then “integrated” them to write constraints on the configuration variables, finally eliminating two of the configuration variables.

Skate constraint. Now consider the same body constrained by a skate or ideal wheel at C instead of a pin. The skate is aligned with the $\hat{\boldsymbol{\imath}}^{\prime}$ so point C can only move in the $\hat{\boldsymbol{\imath}}^{\prime}$ direction. The body is still free to rotate about the point C (to steer). Thus,

$\displaystyle 0$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}\cdot\hat{\boldsymbol{\jmath}}^{\prime}$
$\displaystyle 0$	$\displaystyle=$	$\displaystyle\left({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{O^{\prime}}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/O^{\prime}}}\right)% \cdot\hat{\boldsymbol{\jmath}}^{\prime}$	(18.103)
$\displaystyle 0$	$\displaystyle=$	$\displaystyle\left(\dot{x}_{O^{\prime}}\hat{\boldsymbol{\imath}}+\dot{y}_{O^{% \prime}}\hat{\boldsymbol{\jmath}}+\dot{\theta}\hat{\boldsymbol{k}}\times\left(% x^{\prime}_{C}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}_{C}\hat{% \boldsymbol{\jmath}}^{\prime}\right)\right)\cdot\hat{\boldsymbol{\jmath}}^{\prime}$
$\displaystyle 0$	$\displaystyle=$	$\displaystyle-\dot{x}_{O^{\prime}}\sin\theta+\dot{y}_{O^{\prime}}\cos\theta+% \dot{\theta}x^{\prime}_{C}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle{d\over dt}F(x_{O^{\prime}},y_{O^{\prime}},\theta)\ ?$

As for the hinge where we found 2 constant functions, we might want to find the function $F(x_{O^{\prime}},y_{O^{\prime}},\theta)$ that satisfies the differential equation above, namely

{d\over dt}F(x_{O^{\prime}},y_{O^{\prime}},\theta)=-\dot{x}_{O^{\prime}}\sin% \theta+\dot{y}_{O^{\prime}}\cos\theta+\dot{\theta}x^{\prime}_{C}.

(18.104)

Another math nightmare. How do we find this $F$ ? You can’t find one. This is the crux of the matter. Neither your calculus professor nor Ramanujan could find one either. No computer can find one, or even a numerical approximation of a solution. There is no function $F(x_{O^{\prime}},y_{O^{\prime}},\theta)$ that solves eqn. (18.104). The solution fundamentally does not exist. That is why we say the skate/wheel constraint eqn. (18.103) is “non-integrable”.

Parallel parking We can use physical reasoning to show that no function $F$ can solve eqn. (18.104). If such an $F$ did exist it would mean that only the set of configurations with position $x_{O^{\prime}},y_{O^{\prime}}$ and angles $\theta$ consistent with $F=constant$ would be allowed by the skate constraint (assuming $F$ depends nontrivially on at least one of the variables). This means there would be some angles and positions that the body couldn’t get to. Remember, we are not doing mechanics, just kinematics. So we can see what configurations are geometrically allowed while still respecting the constraint. The simple observation that motivates the answer is this:

Even though the skate constrains ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}$ to not have a sideways component, point C can get to a point that is straight sideways.

How? Like a car can move sideways into a parking space without skidding sideways; by parallel parking. More generally, the body can get to any position and any orientation by the following moves. First rotate the body so the skate aims to its new goal. Then slide the skate to its new goal. And finally rotate the body to its new desired orientation.

Thus, the skate constraint does not disallow any configurations! Yet the constraint does disallow some velocities (the skate can’t go sideways). In this way, the skate constraint is not “whole”. It constrains velocities without constraining configurations.

Counting degrees of freedom. How many degrees of freedom does a body with a skate constraint have? There are two different answers. By counting possible configurations there are three degrees of freedom (it takes three variables to describe all possible configurations). But at any configuration the velocity can be described by 2 numbers ( $\dot{\theta}$ and $v_{\hat{\boldsymbol{\imath}}^{\prime}}$ ). Whenever the number of configuration degrees of freedom is greater than the number of velocity degrees of freedom (for example, 3¿2) there are non-holonomic constraints.

