Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap18 Time-varying basis vectors 20 Fixed-axis rotation in 3D

Chapter 19 Mechanics of constrained particles and rigid objects

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19.0Introductionstudy PDF
19.1Mechanics of a constrained particlestudy PDF + solutions
19.2One-degree-of-freedom 2-D mechanismsstudy PDF + solutions
19.3Multi-degree-of-freedom 2-D mechanismsstudy PDF + solutions

All chapters and individual parts

The dynamics of particles and rigid objects is studied using the relative-motion kinematics ideas from chapter chapter 18. This is the capstone chapter for a two-dimensional dynamics course. After this chapter a good student should be able to navigate through and use most of the skills in the concept map inside the back cover.

We have studied the mechanics of particles and rigid bodies with constraints that require progressively more involved kinematics. We now proceed to study the mechanics of more complex systems: particles with constrained paths, particles moving relative to moving frames, and mechanisms with several parts.

The basic strategy throughout is to use, in combination, the following skills:

1.

Basic modeling. Describe a system in an appropriate way using the language of particle and rigid body mechanics. As described in Chapter 2, the force modeling and kinematic modeling are coupled. Where relative motion is freely allowed there is no force. And where motion is caused or prevented there is a force. Here is where you decide the constitutive (force) laws you are using for springs, contact, gravity, etc.
2.

Draw free-body diagrams of the system of interest and of its parts. These diagrams should show what you do and do not know about the constraint forces (e.g., at a pin connection cut free in a free-body diagram the FBD should show an arbitrary force and no moment). These are exactly the same free-body diagrams that one would draw for statics.
3.

Kinematics calculations. Pick appropriate configuration variables, as many as there are degrees of freedom. Then write the velocities, accelerations, angular velocities and angular accelerations of interest in terms of the configuration variables and their first and second time derivatives, possibly using methods from Chapter 10. Often this is the hardest part of the analysis.
4.

Use appropriate balance equations: linear momentum, angular momentum, or power balance equations.
5.

Solve the balance equations for unknown forces or accelerations of interest. Sometimes this can be done by hand by writing out components and solving simultaneous equations or by using appropriate dot products. And sometimes it is best done by setting up a matrix equation and solving on the computer.
6.

Solve the differential equations to find how the basic configuration variables change with time. For some special problems this can be done by hand, but most often involves computer solution.
7.

Plug the ODE solution from (6) above into the equations from kinematics (3 above) and the balance laws (4 above). This is not a different skill from (3) or (4), it is just applied at a different time in the work.
8.

Make plots of how forces, positions and velocities change with time, or of trajectories. Animations are also often nice.

These skills are used to solve dynamics problems which often fall into one of these 4 categories.

a.

Kinematics. These are problems where only geometry is used, where the kinematics constraints determine what you are interested in, independent of the forces or time history. A classic example is determining the path of a point on a given four-bar linkage. More basic examples include finding position or acceleration from a given velocity history.
b.

Instantaneous dynamics. These are problems where the positions and velocities of all points are given and you need to find forces or accelerations. Often these are “first-motion” problems: what are accelerations and forces immediately after something is released from rest?
c.

“Inverse dynamics.” These problems are called “inverse” because they are backwards of the original hard dynamics problems ((d) below). In these problems the motion is given as a function of time, and you have to calculate the forces. These problems are easier than non “inverse” problems because the differential equations from the balance laws don’t need to be solved. A classic example is a slider-crank where the motion of the crank is known a priori to be at constant rate and you need to find the torque required to keep that motion. Usually in science “inverse” problems are harder. In dynamics this kind of “inverse” problem is easier than the non “inverse” problems.
d.

Dynamics analysis. You are given some information about forces and constraints and you have to find the motion and more about the forces. These are the capstone problems that require use of all the skills.

A flow chart showing how these problem types are solved using the basic skill components ideas is shown in the chart in the inside back cover.

As you solve a problem, at any instant in time you should be able to place your work on this chart.

In the sense of putting all the basic ideas together, this chapter completes the book. But in this chapter we only consider two-dimensional models and motions. Three-dimensional models and motions involve the same assembly of basic ideas, but more difficult kinematics, so are postponed.

19.1 Mechanics of a constrained particle

The kinematics of time-varying base vectors help us deal with some more difficult particle motion mechanics problems. For one point mass it is easy to write balance of linear momentum. It is:

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}.

The mass of the particle $m$ times its vector acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ is equal to the total force on the particle ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ . No problem.

Now, however, we can write this equation in five somewhat distinct ways.

1.

In general abstract vector form: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ .
2.

In cartesian coordinates: $F_{x}\hat{\boldsymbol{\imath}}+F_{y}\hat{\boldsymbol{\jmath}}+F_{z}\hat{% \boldsymbol{k}}=m[\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{y}\hat{\boldsymbol{% \jmath}}+\ddot{z}\hat{\boldsymbol{k}}]$ .
3.

In polar coordinates: $F_{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+F_{\theta}\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!\theta}+F_{z}\hat{\boldsymbol{k}}=m[(\ddot{R}-R{\dot{% \theta}}^{2})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+(2\dot{R}\dot{\theta% }+R\ddot{\theta})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+\ddot{z}% \hat{\boldsymbol{k}}]$ .
4.

In path coordinates:

$F_{t}\hat{\boldsymbol{e}}_{t}+F_{n}\hat{\boldsymbol{e}}_{n}=m[\dot{v}\hat{% \boldsymbol{e}}_{t}+(v^{2}/\rho)\hat{\boldsymbol{e}}_{n}]$

All of these equations are always right. Additionally, for a given particle moving under the action of a given force there are many more correct equations that can be found by shifting the origin and orientation of the coordinate systems. For example for a moving frame $\cal B$ with origin $0^{\prime}$ , rotation rate ${{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}_{\cal B}$ and angular acceleration ${{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}}_{\cal B}$ :

5.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m\left\{{\accentset% {\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{0^{\prime}}}\quad-\omega_{\cal B}^% {2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}\quad+{{\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}\quad+{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{\mathrm{/\cal B}}\quad+2{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{% v}}}_{\mkern 0.5mu\mathrm{/\cal B}}\right\}$

where, to simplify the notation, all motions are relative to $\cal F$ and positions relative to 0 unless explicitly indicated by a $/\cal B$ or $/0^{\prime}$ .

This is quite a collection of kinematic tools. In general we want to choose the best tools for the job. But to get a sense let’s first look at a simple problem using each of these kinematic approaches, some of which are rather inappropriate.

A particle that moves with no net force

In the special case that a particle has no force on it we know intuitively, or from the verbal statement of Newton’s First Law, that the particle travels in a straight line at constant speed. As a first example, let’s try to find this result using the vector equations of motion five different ways: in the general abstract form, in cartesian coordinates, in polar coordinates, in path coordinates, and relative to a moving frame (see fig. 19.1).

Filename:tfigure8-rel-ang-vel

Figure 19.1: A particle P moves. One can track its motion using the general vector form

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

, Cartesian coordinates in the fixed frame

{\cal F}=0\hat{\boldsymbol{\imath}}\hat{\boldsymbol{\jmath}}

, Polar coordinates using

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

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

, path coordinates

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

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

and cartesian coordinates in a rotating frame

{\cal B}=0^{\prime}\hat{\boldsymbol{\imath}}^{\prime}\hat{\boldsymbol{\jmath}}% ^{\prime}

General abstract form. The equation of linear momentum balance is ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ or, if there is no force, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ , which means that $d{\accentset{\rightharpoonup}{\boldsymbol{v}}}/dt={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ . So ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ is a constant. We can call this constant ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{0}$ . So after some time the particle is where it was at $t=0$ , say, ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{0}$ , plus its velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{0}$ times time. That is:

{\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{0}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{0}t.

(19.1)

This vector relation is a parametric equation for a straight line. The particle moves in a straight line, as expected.

Cartesian coordinates. If instead we break the linear momentum balance equation into cartesian coordinates we get

F_{x}\hat{\boldsymbol{\imath}}+F_{y}\hat{\boldsymbol{\jmath}}+F_{z}\hat{% \boldsymbol{k}}=m(\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{y}\hat{\boldsymbol{% \jmath}}+\ddot{z}\hat{\boldsymbol{k}}).

Because the net force is zero and the net mass is not negligible,

\ddot{x}=0,\qquad\ddot{y}=0,\mbox{ and }\qquad\ddot{z}=0.

These equations imply that $\dot{x}$ , $\dot{y}$ , and $\dot{z}$ are all constants, let’s call them $v_{x0}$ , $v_{y0}$ , $v_{z0}$ . So $x$ , $y$ , and $z$ are given by

x=x_{0}+v_{x0}t,\quad y=y_{0}+v_{y0}t,\quad\&\ z=z_{0}+v_{z0}t.

We can put these components into their place in vector form to get:

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=x\hat{\boldsymbol{\imath}}+y\hat% {\boldsymbol{\jmath}}+z\hat{\boldsymbol{k}}=(x_{0}+v_{x0}t)\hat{\boldsymbol{% \imath}}+(y_{0}+v_{y0}t)\hat{\boldsymbol{\jmath}}+(z_{0}+v_{z0}t)\hat{% \boldsymbol{k}}.

(19.2)

Note that there are six free constants in this equation representing the initial position and velocity. Equation 19.2 is a cartesian representation of equation 19.1; it describes a straight line being traversed at constant rate.

Polar/cylindrical coordinates.

When there is no force, in polar coordinates we have:

\underbrace{F_{R}}_{0}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\underbrace% {F_{\theta}}_{0}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+\underbrace{% F_{z}}_{0}\hat{\boldsymbol{k}}=m[(\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}}].

This vector equation leads to the following three scalar differential equations, the first two of which are coupled non-linear equations (neither can be solved without the other).

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

A tedious calculation will show that these equations are solved by the following functions of time:

$\displaystyle R$	$\displaystyle=$	$\displaystyle\sqrt{d^{2}+[v_{0}(t-t_{0})]^{2}}$	(19.3)
$\displaystyle\theta$	$\displaystyle=$	$\displaystyle\theta_{0}+\tan^{-1}[v_{0}(t-t_{0})/d]$
$\displaystyle z$	$\displaystyle=$	$\displaystyle z_{0}+v_{z0}t,$

where $\theta_{0}$ , $d$ , $t_{0}$ , $v_{0}$ , $z_{0}$ , and $v_{z0}$ are constants. Note that, though eqn. (19.1) looks different than eqn. (19.2), there are still $6$ free constants. From the physical interpretation you know that eqn. (19.1) must be the parametric equation of a straight line. And, indeed, you can verify that picking arbitrary constants and using a computer to make a polar plot of eqn. (19.1) does in fact show a straight line. From eqn. (19.1) it seems that polar coordinates’ main function is to obfuscate rather than clarify. For the simple case that a particle moves with no force at all, we have to solve non-linear differential equations whereas using cartesian coordinates we get linear equations which are easy to solve and where the solution is easy to interpret.

But, if we add a central force, a force like earth’s gravity acting on an orbiting satellite (the force on the satellite is directed towards the center of the earth), the equations become almost intolerable in cartesian coordinates. But, in polar coordinates, the solution is almost as easy (which is not all that easy for most of us) as the solution 19.1. So the classic analytic solutions of celestial mechanics are usually expressed in terms of polar coordinates.

Path coordinates.

When there is no force, ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ is expressed in path coordinates as

\underbrace{F_{t}}_{0}\hat{\boldsymbol{e}}_{t}+\underbrace{F_{n}}_{0}\hat{% \boldsymbol{e}}_{n}=m(\dot{v}\hat{\boldsymbol{e}}_{t}+\underbrace{(v^{2}/\rho)% \hat{\boldsymbol{e}}_{n}}_{v^{2}{\accentset{\rightharpoonup}{\boldsymbol{% \kappa}}}}).

That is,

\dot{v}=0\quad\mbox{and}\quad v^{2}/\rho=0.

So the speed $v$ must be constant and the radius of curvature $\rho$ of the path infinite. That is, the particle moves at constant speed in a straight line.