One might like more examples. But besides artificial mathematical ones, there are none. The only smooth non-holonomic constraint in 2D mechanics is the ideal skate or wheel.

Example: Four bar linkage: configuration rates

If you write the relative velocities of the ends of the bars in terms of configuration rates $\dot{\theta}_{1},\dot{\theta}_{2},$ and $\dot{\theta}_{3}$ and then write the chain closure equation you get a linear equation in the rates. Likewise if you write the closure condition in terms of acceleration. The coefficients in these equations are likely to be complex functions of the configuration, so integrating these equations requires numerics. But the constraint is linear.

Thus, as shown in the last sample of Sect. 10.4, one way to calculate the evolving configurations of a closed chain is to integrate the velocity relations numerically.

Rolling of not round objects

When two objects roll on each other they maintain contact and do not slip relative to each other. That is to say rolling of one rigid curve $\cal B$ on another $\cal A$ means:

•

The instantaneous relative motion of $\cal B$ with respect to $\cal A$ is a rotation about the contact point at the common tangent C, and
•

The sequence of points C moves the same distance on both curves.

For simplicity let’s take $\cal A$ to be a curve fixed in space on which rigid curve $\cal B$ rolls. Take a reference point of interest fixed on body $\cal B$ to be O’. So,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{O^{\prime}}% }={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset% {\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O^{\prime}/C}}\qquad\mbox{where % }{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\theta}_{\cal B}\hat{% \boldsymbol{k}}

(18.105)

and $\theta_{\cal B}$ is, say, the rotation of a $\hat{\boldsymbol{\imath}}^{\prime}$ axis fixed in $\cal B$ relative to a $\hat{\boldsymbol{\imath}}$ axis fixed in $\cal A$ . If we use the rotation $\theta_{\cal B}$ of body $\cal B$ as our configuration variable, we now know how to find the velocity of all points in terms of their positions and the rotation rate. Thus we can find the rate of change of the configuration. To proceed as time progresses we also need to know how the position of point C evolves. Not the material point C on either body, but the location of mutual contact.

If we assume that both curves are parameterized by arc-length going counter-clock wise, if we take curvature as positive if directed towards the interior of each curve’s body, then the condition of maintaining contact requires that

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}=\dot{s}% \hat{\boldsymbol{e}}_{t}\qquad\mbox{where $\hat{\boldsymbol{e}}_{t}$ is the % tangent to fixed curve $\cal A$.}

and $s$ is the advance along curve $\cal A$ . To maintain tangency, the angles must be maintained so

\dot{s}={1\over\kappa_{\cal B}+\kappa_{\cal A}}\dot{\theta}.

Altogether this gives

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}={1\over% \kappa_{\cal B}+\kappa_{\cal A}}\dot{\theta}\hat{\boldsymbol{e}}_{t}\quad\mbox% {(not the velocity of any material point).}

To find the acceleration of material point O’ on $\cal B$ we differentiate eqn. (18.105) with respect to time:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{O^{\prime}}}$	$\displaystyle=$	$\displaystyle{d\over dt}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0% .5mu\mathrm{O^{\prime}}}$
	$\displaystyle=$	$\displaystyle{d\over dt}\left({\accentset{\rightharpoonup}{\boldsymbol{\omega}% }}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O^{% \prime}/C}}\right)$
	$\displaystyle=$	$\displaystyle{d\over dt}\left({\accentset{\rightharpoonup}{\boldsymbol{\omega}% }}_{\cal B}\times\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{O^{\prime}}}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{% C}}\right)\right)$
	$\displaystyle=$	$\displaystyle{\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}_{\cal B}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O^{\prime}/C}}+% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{O^{\prime}}}-{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}$
	$\displaystyle=$	$\displaystyle{\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}_{\cal B}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O^{\prime}/C}}-% \omega_{\cal B}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O^% {\prime}/C}}-{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\left({1% \over\kappa_{\cal B}+\kappa_{\cal A}}\dot{\theta}\hat{\boldsymbol{e}}_{t}\right)$
	$\displaystyle=$	$\displaystyle{\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}_{\cal B}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O^{\prime}/C}}-% \omega_{\cal B}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O^% {\prime}/C}}+\left({\dot{\theta}_{\cal B}^{2}\over\kappa_{\cal B}+\kappa_{\cal A% }}\hat{\boldsymbol{e}}_{n}\right)$