Relative to a rotating reference frame Let’s look at the equations using a frame $\cal B$ that shares an origin with $\cal F$ but is rotating at a constant rate ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}=\omega\hat{% \boldsymbol{k}}$ relative to $\cal F$ . Thus

{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}\hbox to36.135pt{\hfil and\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{0^{\prime}/0}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}

and we have that ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ is written as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle m\bigm{\{}\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{a% }}}_{\mathrm{0^{\prime}}}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}% \quad-\omega_{\cal B}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}\quad+% \underbrace{{{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}}_{\cal B}}_{{% \accentset{\rightharpoonup}{\boldsymbol{0}}}}\times{\accentset{\rightharpoonup% }{\boldsymbol{r}}}\quad+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm% {/\cal B}}\quad+2{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}_{\cal B}% \times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{/\cal B% }}\bigm{\}}$	(19.4)
	$\displaystyle=$	$\displaystyle-\omega^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}+{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{/\cal B}}+2\omega\hat{% \boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5% mu\mathrm{/\cal B}}.$	(19.4)

Now, using the rotating base vectors $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ , we have that $F_{x^{\prime}}\hat{\boldsymbol{\imath}}^{\prime}+F_{y^{\prime}}\hat{% \boldsymbol{\jmath}}^{\prime}=0\hat{\boldsymbol{\imath}}^{\prime}+0\hat{% \boldsymbol{\jmath}}^{\prime}$ and

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=x^{\prime}\hat{\boldsymbol{% \imath}}^{\prime}+y^{\prime}\hat{\boldsymbol{\jmath}}^{\prime},\quad{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{\cal B}}=% \dot{x}^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+\dot{y}^{\prime}\hat{% \boldsymbol{\jmath}}^{\prime},\hbox to36.135pt{\hfil and\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{\cal B}}=\ddot{x}^{\prime}\hat{% \boldsymbol{\imath}}^{\prime}+\ddot{y}^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}

so eqn. (19.4) can be rewritten as

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}

\displaystyle=

\displaystyle-\omega^{2}(x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{% \prime}\hat{\boldsymbol{\jmath}}^{\prime})+(\ddot{x}^{\prime}\hat{\boldsymbol{% \imath}}^{\prime}+\ddot{y}^{\prime}\hat{\boldsymbol{\jmath}}^{\prime})+2\omega% \hat{\boldsymbol{k}}\times(\dot{x}^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+% \dot{y}^{\prime}\hat{\boldsymbol{\jmath}}^{\prime})

which in turn can be broken into components and written as:

	$\displaystyle\ddot{x}^{\prime}$	$\displaystyle=$	$\displaystyle\omega^{2}x^{\prime}+2\omega\dot{y}^{\prime}$		(19.5)
	$\displaystyle\ddot{y}^{\prime}$	$\displaystyle=$	$\displaystyle\omega^{2}y^{\prime}-2\omega\dot{x}^{\prime}$

which makes up a pair of second order linear differential equations. With some work, someone good at ODEs can solve this with pencil and paper. But most of us would use a computer for such a system. If, in some consistent units, we had

y^{\prime}(0)=0,\dot{y}^{\prime}(0)=0,x^{\prime}(0)=0,\dot{x}^{\prime}(0)=1,% \omega=1

then the solution turns out to be

	$\displaystyle x^{\prime}$	$\displaystyle=$	$\displaystyle\phantom{-}t\cos(\omega t)$
	$\displaystyle y^{\prime}$	$\displaystyle=$	$\displaystyle-t\sin(\omega t)$

as you can check by substituting into eqn. (19.1). That is, a particle which goes in a straight line away from the origin goes, in spirals as seen in the rotating frame^†^†margin: And the straight line motion could have been more complicated yet, as seen in the rotating frame, with variable rate rotation and acceleration.This is why we are lucky the earth is not spinning fast or at highly variable rates. Otherwise the motion we would see, us humans here on the rotating earth, would be particles with no force on them spiraling around all over the place. And that’s what Isaac Newton would have seen. And he would have written “A particle in motion tends to go in crazy spirals, and a particle that is initially at rest also goes bonkers” and the equations we use through out this book would have been much harder to discover. .

Constrained motion

A particle in a plane has 2 degrees of freedom. There is basically only one kind of constraint — to a path. When constrained to a path the particle has one remaining degree of freedom so its configuration can be described with one variable. For a given problem you must think about

•

What force(s) constrain the motion to the path?
•

What do you want to use for a configuration variable?

You use these ideas to

•

draw an appropriate free-body diagram, and then
•

calculate the velocity and acceleration in terms of the configuration variable and its derivatives.

After these key first steps you plug into equations of motion and solve for what you are interested in. Of course the needed math could be difficult or impossible, but the work is somewhat routine from a mechanics point of view.

Filename:tfigure8-ang-vel-ex

Figure 19.2: A point-mass bead slides on a rigid immobile frictionless wire. The free-body diagram shows that the only force on the bead is in the direction normal to the wire.

Example: Bead on frictionless wire

A bead slides on a frictionless wire with a crazy but smooth shape. No forces are applied to the bead besides the constraint force (see fig. 19.2).

Fig. 19.2b shows a free-body diagram where $\hat{\boldsymbol{n}}$ is the normal to the wire at the point of interest. It doesn’t matter if you use for $\hat{\boldsymbol{n}}=\hat{\boldsymbol{e}}_{n}$ , or $\hat{\boldsymbol{n}}=$ a vector always, say, to the left, just so long as you know what you mean by $\hat{\boldsymbol{n}}$ . The free-body diagram shows that you know that the constraint force is normal to the path (the frictionless wire) but that you don’t know how big it is ( $F$ is an unknown scalar).

For some purposes, especially general problems like this where no specific path is given, the most appropriate configuration variable is $s$ , the arc length along the path. If the path is given we assume we know the position at any given arc length by the functions

x(s)\hbox to36.135pt{\hfil and\hfil}y(s).

{\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}(s),{\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{s}\hat{% \boldsymbol{e}}_{t},\hbox to36.135pt{\hfil and\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}=(\dot{s}^{2}/\rho)\hat{\boldsymbol{e}}_{n}+% \ddot{s}\hat{\boldsymbol{e}}_{t}.

Now we can write linear momentum balance

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	(19.6)
$\displaystyle\bigm{\{}F\hat{\boldsymbol{n}}$	$\displaystyle=$	$\displaystyle m\bigm{(}(\dot{s}^{2}/\rho)\hat{\boldsymbol{e}}_{n}+\ddot{s}\hat% {\boldsymbol{e}}_{t}\bigm{)}\bigm{\}}$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{e}}_{t}\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}\dot{v}$	$\displaystyle=$	$\displaystyle 0\hbox to36.135pt{\hfil and\hfil}$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{e}}_{n}\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}F$	$\displaystyle=$	$\displaystyle mv^{2}/\rho.$

Eqn. 19.6 tells us that the bead moves at constant speed, no matter what the shape of the wire. It also tells us that the more curved the wire, the bigger the constraint force needed to keep the bead on the wire.

Because this is a 1 DOF system, any one equation of motion should give us the result. Instead of linear momentum balance we could have used power balance to get the same result like this:

$\displaystyle P$	$\displaystyle=$	$\displaystyle\dot{E_{\mathrm{K}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\rm tot}\cdot{\accentset{\rightharpoonup}{% \boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{d\over dt}\bigm{\{}{1\over 2}mv^{2}\bigm{\}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F\hat{\boldsymbol{e}}_{n% }\cdot v\hat{\boldsymbol{e}}_{t}$	$\displaystyle=$	$\displaystyle mv\dot{v}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle\dot{v}.$

This is natural enough. The only force on the particle is perpendicular to its motion, so does no work. So the particle must have constant kinetic energy and its speed must be constant.

The example above doesn’t seem that applicable; how often does one see beads on frictionless wires? It is a bit more useful than it appears. For example, if a car is coasting on an open road and tire resistance and sideslip can be ignored, the reasoning above shows that the car is neither speeded nor slowed by turning. Similarly, an idealization of an airplane wing is as something which only causes force perpendicular to motion. So in quick maneuvers where the gravity force is relatively small, the plane maintains its speed.

Filename:tfigure8-ang-vel

Figure 19.3: A bead slides on a frictionless wire on the curve implicitly defined by

y=cs^{2}/2

, where

s

is arc-length along the curve measured from the origin.

Example: The hard brachistochrone problem.

Filename:tfigure8-ang-accel

Figure 19.4: The roller coaster that gets from A to B the fastest is the one with a track in the shape of the brachistochrone

y=cs^{2}/2

Here is a puzzle proposed by Johann Bernoulli in June 1696 ^†^†margin: The original brachistochrone (least time) puzzle:
“I, Johann Bernoulli, greet the most clever mathematicians in the world. Nothing is more attractive to intelligent people than an honest, challenging problem whose possible solution will bestow fame and remain as lasting monument. Following the example set by Pascal, Fermat, etc., I hope to earn the gratitude of the entire scientific community by placing before the finest mathematicians of our time a problem which will test their methods and the strength of their intellect. If someone communicates to me the solution of the proposed problem, I shall then publicly declare him worthy of praise.

“Let two points A and B be given in a vertical plane. Find the curve that a point M, moving on a path AMB must follow such that, starting from A, it reaches B in the shortest time under its own gravity.” Newton. Besides Bernoulli’s brother Jakob, one of the people to solve the puzzle was 55 year old Isaac Newton. He attempted to keep his solution, said to have been worked out in one evening while he also invented the calculus of variations, anonymous. But Bernoulli supposedly saw through this deception, commenting “I recognize the lion by his paw print”, which was presumably not a comment about Newton’s handwriting. :

given that a roller coaster has to coast from rest at one place to another place that is no higher, what shape should the track be to make the trip as quick as possible.

The solution. Finding the solution, or even verifying it, is a problem in the calculus of variations, i.e., too advanced for this book. The solution turns out to be the brachistochrone curve that obeys the following relationship between arc-length $s$ from the origin and vertical position $y$ (drawn accurately in fig. 19.3):

y={1\over 2}cs^{2}\qquad(\mbox{for}|s|\leq 1/c).

(19.7)

Starting at the origin this curve is close to $y=cx^{2}/2$ but gets a bit higher (bigger $y$ ) because for a given value of $x$ , $s$ is greater than $x$ . The curve terminates at vertical tangents at $s=\pm 1/c$ where $y=1/2c$ . (see fig. 19.3). To solve the puzzle this curve is scaled (by choosing a value of $c$ ) and displaced so that it has a vertical tangent at A and also so the curve goes through B. The idea that the hard math seems to be expressing, is that the particle should first build up as much speed as it can (by going straight down) and then head off in the right direction (see fig. 19.4).

On the other hand, here is an easier problem that is a virtual setup for the techniques now at hand.

Example: The easy brachistochrone problem

How long does it take for a particle to slide back and forth on a frictionless wire with $y=cs^{2}/2$ as driven by gravity? (see fig. 19.3)

Let’s use $s$ as our configuration variable. The power balance equation is:

$\displaystyle P$	$\displaystyle=$	$\displaystyle\dot{E_{\mathrm{K}}}$
$\displaystyle(F\hat{\boldsymbol{e}}_{n})\cdot(v\hat{\boldsymbol{e}}_{t})=0% \hbox to36.135pt{\hfil$\Rightarrow$\hfil}-mg\hat{\boldsymbol{\jmath}}\cdot(% \dot{x}\hat{\boldsymbol{\imath}}+\dot{y}\hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle{d\over dt}\bigm{\{}{1\over 2}mv^{2}\bigm{\}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-mg\dot{y}$	$\displaystyle=$	$\displaystyle m{d\over dt}{\dot{s}^{2}\over 2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-mgcs\dot{s}$	$\displaystyle=$	$\displaystyle m\dot{s}\ddot{s}$
$\displaystyle\mbox{Assuming $\dot{s}\neq 0$}\hbox to36.135pt{\hfil$\Rightarrow% $\hfil}\ddot{s}+gcs=0.$

This, remarkably, is the simple harmonic oscillator equation with general solution

s=A\cos(\sqrt{gc}\,t)+B\sin(\sqrt{gc}\,t).

Thus the period of oscillation ( $T$ such that $\sqrt{gc}T=2\pi$ ) is

T={2\pi\over\sqrt{gc}}

which is independent of the amplitude of oscillation

^†^†margin: Actually, the amplitude can’t be arbitrarily large. The solution to the defining eqn. (19.7) only makes sense for

|s|<1/c

. For

|s|>1/c

there is no curve satisfying eqn. (19.7).

The key to this quick solution was using a configuration variable that made the expression for the velocity simple, and using an equation of motion that didn’t involve the unknown reaction force $F$ which we also didn’t care about. We could have got the same equation of motion by writing ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ and eliminated $F\hat{\boldsymbol{e}}_{n}$ by dotting both sides with a convenient vector orthogonal to $F\hat{\boldsymbol{e}}_{n}$ , say ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ .

The brachistochrone is a famous curve that has various interesting properties (e.g., Box 19.1 on page 19.1).

Box 19.1 Some brachistochrone curiosities

The brachistochrone is a cycloid. There is no straightforward way to draw the curve $y=cs^{2}/2$ because the formula doesn’t tell you the $x$ coordinates of the points. You could find them by integrating $dx=\sqrt{ds^{2}-dy^{2}}$ numerically or with calculus tricks. But it turns out (see below) that the curve with $y=cs^{2}/2$ is described by the parametric equations

$x=r(\phi+\sin\phi)$ with $\quad r={1\over 4c}$ .
$y=r(1-\cos\phi)$ .

Filename:tfigure-brachwheel

This is the path of a particle on the perimeter of a wheel that rolls against a horizontal ceiling a distance

2r

above the origin, as you can verify by adding up distances in the picture above (see page 18.1). We will show below that the upside down cycloid and the curve

y=cs^{2}/2

are one and the same.

Note that the osculating circle of this cycloid at its lowest point has radius $4r=1/c$ , just the length of a simple pendulum that, for small oscillations, has the same frequency of oscillation as the bead on the brachistochrone. A point mass swinging on a string is like a bead on a frictionless circular wire and this, in turn, is close to the motion of a bead on a brachistochrone wire for small oscillations.

Galileo (1564-1642). Well before Bernoulli’s challenge, Galileo was interested in things rolling and sliding on ramps. He knew that the shortest distance between two points is a straight line, and had noted that a ball rolling down an appropriately curved ramp gets to its destination faster than a ball traveling the shortest route. A ball going on a straight ramp just doesn’t pick up much speed, and when it finally has its greatest speed the trip is over. Imagine sliding straight sideways; it takes forever on a straight-line route. Better, he must have reasoned, to get the ball rolling fast at the start and then go fast for most of its journey, possibly slowing at the end. Galileo thought the best shape was the bottom of a circle (or fraction thereof), which isn’t far off either in shape or concept, but isn’t quite right. Galileo was apparently obsessed with cycloids for other reasons but didn’t see their connection to this problem.

Filename:tfigure-brachpen

A constant period pendulum. For clock time keeping, a pendulum is better than a bead on a wire because the friction of sliding is avoided. Unfortunately, a simple pendulum has a period which is longer if the amplitude is bigger. Not much longer, 18% if the swinging is $\pm 90^{{}^{\circ}}$ and only 1.7% longer if the amplitude is $\pm 30^{{}^{\circ}}$ , but enough to annoy clock designers. A bead sliding frictionlessly on the path $y=s^{2}/2$ has the nice property that the period does not depend on the amplitude. But any real bead sliding on any real wire has substantial friction. So, at first blush the brachistochrone curve, despite its nice constant-period property, cannot be used to keep time.

But Huygens, one of the smart old timers, looked for a curve that, when a string wraps around it, makes the end follow the brachistochrone curve. To this day you can see fancy old clocks with this wrapping device, a solid piece with two cuspoidal shapes at the hinge of the swinging (“isochronous” or “tautochrone”) pendulum which wraps around it.

Geometry. The two key features discussed above, that the curve $y=cs^{2}/2$ is a cycloid, and that a cycloid can be generated by wrapping a string around another cycloid, can be found from the geometric construction below. Two cycloids are shown, one from wheel 1 rolling under line $\rm L_{1}$ and another from wheel 2 rolling under line $\rm L_{2}$ a distance $2r$ below. Both wheels have radius $r$ . Imagine that the cycloids $\rm A_{1}M_{1}B$ and $\rm A_{2}M_{2}B$ are drawn by wheels always arranged with vertically aligned rolling contact points $\rm C_{1}$ and $\rm C_{2}$ and with points $\rm M_{1}$ and $\rm M_{2}$ initially aligned vertically a distance $4r$ apart at $\rm A_{1}$ and $\rm A_{2}$ .

The two cycloids are thus the same shape but are displaced with one being $2r$ below and $\pi r$ sideways from the other.

Because both wheels have rolled the same distance ( $\rm A_{1}C_{1}$ ) they have rotated the same amount and $\rm M_{2}$ is as far forward of $\rm C_{1}C_{2}D$ as $\rm M_{1}$ is behind. Similarly $\rm M_{1}$ is as far above $\rm L_{2}$ as $\rm M_{2}$ is below. So the line $\rm M_{1}M_{2}$ is bisected by the point $\rm C_{2}$ .

Because $\rm C_{1}C_{2}$ is the diameter of a circle with $\rm M_{1}$ on the perimeter, angle $\rm C_{1}M_{1}C_{2}$ is a right angle. Because material point $\rm C_{1}$ on the wheel has zero velocity the velocity of $\rm M_{1}$ (and thus the tangent to the curve) is orthogonal to $\rm C_{1}M_{1}$ . Thus the line $\rm M_{1}M_{2}$ is tangent to the upper cycloid.

The rolling of wheel 2 instantaneously rotating about $\rm C_{2}$ makes the tangent to the lower cycloid orthogonal to $\rm M_{1}M_{2}$ , the condition for the motion of $\rm M_{2}$ to be from the wrapping of an inextensible line around the curve $\rm A_{1}M_{1}B$ . This shows that cycloid $\rm A_{2}M_{2}B$ is generated by the wrapping of a line anchored at $\rm A_{1}$ about the upper cycloid. And this is Huygen’s wrapping mechanism for making a pendulum bob follow a cycloid. Because of this wrapping generation, the arc-length $s^{\prime}+s$ of $\rm A_{1}M_{1}B$ must be $4r$ and the arc length $s$ of $\rm M_{1}B$ is $\rm M_{1}M_{2}$ so the length $\rm M_{1}C_{2}$ is $s/2$ . By the similarity of the two right triangles that share the length $s/2$ of $\rm M_{1}C_{2}$ :

{y\over s/2}={s/2\over 2r}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}y={1\over 4% r}{s^{2}\over 2}=c{s^{2}\over 2}

Filename:tfigure-brach2wheels

which shows that the upper cycloid is the curve $y=cs^{2}/2$ if $c=1/4r$ , where $s$ is measured from B. This was the equation used to show the constant period nature of the sliding motion of a bead on a frictionless cycloidal curve using power balance.

Example: A collar on two rotating rods

Filename:tfigure8-ang-accel-ex

Figure 19.5: A point-mass collar slides simultaneously on 2 rods.

Consider a pair of collars hinged together as a point mass $m$ at P. Each slides frictionlessly on a rod about whose rotation everything is known (see fig. 19.5. What is the force of rod 1 on the mass? For this 2 degree of freedom system let’s use configuration variables $\theta_{1}$ and $\theta_{2}$ , and two sets of rotating base vectors: $\hat{\boldsymbol{\lambda}}_{1}\hat{\boldsymbol{n}}_{1}$ and $\hat{\boldsymbol{\lambda}}_{2}\hat{\boldsymbol{n}}_{2}$ . These rotating base vectors can be written in terms of the $\theta$ s, $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ in the standard manner. Assume we know $\ell_{1}$ and $\ell_{2}$ in this configuration. First find $\dot{\ell}_{1}$ and $\dot{\ell}_{2}$ by thinking of the velocity of the collar two different ways:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	(19.8)
$\displaystyle\bigm{\{}\dot{\ell}_{1}\hat{\boldsymbol{\lambda}}_{1}+\dot{\theta% }_{1}\ell_{1}\hat{\boldsymbol{n}}_{1}$	$\displaystyle=$	$\displaystyle\dot{\ell}_{2}\hat{\boldsymbol{\lambda}}_{2}+\dot{\theta}_{2}\ell% _{2}\hat{\boldsymbol{n}}_{2}\bigm{\}}$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{n}}_{2}$	$\Rightarrow$	$\displaystyle\dot{\ell}_{1}={\dot{\theta}_{2}\ell_{2}-\dot{\theta}_{1}\ell_{1}% \hat{\boldsymbol{n}}_{1}\cdot\hat{\boldsymbol{n}}_{2}\over\hat{\boldsymbol{% \lambda}}_{1}\cdot\hat{\boldsymbol{n}}_{2}}$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{n}}_{1}$	$\Rightarrow$	$\displaystyle\dot{\ell}_{2}={\dot{\theta}_{1}\ell_{1}-\dot{\theta}_{2}\ell_{2}% \hat{\boldsymbol{n}}_{2}\cdot\hat{\boldsymbol{n}}_{1}\over\hat{\boldsymbol{% \lambda}}_{2}\cdot\hat{\boldsymbol{n}}_{1}}.$

Having found $\dot{\ell}_{1}$ and $\dot{\ell}_{2}$ we can find the velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ by evaluating either side of eqn. (19.8). Now we apply identical reasoning with the acceleration. The result looks messy, but the approach is straightforward:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	(19.9)
$\displaystyle\bigm{\{}\ddot{\ell}_{1}\hat{\boldsymbol{\lambda}}_{1}+\ddot{% \theta}_{1}\ell_{1}\hat{\boldsymbol{n}}_{1}+2\dot{\ell}_{1}\dot{\theta}_{1}% \hat{\boldsymbol{n}}_{1}$	$\displaystyle=$	$\displaystyle\ddot{\ell}_{2}\hat{\boldsymbol{\lambda}}_{2}+\ddot{\theta}_{2}% \ell_{2}\hat{\boldsymbol{n}}_{2}+2\dot{\ell}_{2}\dot{\theta}_{2}\hat{% \boldsymbol{n}}_{2}\bigm{\}}$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{n}}_{2}\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}\ddot{\ell}_{1}={\ddot{\theta}_{2}\ell_{2}+2\dot{\ell}% _{2}\dot{\theta}_{2}-\ddot{\theta}_{1}\ell_{1}\hat{\boldsymbol{n}}_{1}\cdot% \hat{\boldsymbol{n}}_{2}-2\dot{\ell}_{1}\dot{\theta}_{1}\hat{\boldsymbol{n}}_{% 1}\cdot\hat{\boldsymbol{n}}_{2}\over\hat{\boldsymbol{\lambda}}_{1}\cdot\hat{% \boldsymbol{n}}_{2}}$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{n}}_{1}\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}\ddot{\ell}_{2}={\ddot{\theta}_{1}\ell_{1}+2\dot{\ell}% _{1}\dot{\theta}_{1}-\ddot{\theta}_{2}\ell_{2}\hat{\boldsymbol{n}}_{2}\cdot% \hat{\boldsymbol{n}}_{1}-2\dot{\ell}_{2}\dot{\theta}_{2}\hat{\boldsymbol{n}}_{% 2}\cdot\hat{\boldsymbol{n}}_{1}\over\hat{\boldsymbol{\lambda}}_{2}\cdot\hat{% \boldsymbol{n}}_{1}}$

We use the results from eqn. (19.8) for $\dot{\ell}_{1}$ and $\dot{\ell}_{2}$ to evaluate the right hand sides of the expressions for $\ddot{\ell}_{1}$ and $\ddot{\ell}_{2}$ in eqn. (19.9). So now either the left hand side or the right hand side of the second of Eqns. 19.9 can be used to evaluate the acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ , all the terms in both expressions have been found.