where $\hat{\boldsymbol{e}}_{n}$ is normal to the curves and directed towards the interior of $\cal B$ . Thus the acceleration of all points on $\cal B$ is the same as if the body were pinned at C plus an acceleration due to rolling. This rolling acceleration is small if either of the bodies is sharp (has very large $\kappa$ ) and large if the bodies are nearly conformal.

Body curve and space curve

As one rigid body moves arbitrarily on a plane with some non-zero rotation rate we can find a point at relative position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/O^{\prime}}}$ where ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ . That is a place where the velocity due to rotation about O’ exactly cancels the velocity of O’.

{\accentset{\rightharpoonup}{\boldsymbol{0}}}={\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{\mkern 0.5mu\mathrm{O^{\prime}}}+{\accentset{\rightharpoonup% }{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{\!\mathrm{C/O^{\prime}}}

Crossing both sides with $\hat{\boldsymbol{k}}$ and using that ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ we get

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/O^{\prime}}}={\hat{% \boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5% mu\mathrm{O^{\prime}}}\over\omega_{\cal B}}

as the point “on” the body that has no velocity. This point does not literally have to be on the body, rather it is fixed to the reference frame defined by the body.

As motion progresses a sequence of such points C is traced on the ground. Similarly a sequence of points is traced on the body. These two sequences are called the space curve and the body curve (or “polohodie” and “herpolhodie” in older books). The motion of body $\cal B$ is thus a rolling of the body curve on the space curve.

As a machine designer this means you can generate any desired motion by rolling of appropriate shapes. Move the object in the desired manner, draw the space curve and body curve, make parts with those shapes, and the desired motion occurs by a rolling of those shapes.

Ideal wheels and skates, looking down

If we look down on an ideal skate or wheel at point C on a rigid body and assume that the skate is oriented with the positive $\hat{\boldsymbol{\imath}}^{\prime}$ axis at point C on the body then we know that

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}=v_{C}% \hat{\boldsymbol{\imath}}^{\prime}

and hence the velocity of any point G on the body of interest is

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{G}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm% {G/C}}$
		$\displaystyle=$	$\displaystyle v_{c}\hat{\boldsymbol{\imath}}^{\prime}+\dot{\theta}\hat{% \boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{% G/C}}.$

The acceleration is found by differentiating this expression as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G}}={d% \over dt}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{G}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}+{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G/C}}$
	$\displaystyle=$	$\displaystyle{d\over dt}\left(v_{c}\hat{\boldsymbol{\imath}}^{\prime}+\dot{% \theta}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{\!\mathrm{G/C}}\right)$
	$\displaystyle=$	$\displaystyle\dot{v}_{C}\hat{\boldsymbol{\imath}}^{\prime}+v_{C}\dot{\hat{% \boldsymbol{\imath}}^{\prime}}+\ddot{\theta}\hat{\boldsymbol{k}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}-\dot{\theta}^{2}% {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}$
	$\displaystyle=$	$\displaystyle\underbrace{\dot{v}_{C}\hat{\boldsymbol{\imath}}^{\prime}+v_{C}% \dot{\theta}\hat{\boldsymbol{\jmath}}^{\prime}}_{{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{\mathrm{C}}}++\underbrace{\ddot{\theta}\hat{\boldsymbol{k}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}-\dot{% \theta}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}}_{{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G/C}}}.$

It is interesting to note that the Coriolis-like term $v_{C}\dot{\theta}\hat{\boldsymbol{\jmath}}^{\prime}$ does not have the usual factor of 2 one encounters in holonomic problems. To find the trajectory of the point C, say, one needs to integrate the velocity like this:

$\displaystyle\dot{x}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}\cdot\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle v_{C}\hat{\boldsymbol{\imath}}^{\prime}\cdot\hat{\boldsymbol{% \imath}}$
	$\displaystyle=$	$\displaystyle v_{C}\cos\theta$
$\displaystyle\dot{y}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}\cdot\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle v_{C}\hat{\boldsymbol{\imath}}^{\prime}\cdot\hat{\boldsymbol{% \jmath}}$
	$\displaystyle=$	$\displaystyle v_{C}\sin\theta.$

SAMPLE 18.15

Filename:sfig4-5-6a

Figure 18.89:

Kinematics of a four bar linkage. A four bar linkage ABCD is shown in the figure (fourth bar is the ground AD) at some instant $t_{0}$ . The driving bar AB rotates with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=\dot{\theta}(t)\hat{% \boldsymbol{k}}$ . Find the angular velocities of rods BC and CD as a function of $\dot{\theta}$ . How can you solve for the positions of the bars at any $t$ if the initial configuration is as shown in the figure?

Solution Let the angles that rods AB, BC, and CD make with the horizontal ( $x$ -axis) be $\theta,\,\beta$ , and $\phi$ , respectively. Then, we can write ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{BC}=\dot{\beta}\hat{% \boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{CD}=\dot{\phi}\hat{% \boldsymbol{k}}$ . We have to find $\dot{\beta}$ and $\dot{\phi}$ .

Note that the motion of point B is a simple circular motion about point A with given angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}$ . Thus, the velocity of point B is known. Now, we can find the velocity of point C two ways: (i) by considering rod CB: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}+{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C/B}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{BC}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}$ , and (ii) by considering rod CD: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{CD}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/D}}.$ Either way the velocity must be the same. Thus, we have a 2-D vector equation with two unknowns $\dot{\beta}$ and $\dot{\phi}$ . We can get two independent scalar equations from the vector equation and thus we can solve for the desired unknowns.

Let us use the rotating base vectors $(\hat{\boldsymbol{\lambda}}_{1},\hat{\boldsymbol{n}}_{1}),(\hat{\boldsymbol{% \lambda}}_{2},\hat{\boldsymbol{n}}_{2}),\mbox{ and }(\hat{\boldsymbol{\lambda}% }_{3},\hat{\boldsymbol{n}}_{3})$ with rods AB, BC, and CD, respectively. Note that these base vectors are basically the $(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R},\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta})$ pairs; we use $(\hat{\boldsymbol{\lambda}},\hat{\boldsymbol{n}})$ just for the sake of easy subscripting. Now,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}=\ell_{1}\dot{% \theta}\hat{\boldsymbol{n}}_{1}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{BC}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}=\ell_{1}\dot{% \theta}\hat{\boldsymbol{n}}_{1}+\ell_{2}\dot{\beta}\hat{\boldsymbol{n}}_{2}$	(18.106)
$\displaystyle\mbox{also,}\quad{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{C}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{CD}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/D}}=\ell_{3}\dot{% \phi}\hat{\boldsymbol{n}}_{3}$	(18.107)

Thus, from eqn. (18.106) and (18.107), we have,

Filename:sfig4-6-8

Figure 18.90:

\ell_{1}\dot{\theta}\hat{\boldsymbol{n}}_{1}+\ell_{2}\dot{\beta}\hat{% \boldsymbol{n}}_{2}=\ell_{3}\dot{\phi}\hat{\boldsymbol{n}}_{3}

(18.108)

Dotting eqn. (18.108) with $\hat{\boldsymbol{\lambda}}_{2}$ (to eliminate $\dot{\beta}$ term), we get