To find the forces we use linear momentum balance and the free-body diagram

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\rm tot}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	(19.10)
$\displaystyle\bigm{\{}F_{1}\hat{\boldsymbol{n}}_{1}+F_{2}\hat{\boldsymbol{n}}_% {2}$	$\displaystyle=$	$\displaystyle m\{\ddot{\ell}_{1}\hat{\boldsymbol{\lambda}}_{1}+\ddot{\theta}_{% 1}\ell_{1}\hat{\boldsymbol{n}}_{1}+2\dot{\ell}_{1}\dot{\theta}_{1}\hat{% \boldsymbol{n}}_{1}\}\bigm{\}}$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{\lambda}}_{2}\hbox to36.1% 35pt{\hfil$\Rightarrow$\hfil}F_{2}={\{\ddot{\ell}_{1}\hat{\boldsymbol{\lambda}% }_{1}+\ddot{\theta}_{1}\ell_{1}\hat{\boldsymbol{n}}_{1}+2\dot{\ell}_{1}\dot{% \theta}_{1}\hat{\boldsymbol{n}}_{1}\}\cdot\hat{\boldsymbol{\lambda}}_{2}\over% \hat{\boldsymbol{n}}_{1}\cdot\hat{\boldsymbol{\lambda}}_{2}}$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{\lambda}}_{1}\hbox to36.1% 35pt{\hfil$\Rightarrow$\hfil}F_{1}={\{\ddot{\ell}_{1}\hat{\boldsymbol{\lambda}% }_{1}+\ddot{\theta}_{1}\ell_{1}\hat{\boldsymbol{n}}_{1}+2\dot{\ell}_{1}\dot{% \theta}_{1}\hat{\boldsymbol{n}}_{1}\}\cdot\hat{\boldsymbol{\lambda}}_{1}\over% \hat{\boldsymbol{n}}_{2}\cdot\hat{\boldsymbol{\lambda}}_{1}}$

When actually evaluating the expressions above one can write the base vectors in terms of $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ or use geometry.

Often when working out a problem it is best to not substitute numbers until the end of a problem. This example shows the opposite. If we left the expressions for $\dot{\ell}_{1}$ and $\dot{\ell}_{2}$ with letters and substituted that into the expressions for $\ddot{\ell}_{1}$ and $\ddot{\ell}_{2}$ and left those expressions intact while substituting for the acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ we would have large expressions for the force components $F_{1}$ and $F_{2}$ . On the other hand, by using numbers as the calculation progresses the formulas do not grow so much in complexity.

As is the case with most mechanism-mechanics problems, the hard work in getting the dynamics equations is in the kinematics. Generally there are no great short-cuts. There are alternative methods. In this case the location of the base points and the two angles determine the base and two angles of a triangle. This triangle can be solved for the location of the point P. Once that position is known in terms of $\theta_{1}$ and $\theta_{2}$ the velocity and acceleration can be found by differentiation.

As a robot manipulator, this design has the advantage that no motors need to be displaced. It has the disadvantage of requiring good sliding joints.

An alternative solution of the kinematics of this problem would be to use trigonometry to find the position of point P in terms of the angles $\theta_{1}$ and $\theta_{2}$ . Then the acceleration of point P is found by taking two time derivatives. The result is approximately equal in the complexity of its appearance to the results used above. That method requires more cleverness at the start (solving an angle-side-angle triangle) and then just brute force differentiation using the chain rule and the product rule.

Filename:tfigure8-ang-accel1

Figure 19.6: The base of a pendulum is vertically vibrated.

Example: Inverted pendulum with a vibrating base

Assume that the base $0^{\prime}$ of an inverted point-mass pendulum of length $\ell$ is vibrated according to (see fig. 19.6)

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0^{\prime}}}=d\sin% \omega t\hat{\boldsymbol{\imath}}.

The point P thus has acceleration

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{0^{\prime}% }}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/0^{\prime}}}$
		$\displaystyle=$	$\displaystyle-d\omega^{2}\sin\omega t\hat{\boldsymbol{\imath}}+\ddot{\theta}% \ell\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}-\dot{\theta}^{2}\ell\hat% {\boldsymbol{e}}_{\scriptscriptstyle\!R}$

Now apply linear momentum balance as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{tot}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}$
$\displaystyle\bigm{\{}-mg\hat{\boldsymbol{\imath}}+-T\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}$	$\displaystyle=$	$\displaystyle m\{-d\omega^{2}\sin\omega t\hat{\boldsymbol{\imath}}+\ddot{% \theta}\ell\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}-\dot{\theta}^{2}% \ell\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\}\bigm{\}}$
$\displaystyle\bigm{\{}\bigm{\}}\cdot\hat{\boldsymbol{e}}_{\scriptscriptstyle\!% \theta}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-g\hat{\boldsymbol{\imath}}% \cdot\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle-d\omega^{2}\sin\omega t\hat{\boldsymbol{\imath}}\cdot\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}+\ddot{\theta}\ell$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}g\sin\theta$	$\displaystyle=$	$\displaystyle d\omega^{2}\sin\omega t\sin\theta+\ddot{\theta}\ell$

which you write as

0=\ddot{\theta}+(d\omega^{2}\sin\omega t-g)\sin\theta/\ell\quad\mbox{or}\quad% \ddot{\theta}=(g-d\omega^{2}\sin\omega t)\sin\theta/\ell

depending on whether you are analytically or numerically inclined. This is a second order non-linear ordinary differential equation. If $\omega=0$ or $d=0$ then this is the classic inverted pendulum equation and has solutions that show that the pendulum doesn’t stay near upright. But, you can find by analytic cleverness or numerical integration that for some values of $d$ and $\omega$ that the pendulum does not fall down! Just shaking the base keeps the pendulum up ( $\omega^{2}d>g$ for all cases where this is possible). This isn’t just academic nonsense, the device can be built and the balancing demonstrated.

One alternative to using linear momentum balance in the equations above would be to use angular momentum balance about the point 0’. The resulting vector equation

\ell\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times(-mg\hat{\boldsymbol{% \imath}})={\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0^{\prime% }}}\times(m{\accentset{\rightharpoonup}{\boldsymbol{a}}})

yields the same second order scalar ODE.

The vibrating mechanism shown is a “Scotch yoke”. An eccentric disk is mounted to the shaft of a constant angular velocity motor. The rectangular slot moves up and down sinusoidally as the disk wobbles.

SAMPLE 19.1

Filename:sfig8-2-3

Figure 19.7:

A bead on a straight wire. A straight wire is hung between points A and B in the $x y$ plane as shown in the figure. A bead slides down the wire from point A. Write the geometric constraint equation for the bead’s motion and derive the conditions on velocity and acceleration components of the bead due to the constraint.

Solution The constraint on the bead’s motion is that its path must be along the wire, i.e., a straight line between points A and B. Thus the geometric constraint on the motion is expressed by the equation of the path which is

y=h-\frac{h}{\ell}x.

Since the bead is constrained to move on this path, its velocity and acceleration vectors are also constrained to be directed along AB. This imposes conditions on their $x$ and $y$ components that are easily derived by differentiating the geometric constraint equation with respect to $t$ . Thus,

	$\displaystyle\dot{y}$	$\displaystyle=$	$\displaystyle-\frac{h}{\ell}\dot{x},$
	$\displaystyle\ddot{y}$	$\displaystyle=$	$\displaystyle-\frac{h}{\ell}\ddot{x}.$

Answer: $y=h-\frac{h}{\ell}x,\quad\dot{y}=-\frac{h}{\ell}\dot{x},\quad\ddot{y}=-\frac{h% }{\ell}\ddot{x}$

SAMPLE 19.2

Filename:sfig8-2-3a

Figure 19.8:

A particle sliding on a parabolic path. A particle slides on a parabolic trough given by $y=ax^{2}$ where $a$ is a constant. Write the geometric constraints of motion (on the path, velocity, and acceleration) of the particle. Write the velocity and acceleration of the particle at a generic location $(x,y)$ on its path.

Solution The geometric constraint on the path of the particle is already given, $y=ax^{2}$ . Differentiating the path constraint with respect to time, we get the constraint on velocity and acceleration components.

	$\displaystyle\dot{y}$	$\displaystyle=$	$\displaystyle 2ax\dot{x},$
	$\displaystyle\ddot{y}$	$\displaystyle=$	$\displaystyle 2ax\ddot{x}+2a\dot{x}^{2}.$

Now, at a point $(x,y)$ , we can write the velocity and acceleration of the particle as

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle\dot{x}\hat{\boldsymbol{\imath}}+\dot{y}\hat{\boldsymbol{\jmath}}% =\dot{x}\hat{\boldsymbol{\imath}}+2ax\dot{x}\hat{\boldsymbol{\jmath}},$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{y}\hat{\boldsymbol{\jmath% }}=\ddot{x}\hat{\boldsymbol{\imath}}+(2ax\ddot{x}+2a\dot{x}^{2})\hat{% \boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{x}\hat{\boldsymbol{\imath}}% +2ax\dot{x}\hat{\boldsymbol{\jmath}},\quad{\accentset{\rightharpoonup}{% \boldsymbol{a}}}=\ddot{x}\hat{\boldsymbol{\imath}}+(2ax\ddot{x}+2a\dot{x}^{2})% \hat{\boldsymbol{\jmath}}$

SAMPLE 19.3

Filename:sfig8-2-3b

Figure 19.9:

Circular motion of a particle. A particle is constrained to move on a frictionless circular path of radius $R_{0}$ with constant angular speed $\dot{\theta}$ . There is no gravity. Find the equation of motion of the particle in the $x$ -direction and show that this motion is simple harmonic.

Solution This is a simple problem that you have solved before, probably a few times. Here, we do this problem again just to show how it works out with the constraint machinery in evidence. The geometric constraint on the path of the particle is $R=R_{0}$ (in polar coordinates). This constraint gives us $\dot{R}=0$ and $\ddot{R}=0$ . Then the acceleration of the particle (in polar coordinates), ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=(\ddot{R}-R\dot{\theta}^{2})\hat% {\boldsymbol{e}}_{\scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ reduces to ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-R\dot{\theta}^{2}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}$ , (of course).

Filename:sfig8-3-1

Figure 19.10:

The free-body diagram of the particle shows that there is only one force acting on the particle, the normal reaction $N$ of the path acting in the $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ direction. Therefore, the linear momentum balance gives,

N\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}=m(-R\dot{\theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle% \!R})\hbox to36.135pt{\hfil$\Rightarrow$\hfil}N=-mR\dot{\theta}^{2}.

But, to write the equation of motion in the $x$ -direction, we need to write the linear momentum balance in the $x$ -direction. We can write $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ using mixed basis vectors as $N\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=m(\ddot{x}\hat{\boldsymbol{% \imath}}+\ddot{y}\hat{\boldsymbol{\jmath}})$ . Dotting this equation with $\hat{\boldsymbol{\imath}}$ , we get

\ddot{x}=\frac{N}{m}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\cdot\hat{% \boldsymbol{\imath}})=\frac{-mR\dot{\theta}^{2}}{m}\cos\theta=-\dot{\theta}^{2% }(R\cos\theta)=-\dot{\theta}^{2}x

or, $\ddot{x}+\dot{\theta}^{2}x=0$ , which is the equation of simple harmonic motion in $x$ . You can easily show that the motion in the $y$ -direction is also simple harmonic ( $\ddot{y}+\dot{\theta}^{2}y=0$ ).

Answer:

SAMPLE 19.4

A bead slides on a straight wire. Consider the problem of the bead sliding on a straight, inclined, frictionless wire of Sample 19.6 again. Find the position of the bead $x(t)$ and $y(t)$ assuming it slides under gravity starting from rest at A.

Solution To find the position of the bead, we need to write the equation of motion and solve it. This is a single DOF system and, therefore, one scalar equation of motion should suffice.

Filename:sfig8-3-1a

Figure 19.11:

The free-body diagram of the bead is shown in fig. 19.11. Using basis vectors $(\hat{\boldsymbol{\lambda}},\hat{\boldsymbol{n}})$ and $(\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}})$ we write the LMB for the bead as

-mg\hat{\boldsymbol{\jmath}}+N\hat{\boldsymbol{n}}=m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}=m(\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{y}% \hat{\boldsymbol{\jmath}}).