	$\displaystyle\ell_{1}\dot{\theta}(\hat{\boldsymbol{n}}_{1}\cdot\hat{% \boldsymbol{\lambda}}_{2})$	$\displaystyle=$	$\displaystyle\ell_{3}\dot{\phi}(\hat{\boldsymbol{n}}_{3}\cdot\hat{\boldsymbol{% \lambda}}_{2})$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\phi}$	$\displaystyle=$	$\displaystyle\frac{\ell_{1}}{\ell_{3}}\frac{(\hat{\boldsymbol{n}}_{1}\cdot\hat% {\boldsymbol{\lambda}}_{2})}{(\hat{\boldsymbol{n}}_{3}\cdot\hat{\boldsymbol{% \lambda}}_{2})}\,\dot{\theta}.$		(18.109)

Similarly, dotting eqn. (18.108) with $\hat{\boldsymbol{\lambda}}_{3}$ (to eliminate $\dot{\phi}$ term), we get

\displaystyle\dot{\beta}

\displaystyle=

\displaystyle-\frac{\ell_{1}}{\ell_{2}}\frac{(\hat{\boldsymbol{n}}_{1}\cdot% \hat{\boldsymbol{\lambda}}_{3})}{(\hat{\boldsymbol{n}}_{2}\cdot\hat{% \boldsymbol{\lambda}}_{3})}\,\dot{\theta}.

(18.110)

We are practically done at this point with the kinematics — we have found $\dot{\beta}$ and $\dot{\phi}$ as functions of $\dot{\theta}$ . The various dot products are just geometry and vector algebra. To write them explicitly, we note that $\hat{\boldsymbol{\lambda}}_{1}=\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta% \hat{\boldsymbol{\jmath}},\;\hat{\boldsymbol{n}}_{1}=-\sin\theta\hat{% \boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{\jmath}}$ , $\hat{\boldsymbol{\lambda}}_{2}~{}=~{}\cos\beta\hat{\boldsymbol{\imath}}+\sin% \beta\hat{\boldsymbol{\jmath}}$ , etc. Thus,

$\displaystyle\hat{\boldsymbol{n}}_{1}\cdot\hat{\boldsymbol{\lambda}}_{2}$	$\displaystyle=$	$\displaystyle-\sin\theta\cos\beta+\cos\theta\sin\beta=\sin(\beta-\theta)$
$\displaystyle\hat{\boldsymbol{n}}_{3}\cdot\hat{\boldsymbol{\lambda}}_{2}$	$\displaystyle=$	$\displaystyle\sin(\beta-\phi)$
$\displaystyle\hat{\boldsymbol{n}}_{1}\cdot\hat{\boldsymbol{\lambda}}_{3}$	$\displaystyle=$	$\displaystyle\sin(\phi-\theta),\quad\hat{\boldsymbol{n}}_{2}\cdot\hat{% \boldsymbol{\lambda}}_{3}=\sin(\phi-\beta).$

Substituting the appropriate expressions in eqn. (18.110) and (18.109), we get

\dot{\beta}=-\frac{\ell_{1}}{\ell_{2}}\frac{\sin(\phi-\theta)}{\sin(\phi-\beta% )}\dot{\theta},\qquad\dot{\phi}=\frac{\ell_{1}}{\ell_{3}}\frac{\sin(\beta-% \theta)}{\sin(\beta-\phi)}\dot{\theta}.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{BC}=-\frac{\ell_{1}}{\ell_% {2}}\frac{\sin(\phi-\theta)}{\sin(\phi-\beta)}\dot{\theta}\,\hat{\boldsymbol{k% }},\qquad{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{CD}=\frac{\ell_{1% }}{\ell_{3}}\frac{\sin(\beta-\theta)}{\sin(\beta-\phi)}\dot{\theta}\,\hat{% \boldsymbol{k}}$

Note that the expressions for $\dot{\beta}$ and $\dot{\phi}$ are coupled, nonlinear, first order ordinary differential equations. To be able to find $\dot{\theta},\beta(t)$ and $\phi(t)$ , we need to integrate $\dot{\theta},\dot{\beta}$ , and $\dot{\phi}$ . Here, we set up these differential equations for numerical integration. Although, we can use any given $\dot{\theta}(t)$ (e.g., $\alpha t$ or $\dot{\theta}_{0}\sin(\Omega t)$ or whatever), for definiteness in our numerical integration, let us take a constant $\dot{\theta}$ , that is, let $\dot{\theta}=C=10\,\mathrm{rad}/\mathrm{s}$ (say). So, our equations are,