We can easily eliminate the constraint force $N$ from this equation by dotting this equation with $\hat{\boldsymbol{\lambda}}$ , which gives

	$\displaystyle-\cancel{m}g(\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{% \lambda}})$	$\displaystyle=$	$\displaystyle\cancel{m}[\ddot{x}(\hat{\boldsymbol{\imath}}\cdot\hat{% \boldsymbol{\lambda}})+\ddot{y}(\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol% {\lambda}})]$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}g\sin\theta$	$\displaystyle=$	$\displaystyle\ddot{x}\cos\theta-\ddot{y}\sin\theta.$

But, from the geometric constraint $y=h-(h/\ell)x$ , we have $\ddot{y}=-(h/\ell)\ddot{x}=-(\tan\theta)\ddot{x}$ . Therefore,

g\sin\theta=\ddot{x}\cos\theta+\ddot{x}\tan\theta\sin\theta\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}\ddot{x}=g\sin\theta\cos\theta.

Since $g\sin\theta\cos\theta$ is constant, we integrate the equation of motion easily to find $x(t)=\frac{1}{2}g\sin\theta\cos\theta t^{2}$ since $x(0)=0,\dot{x}(0)=0$ . And, since $y=h-x\tan\theta$ , we have $y(t)=h-\frac{1}{2}g\sin^{2}\theta t^{2}.$

Answer: $x(t)=\frac{1}{2}ght^{2}\sin\theta\cos\theta,\quad y(t)=h-\frac{1}{2}gt^{2}\sin% ^{2}\theta$

SAMPLE 19.5

Filename:sfig8-5-wiper

Figure 19.12:

A bead sliding down a parabolic trough. Consider the problem of Sample 19.7 again. Find the equation of motion of the bead.

Solution This is, again, a one DOF system. Therefore, we will get a single scalar equation of motion. The free-body diagram shown in fig. 19.13 shows two forces acting on the bead. The constraint force $N$ acts normal to the path. Let $\hat{\boldsymbol{e}}_{t}$ and $\hat{\boldsymbol{e}}_{n}$ be unit vectors tangential and normal to the path, respectively. Then the linear momentum balance gives

Filename:sfig8-5-wiper-a

Figure 19.13:

-mg\hat{\boldsymbol{\jmath}}+N\hat{\boldsymbol{e}}_{n}=m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}=m(\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{y}% \hat{\boldsymbol{\jmath}}).

To eliminate the unknown constraint force $N$ from this equation, we can take a dot product of this equation with $\hat{\boldsymbol{e}}_{t}$ . However, we must first find $\hat{\boldsymbol{e}}_{t}$ . Now $\hat{\boldsymbol{e}}_{t}$ is the unit tangent vector. So, we can find it by finding a tangent vector to the path (remember gradient of a function $\nabla f$ ?) and then dividing it by the length of the vector. That is doable but a little complicated. All we need here is the dot product with a vector normal to $\hat{\boldsymbol{e}}_{n}$ . Why not use the velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{x}\hat{\boldsymbol{\imath}}% +\dot{y}\hat{\boldsymbol{\jmath}}$ ? The velocity vector is always tangential to the path. Furthermore, we know that from the geometric constraint ( $y=ax^{2}$ ), $\dot{y}=2ax\dot{x}$ and $\ddot{y}=2ax\ddot{x}+2a\dot{x}^{2}$ . Therefore, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{x}\hat{\boldsymbol{\imath}}% +2ax\dot{x}\hat{\boldsymbol{\jmath}}$ . Now, dotting the LMB equation with ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ we get,

$\displaystyle-\cancel{m}g\overbrace{(\hat{\boldsymbol{\jmath}}\cdot{\accentset% {\rightharpoonup}{\boldsymbol{v}}})}^{2ax\dot{x}}$	$\displaystyle=$	$\displaystyle\cancel{m}\bigm{[}\ddot{x}\overbrace{(\hat{\boldsymbol{\imath}}% \cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}})}^{\dot{x}}+\ddot{y}% \overbrace{(\hat{\boldsymbol{\jmath}}\cdot{\accentset{\rightharpoonup}{% \boldsymbol{v}}})}^{2ax\dot{x}}\bigm{]}$
$\displaystyle-2gax\cancel{\dot{x}}$	$\displaystyle=$	$\displaystyle\ddot{x}\cancel{\dot{x}}+2ax\ddot{y}\cancel{\dot{x}}$
	$\displaystyle=$	$\displaystyle\ddot{x}+2ax\underbrace{(2ax\ddot{x}+2a\dot{x}^{2})}_{\ddot{y}}$
	$\displaystyle=$	$\displaystyle\ddot{x}(1+4a^{2}x^{2})+4a^{2}x\dot{x}^{2}.$

Rearranging the terms above, we get the required equation of motion:

\ddot{x}+\frac{4a^{2}x}{1+4a^{2}x^{2}}\dot{x}^{2}+\frac{2gax}{1+4a^{2}x^{2}}=0.

As you can see, this is a nonlinear ODE. Analytical solution of this equation is rather difficult. We can, however, always solve it numerically. Note that a solution of this equation only gives you $x(t)$ , i.e., the $x$ coordinate of the position of the bead. But, you can always find the $y$ coordinate since $y=ax^{2}$ .

Answer: $\ddot{x}+\frac{4a^{2}x}{1+4a^{2}x^{2}}\dot{x}^{2}+\frac{2gax}{1+4a^{2}x^{2}}=0$

Comment: Note that if we consider $x$ and $\dot{x}$ to be very small so that we can ignore the $\dot{x}^{2}$ terms completely, and take $1+4a^{2}x^{2}\approx 1$ , then the equation of motion becomes

\ddot{x}+(2ga)\,x=0

which is the equation of simple harmonic motion with frequency $\sqrt{2ga}$ . Thus, if we consider a shallow parabola, and release the bead close to the origin, it executes simple harmonic motion, much like a simple pendulum. This is an intuitively realizable motion.

SAMPLE 19.6

Filename:sfig8-5-2disks

Figure 19.14: A pin is constrained to move in a groove and a slotted arm.

Constrained motion of a pin. During a small interval of its motion, a pin of 100 grams is constrained to move in a groove described by the equation $R=R_{0}+k\theta$ where $R_{0}=0.3\,\mathrm{m}$ and $k=0.05\,\mathrm{m}$ . The pin is driven by a slotted arm AB and is free to slide along the arm in the slot. The arm rotates at a constant speed $\omega=6\,\mathrm{rad}/\mathrm{s}$ . Find the magnitude of the force on the pin at $\theta=60^{{}^{\circ}}$ .

Solution Let ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ denote the net force on the pin. Then from the linear momentum balance

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}

where ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ is the acceleration of the pin. Therefore, to find the force at $\theta=60^{{}^{\circ}}$ we need to find the acceleration at that position.

From the given figure, we assume that the pin is in the groove at $\theta=60^{{}^{\circ}}$ . Since the equation of the groove (and hence the path of the pin) is given in polar coordinates, it seems natural to use polar coordinate formula for the acceleration. For planar motion, the acceleration is

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}.

We are given that $\dot{\theta}\equiv\omega=6\,\mathrm{rad}/\mathrm{s}$ and the radial position of the pin $R~{}=~{}R_{0}~{}+~{}k~{}\theta$ . Therefore, ^†^†margin: Note that $R$ is a function of $\theta$ and $\theta$ is a function of time, therefore $R$ is a function of time. Although we are interested in finding $\dot{R}$ and $\ddot{R}$ at $\theta=60^{{}^{\circ}}$ , we cannot first substitute $\theta=60^{{}^{\circ}}$ in the expression for $R$ and then take its time derivatives (which will be zero).

$\displaystyle\ddot{\theta}$	$\displaystyle=$	$\displaystyle\frac{d\dot{\theta}}{dt}=0\quad\mbox{( since $\dot{\theta}=$ constant) }$
$\displaystyle\dot{R}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}(R_{0}+k\theta)=k\dot{\theta}\hbox to36.135pt{\hfil and\hfil}$
$\displaystyle\ddot{R}$	$\displaystyle=$	$\displaystyle k\ddot{\theta}=0.$

Substituting these expressions in the acceleration formula and then substituting the numerical values at $\theta=60^{{}^{\circ}}$ , (remember, $\theta$ must be in radians!), we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle-\overbrace{(R_{0}+k\theta)\dot{\theta}^{2}}^{R\dot{\theta}^{2}}% \,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\overbrace{2k\dot{\theta}^{2}}^% {2\dot{R}\dot{\theta}}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
	$\displaystyle=$	$\displaystyle-(0.3\,\mathrm{m}+0.05\,\mathrm{m}\cdot\frac{\pi}{3})\cdot(6\,% \mathrm{rad}/\mathrm{s})^{2}\,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+2% \cdot 0.05\,\mathrm{m}\cdot(6\,\mathrm{rad}/\mathrm{s})^{2}\,\hat{\boldsymbol{% e}}_{\scriptscriptstyle\!\theta}$
	$\displaystyle=$	$\displaystyle-13.63\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+3.60\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}.$

Therefore the net force on the pin is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
	$\displaystyle=$	$\displaystyle 0.1\,\mathrm{kg}\cdot(-13.63\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+3.60\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta})% \,\mathrm{m}/\mathrm{s}^{2}$
	$\displaystyle=$	$\displaystyle(-1.36\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+0.36\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta})\,{\mbox{N}}$

and the magnitude of the net force is

F=|{\accentset{\rightharpoonup}{\boldsymbol{F}}}|=\sqrt{(1.36\,{\mbox{N}})^{2}% +(0.36\,{\mbox{N}})^{2}}=1.41\,{\mbox{N}}.

Answer: $F=1.41\,{\mbox{N}}$

SAMPLE 19.7

Filename:sfig8-4-1

Figure 19.15:

A puck sliding on a frictional rotating table. A horizontal turntable rotates with constant angular speed $\omega=100\,{\rm rpm}$ . A puck of mass $m=0.1\,\mathrm{kg}$ is gently placed on the rotating turntable. The puck begins to slide. The coefficient of friction between the puck and the turntable is 0.25. Find the equation of motion of the puck using

1.

A fixed reference frame with cartesian coordinates
2.

A rotating reference frame with cartesian coordinates

Solution

Equation of motion using a fixed reference frame: The puck has two DOF on the turntable. So, we will need two configuration variables, say $x$ and $y$ , and we will have to find equation of motion for each variable.

Let us use a fixed cartesian coordinate system with the origin at the center of the turntable. Let ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}=x\hat{\boldsymbol% {\imath}}+y\hat{\boldsymbol{\jmath}}$ be the position of the puck at some instant $t$ , so that its velocity is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}=\dot{x}% \hat{\boldsymbol{\imath}}+\dot{y}\hat{\boldsymbol{\jmath}}$ and acceleration is ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}=\ddot{x}\hat{% \boldsymbol{\imath}}+\ddot{y}\hat{\boldsymbol{\jmath}}$ .

Filename:sfig8-4-1a

Figure 19.16: A partial free-body diagram of the puck. For linear momentum balance we need to consider only the forces acting in the plane of motion.

The free-body diagram of the puck should show three forces — the force of gravity (in $-\hat{\boldsymbol{k}}$ direction), the normal reaction of the turntable (in $\hat{\boldsymbol{k}}$ direction) and the friction force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ . Since, there is no motion in the vertical direction, we know that $N=mg$ and that $F=|{\accentset{\rightharpoonup}{\boldsymbol{F}}}|=\mu N=\mu mg$ . But, what is the direction of the friction force? Well, we know that it acts in the opposite direction of the relative slip, so that

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=-\mu N\frac{{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{rel}}}{|{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{rel}}|}.

So, we need to find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{rel}}$ . Now ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{rel}}$ is the velocity of the puck relative to the turntable, or more precisely, relative to the point on the turntable just underneath the puck. Let us denote that point by $P^{\prime}$ . Clearly, $P^{\prime}$ goes in circles with constant speed, so that its velocity is

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P^{\prime}}% }={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P^{\prime}}}=\dot{\theta}\hat{% \boldsymbol{k}}\times(x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}})=% \dot{\theta}x\hat{\boldsymbol{\jmath}}-\dot{\theta}y\hat{\boldsymbol{\imath}}.

Therefore, the relative velocity, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{rel}}$ is

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{rel}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}-{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P^{\prime}}}% =(\dot{x}+\dot{\theta}y)\hat{\boldsymbol{\imath}}+(\dot{y}-\dot{\theta}x)\hat{% \boldsymbol{\jmath}}

Now the linear momentum balance for the puck in the $x y$ plane gives

$\displaystyle-\mu\,\cancel{m}g\frac{{\accentset{\rightharpoonup}{\boldsymbol{v% }}}_{\mkern 0.5mu\mathrm{rel}}}{\|{\accentset{\rightharpoonup}{\boldsymbol{v}}}% _{\mkern 0.5mu\mathrm{rel}}\|}$	$\displaystyle=$	$\displaystyle\cancel{m}(\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{y}\hat{% \boldsymbol{\jmath}})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{x}$	$\displaystyle=$	$\displaystyle\frac{-\mu g}{\|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{rel}}\|}\,({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{rel}}\cdot\hat{\boldsymbol{\imath}})=-\frac{\mu g(\dot{x}+% \dot{\theta}y)}{\sqrt{(\dot{x}+\dot{\theta}y)^{2}+(\dot{y}-\dot{\theta}x)^{2}}}$
$\displaystyle\hbox to36.135pt{and\hfil}\ddot{y}$	$\displaystyle=$	$\displaystyle\frac{-\mu g}{\|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{rel}}\|}\,({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{rel}}\cdot\hat{\boldsymbol{\jmath}})=-\frac{\mu g(\dot{y}-% \dot{\theta}x)}{\sqrt{(\dot{x}+\dot{\theta}y)^{2}+(\dot{y}-\dot{\theta}x)^{2}}}.$

These are coupled nonlinear ODEs that represent the equations of motion of the puck.

Answer: $\ddot{x}=-\frac{\mu g(\dot{x}+\dot{\theta}y)}{\sqrt{(\dot{x}+\dot{\theta}y)^{2% }+(\dot{y}-\dot{\theta}x)^{2}}},\quad\ddot{y}=-\frac{\mu g(\dot{y}-\dot{\theta% }x)}{\sqrt{(\dot{x}+\dot{\theta}y)^{2}+(\dot{y}-\dot{\theta}x)^{2}}}$

Note that these equations are valid only as long as there is relative slip between the puck and the turntable. If the puck stops sliding due to friction, it simply goes in circles with the turntable and, therefore, its equations of motion then are $\ddot{x}=-\dot{\theta}^{2}x,\,\ddot{y}=-\dot{\theta}^{2}y$ .

Equation of motion using a rotating reference frame: Now we derive the equations of motion using a rotating reference frame, $\mathcal{B}$ , with $(x^{\prime},y^{\prime})$ coordinate axes, fixed to the rotating turntable. Let the position of the puck in the rotating frame be ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/O^{\prime}}}=x^{% \prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{\boldsymbol{\jmath}}^% {\prime}$ . Note that the velocity of the puck in the rotating reference frame is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\mathcal{% B}}}=\dot{x}^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+\dot{y}^{\prime}\hat{% \boldsymbol{\jmath}}^{\prime}$ , and acceleration is ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\mathcal{B}}}=\ddot{x% }^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+\ddot{y}\hat{\boldsymbol{\jmath}}% ^{\prime}$ .

Now, from the linear momentum balance ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ ) for the puck, we get

Filename:sfig8-6-2

Figure 19.17: Axes

(x^{\prime},y^{\prime})

represent the rotating frame

\mathcal{B}

fixed to the rotating turntable; ie

(x^{\prime},y^{\prime})

rotate with angular velocity

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\dot{\theta}% \hat{\boldsymbol{k}}

-\mu\cancel{m}g\frac{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5% mu\mathrm{rel}}}{|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{rel}}|}=\cancel{m}({\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \mathrm{P^{\prime}}}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/% \mathcal{B}}}+2{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}% }\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/% \mathcal{B}}})

where we have used the three term acceleration formula for ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}$ . Here,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P^{\prime}}}$	$\displaystyle=$	$\displaystyle-\dot{\theta}^{2}(x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+y^% {\prime}\hat{\boldsymbol{\jmath}}^{\prime})$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P/\mathcal% {B}}}$	$\displaystyle=$	$\displaystyle\ddot{x}^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+\ddot{y}^{% \prime}\hat{\boldsymbol{\jmath}}^{\prime}$
$\displaystyle 2{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}% }\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/% \mathcal{B}}}$	$\displaystyle=$	$\displaystyle-2\dot{\theta}\dot{y}^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+% 2\dot{\theta}\dot{x}^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}$

Note that the point $P^{\prime}$ , coincident with P and fixed on the turntable, is stationary with respect to the rotating frame. Therefore, the relative velocity of P as observed in the rotating frame is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{rel}}={\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/\mathcal{B}}}=\dot{x}% ^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+\dot{y}^{\prime}\hat{\boldsymbol{% \jmath}}^{\prime}$ . Substituting these terms in the LMB equation above, we have

-\mu g\frac{\dot{x}^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+\dot{y}^{\prime% }\hat{\boldsymbol{\jmath}}^{\prime}}{\sqrt{\dot{x}^{\prime 2}+\dot{y}^{\prime 2% }}}=(-\dot{\theta}^{2}x^{\prime}+\ddot{x}^{\prime}-2\dot{\theta}\dot{y}^{% \prime})\hat{\boldsymbol{\imath}}^{\prime}+(-\dot{\theta}^{2}y^{\prime}+\ddot{% y}^{\prime}+2\dot{\theta}\dot{x}^{\prime})\hat{\boldsymbol{\jmath}}^{\prime}

Dotting this equation with $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ , respectively, we get

	$\displaystyle\ddot{x}^{\prime}$	$\displaystyle=$	$\displaystyle\dot{\theta}^{2}x^{\prime}-\frac{\mu g\dot{x}^{\prime}}{\sqrt{% \dot{x}^{\prime 2}+\dot{y}^{\prime 2}}}+2\dot{\theta}\,\dot{y}^{\prime}$
	$\displaystyle\ddot{y}^{\prime}$	$\displaystyle=$	$\displaystyle\dot{\theta}^{2}y^{\prime}-\frac{\mu g\dot{y}^{\prime}}{\sqrt{% \dot{x}^{\prime 2}+\dot{y}^{\prime 2}}}-2\dot{\theta}\,\dot{x}^{\prime}$

These are the required equations of motion for the puck in the rotating frame.

Answer: $\ddot{x}^{\prime}=\dot{\theta}^{2}x^{\prime}-\frac{\mu g\dot{x}^{\prime}}{% \sqrt{\dot{x}^{\prime 2}+\dot{y}^{\prime 2}}}+2\dot{\theta}\,\dot{y}^{\prime},% \quad\ddot{y}^{\prime}=\dot{\theta}^{2}y^{\prime}-\frac{\mu g\dot{y}^{\prime}}% {\sqrt{\dot{x}^{\prime 2}+\dot{y}^{\prime 2}}}-2\dot{\theta}\,\dot{x}^{\prime}$

Once we find a solution $x^{\prime}(t)$ and $y^{\prime}(t)$ of these equations, we can find the solution in the fixed frame by transforming ( $x^{\prime},y^{\prime}$ ) to ( $x, y$ ) through

\left\{\begin{array}[]{l}x(t)\\ y(t)\end{array}\right\}=\left[\begin{array}[]{rr}\cos(\dot{\theta}t)&-\sin(% \dot{\theta}t)\\ \sin(\dot{\theta}t)&\cos(\dot{\theta}t)\end{array}\right]\left\{\begin{array}[% ]{l}x^{\prime}(t)\\ y^{\prime}(t)\end{array}\right\}

Also, note that when the solution of the equations of motion in the rotating reference frame brings the puck to halt, the puck stops with respect to the rotating turntable. To an observer in the fixed frame, the puck will be going in circles with a constant $\dot{\theta}$ .

SAMPLE 19.8

Filename:sfig8-6-2a

Figure 19.18: A collar slides on a frictional bar and finally shoots off the end. The bar rotates with constant angular speed.

A collar sliding on a frictional rod. A collar of mass $m=0.5\,{\mathrm{lbm}}$ slides on a massless rigid rod OA of length $\ell=8\,\mathrm{ft}$ . The rod rotates counterclockwise with a constant angular speed $\dot{\theta}=5\,\mathrm{rad}/\mathrm{s}$ . The coefficient of friction between the rod and the collar is $\mu=0.3$ . At time $t=0\,\mathrm{s}$ , the bar is horizontal, the collar is at $R_{0}=1\,\mathrm{ft}$ and has radial speed $\dot{R}_{0}=\mu\dot{\theta}R_{0}$ towards the pivot O. Ignore gravity.

1.

How does the position of the collar change with time (i.e., what is the equation of motion of the collar)?
2.

Plot the path of the collar starting from $t=0\,\mathrm{s}$ till the collar shoots off the end of the bar.
3.

How long does it take for the collar to leave the bar?

Solution

Filename:efig1-2-28

Figure 19.19: Free-Body Diagram of the collar. The only forces on the collar are from the bar: the normal force

N

and the friction force

F_{s}=\mu N

A Free-Body Diagram of the collar at a general position $(R,\,\theta)$ is shown in Fig. 19.19. The geometry of the position vector and basis vectors is shown in Fig. 19.20. In vector notation, the forces on the collar are ${\accentset{\rightharpoonup}{\boldsymbol{N}}}=N\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}$ acting normal to the rod and the force of friction ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{s}=-\mu N\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}$ acting along the rod. Thus linear momentum balance for the collar is:

Filename:pfigure-blue-118-2

Figure 19.20: Geometry of the collar position at an arbitrary time during its slide on the rod.

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
	$\displaystyle-\mu N\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+N\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle m[(\ddot{R}-R\dot{\theta}^{2})\;\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\underbrace{\ddot{\theta}}_{0})% \;\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$		(19.11)

Note that $\ddot{\theta}=0$ because the rod is rotating at a constant rate. Dotting both sides of eqn. (19.11) with $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ we get

$\displaystyle[\mbox{Eqn.~{}(\ref{eq6.5.1.1})}]\cdot\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-\mu N$	$\displaystyle=$	$\displaystyle m(\ddot{R}-R\dot{\theta}^{2})$
$\displaystyle\hbox to36.135pt{\hfil or\hfil}\ddot{R}-R\dot{\theta}^{2}$	$\displaystyle=$	$\displaystyle-\frac{\mu N}{m}$
$\displaystyle[\mbox{Eqn.~{}(\ref{eq6.5.1.1})}]\cdot\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\qquad N$	$\displaystyle=$	$\displaystyle 2m\dot{R}\dot{\theta}.$

Eliminating $N$ from the last two equations we get

\ddot{R}+2\mu\dot{\theta}\;\dot{R}-\dot{\theta}^{2}\;R=0.

Since $\dot{\theta}=\omega$ is constant, the above equation is of the form

\ddot{R}+C\;\dot{R}-\omega^{2}\;R=0

(19.12)

where $C=2\mu\omega$ and $\omega=\dot{\theta}$ .

Note that we could derive eqn. (19.12) from eqn. (19.11) in a single step by taking a dot product of eqn. (19.11) with $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\mu\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}$ . Why? Look at the left hand side of eqn. (19.11). The net reaction force is $N(-\mu\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta})$ acting in the direction $(-\mu\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta})$ . If we dot the reaction force with a vector normal to its direction, we get rid of the reaction force. A vector normal to $(-\mu\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta})$ is $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\mu\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}$ .

	$\displaystyle[\mbox{Eqn.~{}(\ref{eq6.5.1.1})}]\cdot(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+\mu\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta})% \hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle m[(\ddot{R}-R\dot{\theta}^{2})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+2\dot{R}\dot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}]\cdot(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}% +\mu\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta})$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle\ddot{R}+2\mu\dot{\theta}\;\dot{R}-\dot{\theta}^{2}\;R$

which is the same equation as eqn. (19.12).