\dot{\theta}=C,\qquad\dot{\beta}=\frac{\ell_{1}}{\ell_{2}}\frac{\sin(\phi-% \theta)}{\sin(\beta-\phi)}C,\qquad\dot{\phi}=\frac{\ell_{1}}{\ell_{3}}\frac{% \sin(\beta-\theta)}{\sin(\beta-\phi)}C,

and the initial conditions are $\theta(0)=\pi/2,\quad\beta(0)=\pi/4,\quad\phi(0)=3\pi/4.$

Here is a pseudocode that we use to integrate these equations numerically for a period of $2\pi/C=\pi/5$ seconds (one complete revolution of AB).

ODEs = {thetadot = C,
        betadot  = (l1/l2)*sin(phi-theta)/sin(beta-phi)*C,
        phidot   = (l1/l3)*sin(beta-theta)/sin(beta-phi)*C}
IC   = {theta(0) = pi/2, beta(0) = pi/4, phi(0) = 3*pi/4}
Set  C=10, l1=.4, l2=.4*sqrt(2), l3=.8*sqrt(2)
Solve ODEs with IC for t=0 to t=pi/5

After we get the angles, we can compute the $x y$ coordinates of points B and C at each instant and plot the mechanism at those instants. Plots thus obtained from our numerical solution are shown in fig. 18.91 where the configuration of the mechanism is shown at 9 equally spaced times between $t=0$ to $t=\pi/5$ .

Filename:sfig4-6-8a

Figure 18.91: Several configurations of the mechanism at equal intervals of time during one complete revolution of the driving link. After

t_{8}

the mechanism returns to the initial configuration.

Problems for 18.5 Advanced kinematics of planar motion

18.5.1 Double pendulum. The double pendulum shown is made up of two uniform bars, each of length $\ell$ and mass $m$ . At the instant shown, $\phi_{1}$ , $\dot{\phi}_{1}$ , $\phi_{2}$ , and $\dot{\phi}_{2}$ are known. For the instant shown, answer the following questions in terms of $\ell$ , $\phi_{1}$ , $\dot{\phi}_{1}$ , $\phi_{2}$ , and $\dot{\phi}_{2}$ .

(a)

What is the absolute velocity of point $A$ ?
(b)

What is the velocity of point $B$ relative to point $A$ ?
(c)

What is the absolute velocity of point $B$ ?

Filename:sfig4-6-8b

Figure 18.92

18.5.2 Double pendulum, Again. For the double pendulum in problem 18.5, $\ddot{\phi}_{1}$ and $\ddot{\phi}_{2}$ are also known at the instant shown. For the instant shown, answer the following questions in terms of $\ell$ , $\phi_{1}$ , $\dot{\phi}_{1}$ , $\ddot{\phi}_{1}$ , $\phi_{2}$ , $\dot{\phi}_{2}$ , and $\ddot{\phi}_{2}$ .

(a)

What is the absolute acceleration of point $A$ ?
(b)

What is the acceleration of point $B$ relative to point $A$ ?
(c)

What is the absolute acceleration of point $B$ ?

	$\displaystyle a_{n}$	$\displaystyle=$	$\displaystyle\frac{v^{2}}{\rho}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\rho$	$\displaystyle=$	$\displaystyle\frac{v^{2}}{a_{n}}=\frac{45(\,{\mathrm{ft/s}})^{2}}{9{\sqrt{5}}% \,{\mathrm{ft}}/{\mathrm{s}}^{2}}=\sqrt{5}\,\mathrm{ft}.$

	$\displaystyle[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}]_% {\cal F}$	$\displaystyle\neq$	$\displaystyle[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}]_% {\cal B}$
	$\displaystyle\mbox{because}\quad\left[\begin{array}[]{c}4\,\mathrm{m}\\ 5\,\mathrm{m}\\ \end{array}\right]$	$\displaystyle\neq$	$\displaystyle\left[\begin{array}[]{c}6\,\mathrm{m}\\ -2.24\,\mathrm{m}\\ \end{array}\right]$

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}={% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}$	and	$\displaystyle{{}^{\cal F}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}\over dt}={{}^{\cal B}d{\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{\!\mathrm{P/A}}\over dt}$
	$\displaystyle\hbox{but}\quad[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}]_{\cal F}\neq[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}]_{\cal B}$	and	$\displaystyle[{{}^{\cal F}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/A}}\over dt}]_{\cal F}\neq[{{}^{\cal F}d{\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{\!\mathrm{P/A}}\over dt}]_{\cal B}.$

	$\displaystyle n_{\rm DOF}$	$\displaystyle=3*n_{\rm bod}-n_{\rm con}$		(18.76)
		$\displaystyle=3*n_{\rm bod}-2\cdot n_{\rm hinges}=30-2\cdot 10=10$		(18.77)

	$\displaystyle\dot{\phi}$	$\displaystyle=$	$\displaystyle-\frac{\dot{\theta}\ell}{R\tan\theta}=-\frac{\overbrace{-\frac{v_% {A}}{\ell}\sin\theta}^{\dot{\theta}}\ell}{R\frac{\sin\theta}{\cos\theta}}=% \frac{v_{A}}{R}\cos\theta$
		$\displaystyle=$	$\displaystyle-\frac{2\,\mathrm{m}/\mathrm{s}}{0.5\,\mathrm{m}}\frac{\sqrt{3}}{% 2}=2\sqrt{3}\,\mathrm{rad}/\mathrm{s}.$

Chapter 18 Kinematics using time-varying basis vectors

Download this chapter by section

18.1 Polar coordinates and path coordinates

Polar coordinates for general (non-circular) motion

Velocity in polar coordinates

Acceleration in polar coordinates

Path coordinates

The geometry of the path basis vectors.

Formal definition of path basis vectors

Velocity and acceleration in path coordinates.

Recipes for path coordinates.

Summary of polar(cylindrical) coordinates

Summary of path coordinates

Problems for 18.1 Polar coordinates and path coordinates

18.2 Rotating reference frames and their time-varying base vectors

The fixed base-vector method

Reference frames

Using more than one reference frame

Translating and rotating reference frames

Rate of change of a vector relative to a rotating frame: the 𝑸⇀˙ formula

Using the moving-frame derivative to calculate the fixed-frame derivative

Summary of the 𝑸⇀˙ formula

The varying base-vectors method of computing velocity and acceleration

Summary of the varying base-vector method

Geometric ‘derivation’ of the 𝑸⇀˙ formula

Problems for 18.2 Rotating reference frames

18.3 General expressions for velocity 𝒗⇀ and acceleration 𝒂⇀ of a moving point

Absolute velocity of a point moving relative to a moving frame

Acceleration

Acceleration relative to an object or frame

Absolute acceleration of a point P′ glued to a moving frame

Absolute acceleration of a point moving relative to a moving frame

Motion relative to a point versus motion relative to a frame

Warning!

Problems for 18.3 General expressions for velocity and acceleration

18.4 Kinematics of 2-D mechanisms

Degrees of freedom (DOF)

Examples of connections and their effect on nDOF

Some simple mechanisms

Configuration variables

Adding relative angular velocities

Kinematics of mechanisms

Problems for 18.4 Kinematics of 2-D mechanisms

18.5 Advanced kinematics of planar motion

Closed kinematic chains

Rolling of not round objects

Body curve and space curve

Ideal wheels and skates, looking down

Problems for 18.5 Advanced kinematics of planar motion

Rate of change of a vector relative to a rotating frame: the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula

Summary of the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula

Geometric ‘derivation’ of the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula

18.3 General expressions for velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ of a moving point

Absolute acceleration of a point P^′ glued to a moving frame

Examples of connections and their effect on $n_{\rm DOF}$