Solution of equation (19.12): The characteristic equation^†^†margin: If we assume a solution of the form $R(t)=e^{\lambda t}$ with some unknown $\lambda$ and plug back into the given differential equation, we get an algebraic equation in $\lambda$ which is called the characteristic equation. For more information, see box 11.1 on page 11.1, eqn. (11.9). associated with Eqn. (19.12) is

$\displaystyle\lambda^{2}+C\lambda-\omega^{2}$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\lambda$	$\displaystyle=$	$\displaystyle\frac{-C\pm\sqrt{C^{2}+4\omega^{2}}}{2}$
	$\displaystyle=$	$\displaystyle\omega(-\mu\pm\sqrt{\mu^{2}+1}).$

Therefore, the solution of Eqn. (19.12) is

	$\displaystyle R(t)$	$\displaystyle=$	$\displaystyle Ae^{\lambda_{1}t}+Be^{\lambda_{2}t}$
		$\displaystyle=$	$\displaystyle Ae^{(-\mu+\sqrt{\mu^{2}+1})\omega t}+Be^{(-\mu-\sqrt{\mu^{2}+1})% \omega t}.$

Substituting the given initial conditions: $R(0)=R^{0}=1\,\mathrm{ft}$ and $\dot{R}(0)=-\mu\dot{\theta}R^{0}$ , we get^†^†margin: From the general solution and the given initial conditions, we have: $\displaystyle R_{0}$ $\displaystyle=$ $\displaystyle A+B$ $\displaystyle-\mu\omega R_{0}$ $\displaystyle=$ $\displaystyle A\lambda_{1}+B\lambda_{2}$ Solving these two equations simultaneously, we get $A=\frac{\lambda_{2}+\mu\omega}{\lambda_{2}-\lambda_{1}}R_{0}$ , and $B=-\frac{\lambda_{1}+\mu\omega}{\lambda_{2}-\lambda_{1}}R_{0}$ . Substituting the values of $\lambda_{1}$ and $\lambda_{2}$ , we get $A=B=\frac{R_{0}}{2}=\frac{1\,\mathrm{ft}}{2}.$

R(t)=\frac{1\,\mathrm{ft}}{2}\left[e^{(-\mu+\sqrt{\mu^{2}+1})\omega t}+e^{(-% \mu-\sqrt{\mu^{2}+1})\omega t}\right].

(19.13)

Answer: $R(t)=\frac{1\,\mathrm{ft}}{2}\left[e^{(-\mu+\sqrt{\mu^{2}+1})\omega t}+e^{(-% \mu-\sqrt{\mu^{2}+1})\omega t}\right].$

2.

To draw the path of the collar we need both $R$ and $\theta$ . Because $\dot{\theta}=5\,\mathrm{rad}/\mathrm{s}=$ constant,

$\theta=\dot{\theta}\,t=(5\,\mathrm{rad}/\mathrm{s})\,t.$

Filename:pfigure-s95f3a

Figure 19.21: Finding the final time $t_{f}$ from the graph of $R(t)$ such that $R(t_{f})=R_{f}=8\,\mathrm{ft}$ .

Now we can take various values of $t$ from $0\,\mathrm{s}$ to, say, $1\,\mathrm{s}$ , and calculate values of $\theta$ and $R$ . Plotting all these values of $R$ and $\theta$ , however, does not give us the correct path after the collar reaches the end of the bar, at $R=\mbox{length of the bar}=8\,\mathrm{ft}$ . So we need to find the time $t_{f}$ when $R(t_{f})=8\,\mathrm{ft}$ . Equation (19.13) is a nonlinear algebraic equation for $t$ which we can solve iteratively on a computer, or with some patience, even on a calculator. One way to find $t_{f}$ would be to simply plot $R(t)$ and find the intersection with $R=8$ (see fig. 19.21) and read the corresponding value of $t$ . Following either of these methods we find that $t_{f}\approx 0.74518\,\mathrm{s}$ here. Now, we can plot the path of the collar by computing $R$ and $\theta$ from $t=0$ to $t=t_{f}$ and making a polar plot on a computer as follows (pseudocode).
```
tf = 0.74518                % final value of t
t = 0:tf/100:tf;            % take 101 points in [0 tf]
R0 = 1; w = 5; mu = .3;     % initialize variables
f1 = -mu + sqrt(mu^2 +1);   % first partial exponent
f2 = -mu - sqrt(mu^2 +1);   % second partial exponent
R = 0.5*R0*(exp(f1*w*t) + exp(f2*w*t));  % calculate R
theta = w*t;                % calculate theta
polarplot(theta,r)
```
Filename:Danef94s1q2

Figure 19.22: Plot of the path of the collar until it leaves the rod.

The plot produced thus is shown in Fig. 19.22.
3.

The time $t_{f}$ computed above was

$t_{f}=0.7452\,\mathrm{s}.$

By plugging this value in the expression for $R(t)$ (Eqn. (19.13) we get, indeed,

$R=8\,\mathrm{ft}.$

Answer: $t_{f}=0.7452\,\mathrm{s}$

Note, although we have not checked it explicitly, the increase in kinetic energy of the particle comes from the work of the force from the rod on the particle:

\Delta{E_{\mathrm{K}}}=\int{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{v}}}\,dt

SAMPLE 19.9

Filename:pfigure-blue-123-1

Figure 19.23: A collar of mass

m

slides on a massless and frictionless bar, bent at an angle

\phi

with the vertical axis. The bar is rotating at a constant angular speed

\omega

about the

z

-axis.

A collar sliding on a rotating rod in 3-D. A massless and frictionless rod AB is rotating about the vertical axis through point A. The rod is bent at an angle $\phi$ from the vertical and is rotating with a constant angular speed $\omega$ about the vertical axis. A small collar of mass $m$ slides on the rod. Assume that at time $t=0$ the collar is released from a rest position with respect to the rod at a distance $R_{0}$ from the axis of rotation. There is no gravity.

1.

Find the equation of motion for the collar (a differential equation for the position of the collar).
2.

How does the distance of the collar from the vertical axis change with time?
3.

For $\phi=\pi/2$ , show that the solution obtained in (ii) above is the same as that obtained in Sample 19.15 for $\mu=0$ .

Solution Since the bar is bent and it rotates about the vertical axis, it sweeps a conical surface about the axis of rotation. As the collar slides on the rod, it traces a path on this surface. Let $(R,\,\theta,\,z)$ be the cylindrical coordinates of the collar at any general time $t$ (Fig. 19.24(a)). Since the rod is frictionless and there is no gravity, the only force acting on the collar is the normal reaction from the rod. This force is shown in the free-body diagram of the collar in Fig. 19.24(b). Note that there are a lot of possible directions for a normal to the rod at the collar. In fact, any vector in the plane perpendicular to the rod at the collar is normal to the rod. So, at this point let us write

{\accentset{\rightharpoonup}{\boldsymbol{N}}}=N\hat{\boldsymbol{n}}

where $\hat{\boldsymbol{n}}$ is a unit normal to the rod. Thus, if $\hat{\boldsymbol{\lambda}}$ is a unit vector along the rod, then

\hat{\boldsymbol{\lambda}}\cdot\hat{\boldsymbol{n}}=0

(19.14)

where

\hat{\boldsymbol{\lambda}}=\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{\rm AB}}{|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AB}|}=\sin\phi% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\cos\phi\hat{\boldsymbol{k}}.

Filename:pfigure-blue-119-2

Figure 19.24: (a) Cylindrical coordinates

R

and

z

of the collar (

\theta

is not shown) and the orientation of the cylindrical basis vectors. (b) Free-body diagram of the collar (there is no gravity). (c) Instantaneous

R

t h e t a

plane of the collar.

Equation of Motion: We now write the linear momentum balance for the collar:

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}

where

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle N\hat{\boldsymbol{n}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle(\ddot{R}-R\dot{\theta}^{2})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\underbrace{\ddot{\theta}}_{0}\;% \;)\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+\ddot{z}\hat{\boldsymbol{% k}}$

Dotting both sides of the equation $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ by $\hat{\boldsymbol{\lambda}}$ we get

	$\displaystyle N\underbrace{\hat{\boldsymbol{n}}\cdot\hat{\boldsymbol{\lambda}}% }_{0}$	$\displaystyle=$	$\displaystyle m[(\ddot{R}-R\dot{\theta}^{2})\underbrace{\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}\cdot\hat{\boldsymbol{\lambda}}}_{\sin\phi}+2\dot{R}\dot% {\theta}\underbrace{\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}\cdot\hat% {\boldsymbol{\lambda}}}_{0}+\ddot{z}\underbrace{\hat{\boldsymbol{k}}\cdot\hat{% \boldsymbol{\lambda}}}_{\cos\phi}]$
	$\Rightarrow$		$\displaystyle(\ddot{R}-R\dot{\theta}^{2})\sin\phi+\ddot{z}\cos\phi=0.$

This expression is an equation of motion of the collar. However, it is only a single equation in terms of derivatives of two variables $R$ and $z$ . Now, from geometry

R=z\tan\phi\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\ddot{z}=\ddot{R% }/\tan\phi.

Substituting this relationship in the equation of motion we get

\ddot{R}(\sin\phi+\frac{\cos\phi}{\tan\phi})-R\dot{\theta}^{2}=0.

Noting that $\dot{\theta}=\omega=$ a constant, we may write the above equation, with some trigonometric simplifications, as

\ddot{R}+(\omega\,\sin\phi)^{2}\,R=0

(19.15)

which is an equation of motion of the collar in terms of its distance $R$ from the vertical axis.

2.

Solution of the equation of motion: The solution of Eqn. (19.15) is given by ^†^†margin: You may find the solution by solving the corresponding characteristic equation (see the solution of Eqn. (19.15) in Sample 19.15, for example), or by looking up the table of “The Simplest ODEs and Their Solutions” on page 22 of the text, or by guessing a solution yourself if you have some experience with ODEs.

$R(t)=C_{1}e^{(\omega\,\sin\phi)t}+C_{2}e^{-(\omega\,\sin\phi)t}$

where $C_{1}$ and $C_{2}$ are arbitrary constants to be determined from the initial conditions. Substituting the given initial conditions: $R(0)=R_{0}$ and $\dot{R}(0)=0$ we get

$C_{1}=C_{2}=\frac{R_{0}}{2}.$

Therefore, the solution may be written as

$R(t)=\frac{R_{0}}{2}\left[e^{(\omega\,\sin\phi)t}+e^{-(\omega\,\sin\phi)t}% \right].$ (19.16)
3.

Special case, $\phi=\pi/2$ : Substituting $\phi=\pi/2$ in Eqn. (19.16) we get

$R(t)=\frac{R_{0}}{2}\left[e^{\omega t}+e^{-\omega t}\right]$

which is the same solution as obtained in Eqn. (19.16) in Sample 19.15 for $\mu=0$ and $R_{0}=1\,\mathrm{ft}$ .

SAMPLE 19.10

Filename:pfigure-s95q14

Figure 19.25: A motor driven bead slides down a helical wire-frame described by the equations

R=R_{0}\theta,\;z=2R_{0}\theta

Osculating circle in 3-D. A small bead is driven down a a wire-frame bent in the shape of a conical helix by a tiny motor imbedded in the bead. The combined mass of the bead and the motor is $m=200$ gm. The shape of the helix is given: $R=R_{0}\theta,\ z=2R_{0}\theta$ and where $R_{0}=0.3\,\mathrm{m}$ . At the instant when $\theta=2\,{\rm radians}$ , the angular speed and the angular acceleration of the bead are $\dot{\theta}=1\,\mathrm{rad}/\mathrm{s}$ and $\ddot{\theta}=2\,\mathrm{rad}/\mathrm{s}^{2}$ . Find

1.

the net normal force on the bead and
2.

the radius of the osculating circle at the instant given.

^†^†margin:

Solution

We can find the net normal force on the bead from the linear momentum balance of the bead (see the free-body diagram of the bead):

Filename:Danef94s3q2

Figure 19.26: Free-body diagram of the bead. Note that the net normal force is

F_{n}=\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot\hat{\boldsymbol{e% }}_{n}

where

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=N\hat{\boldsymbol{e}}_{n}+T% \hat{\boldsymbol{e}}_{t}-mg\hat{\boldsymbol{k}}

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
$\displaystyle N\hat{\boldsymbol{e}}_{n}+T\hat{\boldsymbol{e}}_{t}-mg\hat{% \boldsymbol{k}}$	$\displaystyle=$	$\displaystyle m(a_{t}\hat{\boldsymbol{e}}_{t}+a_{n}\hat{\boldsymbol{e}}_{n})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F_{n}$	$\displaystyle\equiv$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot\hat{% \boldsymbol{e}}_{n}=ma_{n}.$

Thus we need to find the normal acceleration of the bead. We can write the position of the bead as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=$	$\displaystyle\underbrace{R}_{R_{0}\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+\underbrace{z}_{2R_{0}\theta}\hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle\equiv$	$\displaystyle\dot{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R\dot{\theta}% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+\dot{z}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle R_{0}\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R_% {0}\theta\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+2R_{0}% \dot{\theta}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle R_{0}\dot{\theta}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+% \theta\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+2\hat{\boldsymbol{k}}),$
$\displaystyle\hbox to36.135pt{and\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$	$\displaystyle\equiv$	$\displaystyle(\ddot{R}-R\dot{\theta}^{2})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!\theta}+\ddot{z}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(R_{0}\ddot{\theta}-R_{0}\theta\dot{\theta}^{2})\hat{\boldsymbol{% e}}_{\scriptscriptstyle\!R}+(2R_{0}\dot{\theta}^{2}+R_{0}\theta\ddot{\theta})% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+2R_{0}\ddot{\theta}\hat{% \boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle R_{0}[(\ddot{\theta}-\theta\dot{\theta}^{2})\hat{\boldsymbol{e}}% _{\scriptscriptstyle\!R}+(\dot{\theta}^{2}+\theta\ddot{\theta})\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}+2\ddot{\theta}\hat{\boldsymbol{k}}]$

Substituting the given numerical values for $\theta,\dot{\theta},R_{0}$ and $\ddot{\theta}$ in the above expressions for ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ we get the velocity and the acceleration of the bead at the instant of interest:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle 0.3\,\mathrm{m}/\mathrm{s}(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+2\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+2% \hat{\boldsymbol{k}})$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle 1.2\,\mathrm{m}/\mathrm{s}^{2}(2\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}+\hat{\boldsymbol{k}}).$

In path coordinates,

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

where

$\displaystyle\hat{\boldsymbol{e}}_{t}$	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{v}}}}{\|{\accentset% {\rightharpoonup}{\boldsymbol{v}}}\|}$	(19.17)
	$\displaystyle=$	$\displaystyle\frac{0.3\,\mathrm{m}/\mathrm{s}(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+2\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+2% \hat{\boldsymbol{k}})}{0.3\sqrt{1+4+4}\;\,\mathrm{m}/\mathrm{s}}$	(19.18)
	$\displaystyle=$	$\displaystyle\frac{1}{3}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+2\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}+2\hat{\boldsymbol{k}}).$	(19.19)

Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}$	$\displaystyle=$	$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot\hat{% \boldsymbol{e}}_{t})\hat{\boldsymbol{e}}_{t}$
	$\displaystyle=$	$\displaystyle(\frac{4.8+2.4}{3}\,\mathrm{m}/\mathrm{s}^{2})\hat{\boldsymbol{e}% }_{t}$
	$\displaystyle=$	$\displaystyle 0.8\,\mathrm{m}/\mathrm{s}^{2}(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+2\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+2% \hat{\boldsymbol{k}})$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}-{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{t}$
	$\displaystyle=$	$\displaystyle[2.4\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+1.2\hat{% \boldsymbol{k}}-0.8\,\mathrm{m}/\mathrm{s}^{2}(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+2\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}+2% \hat{\boldsymbol{k}})]$
	$\displaystyle=$	$\displaystyle(-0.8\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+0.8\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}-0.4\hat{\boldsymbol{k}})\,\mathrm% {m}/\mathrm{s}^{2}$
$\displaystyle a_{n}$	$\displaystyle=$	$\displaystyle\|{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}\|=1.2\,\mathrm{% m}/\mathrm{s}^{2}$
$\displaystyle\hat{\boldsymbol{e}}_{n}$	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}}{a_{n}}$
	$\displaystyle=$	$\displaystyle-\frac{2}{3}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\frac{2}% {3}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}-\frac{1}{3}\hat{% \boldsymbol{k}}.$

Hence, the net normal force on the bead

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{n}$	$\displaystyle=$	$\displaystyle ma_{n}\hat{\boldsymbol{e}}_{n}$
	$\displaystyle=$	$\displaystyle 0.2\,\mathrm{kg}\cdot 1.2\,\mathrm{m}/\mathrm{s}^{2}\cdot\frac{1% }{3}(-2\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+2\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}-\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle(-0.16\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+0.16\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}-0.08\hat{\boldsymbol{k}})\,{\mbox% {N}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{n}=0.08\,{\mbox{N}}(-2\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}+2\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}+\hat{\boldsymbol{k}})$

2.

For calculating the radius $\rho$ of the osculating circle, we note that

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

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\rho$ $\displaystyle=$ $\displaystyle\frac{v^{2}}{a_{n}}=\frac{(0.9\,\mathrm{m}/\mathrm{s})^{2}}{1.2\,% \mathrm{m}/\mathrm{s}^{2}}=0.675\,\mathrm{m}.$

Answer: $\rho=0.675\,\mathrm{m}$

Problems for 19.1 Mechanics of a constrained particle

19.1.1

A bead slides on a frictionless circular hoop . The mass of the bead $m_{\rm bead}=2\,{\mathrm{grams}}$ ,

mass of hoop $m_{\rm hoop}=1\,\mathrm{kg}$ and

radius of hoop $R_{\rm hoop}=3\,\mathrm{m}$ .

Neglect gravity. The center of the circular hoop is the origin $O$ of a fixed (Newtonian) coordinate system $O x y z$ . The hoop is on the $x y$ plane. The hoop is kept from moving by little angels who let the bead slide by unimpeded. At $t=0$ , the speed of the bead is $4\,\mathrm{m}/\mathrm{s}$ , it is traveling counter-clockwise (looking down the $z$ -axis), and it is on the $+x$ -axis. There are no other external forces applied.

(a)

At $t=0$ what is the bead’s kinetic energy ?
(b)

At $t=0$ what is the bead’s linear momentum ?
(c)

At $t=0$ what is the bead’s angular momentum about the origin?
(d)

At $t=0$ what is the bead’s acceleration?
(e)

At $t=0$ what is the radius of the osculating circle of the bead’s path $\rho$ ?
(f)

At $t=0$ what is the force of the hoop on the bead?
(g)

At $t=0$ what is the net force of the angel’s hands on the hoop?
(h)

At $t=27s$ what is the $x$ -component of the bead’s linear momentum?

19.1.2 A warehouse operator wants to move a crate of weight $W=100\,{\rm lb}$ from a $6\,\mathrm{ft}$ high platform to ground level by means of a roller conveyer as shown. The rollers on the inclined plane are well lubricated and thus assumed frictionless; the rollers on the horizontal conveyer are frictional, thus providing an effective friction coefficient $\mu$ . Assume the rollers are massless.

(a)

What are the kinetic and potential energies(pick a suitable datum) of the crate at the elevated platform, point A?
(b)

What are the kinetic and potential energies of the crate at the end of the inclined plane just before it moves onto the horizontal conveyor, point B?
(c)

Using conservation of energy, calculate the maximum speed attained by the crate at point B, assuming it starts moving from rest at point A.
(d)

If the crate slides a distance $x$ , say, on the horizontal conveyor, what is the energy lost to sliding in terms of $\mu$ , $W$ , and $x$ ?
(e)

Calculate the value of the coefficient of friction, $\mu$ , such that the crate comes to rest at point C.

Filename:bikefork1-ang-accel

Figure 19.27

19.1.3 On a wintry evening, a student of mass $m$ starts down the steepest street in town, Steep Street, which is of height $h$ and slope $\theta$ . At the top of the slope she starts to slide (coefficient of dynamic friction $\mu$ ).

(a)

What is her initial kinetic and potential energy?
(b)

What is the energy lost to friction as she slides down the hill?
(c)

What is her velocity on reaching the bottom of the hill? Ignore air resistance and all cross streets (i.e. assume the hill is of constant slope).
(d)

If, upon reaching the bottom of the street, she collides with another student of mass $M$ and they embrace, what is their instantaneous mutual velocity just after the embrace?
(e)

Assuming that friction still acts on the level flats on the bottom, how much time will it take before they come to rest?

Filename:bikefork-ang-accel

Figure 19.28

19.1.4 Reconsider the system of blocks in problem 2.3, this time with equal mass, $m_{1}=m_{2}=m$ . Also, now, both blocks are frictional and sitting on a frictional surface. Assume that both blocks are sliding to the right with the top block moving faster. The coefficient of friction between the two blocks is $\mu_{1}$ . The coefficient of friction between the lower block and the floor is $\mu_{2}$ . It is known that $\mu_{1}>2\mu_{2}$ .

(a)

Draw free-body diagrams of the blocks together and separately.
(b)

Write the equations of linear momentum balance for each block.
(c)

Find the acceleration of each block for the case of frictionless blocks. For the frictional blocks, find the acceleration of each block. What happens if $\mu_{1}>>\mu_{2}$ ?

19.1.5 An initially motionless roller-coaster car of mass 20 $\,\mathrm{kg}$ is given a horizontal impulse $\int{\accentset{\rightharpoonup}{\boldsymbol{F}}}\,dt=P\hat{\boldsymbol{\imath}}$ at position A, causing it to move along the track, as shown below.

(a)

Assuming that the track is perfectly frictionless from point A to C and that the car never leaves the track, determine the magnitude of the impulse $I$ so that the car “just makes it” over the hill at B.
(b)

In the ensuing motion, assume that the horizontal track C-D is frictional and determine the coefficient of friction required to bring the car to a stop at D.

Filename:bikefork1-alt

Figure 19.29

19.1.6 Masses $m_{1}=1\,\mathrm{kg}$ and $m_{2}\,\mathrm{kg}$ move on the frictionless varied terrain shown. Initially, $m_{1}$ has speed $v_{1}=12\,\mathrm{m}/\mathrm{s}$ and $m_{2}$ is at rest. The two masses collide on the flat section. The coefficient of restitution in the collision is $e=0.5$ . Find the speed of $m_{2}$ at the top of the first hill of elevation $h_{1}=1\,\mathrm{m}$ . Does $m_{2}$ make it over the second hill of elevation $h_{2}=4\,\mathrm{m}$ ?

Filename:bikefork-alt

Figure 19.30

19.1.7 A Scotch yoke is a device for converting rotary motion into linear motion. In this case it is used to make a horizontal platform go up and down. The motion of the platform is $y=A\sin(\omega t)$ with constant $\omega$ . A dead bug with mass m is standing on the platform. Her feet have no glue on them. There is gravity.

(a)

Draw a free-body diagram of the bug.
(b)

What is the acceleration of the platform and, hence, the bug?
(c)

Write the equation of linear momentum balance for the bug.
(d)

What condition must be met so that the bug does not bounce the platform?
(e)

What is the maximum spin rate $\omega$ that can be used if the bug is not to bounce off the platform?

Filename:pfigure-blue-128-1

Figure 19.31

19.1.8 A bug of mass $m$ walks straight-forwards with speed $v_{A}$ and rate of change of speed ${\dot{v}}_{A}$ on a straight light (assumed to be massless) stick. The stick is hinged at the origin so that it is always horizontal but is free to rotate about the $z$ axis. Assume that the distance the bug is from the origin, $\ell$ , the angle the stick makes with the $x$ axis, $\phi$ , and its rate of change, ${\dot{\phi}}$ , are known at the instant of interest. Ignore gravity. Answer the following questions in terms of $\phi$ , ${\dot{\phi}}$ , $\ell$ , $m$ , $v_{A}$ , and ${\dot{v}}_{A}$ .

(a)

What is ${\ddot{\phi}}$ ?
(b)

What is the force exerted by the rod on the support?
(c)

What is the acceleration of the bug?

Filename:tfigure8-syst-bods

Figure 19.32

19.1.9 A new kind of gun. Assume $\omega=\omega_{0}$ is a constant for the rod in the figure. Assume the mass is free to slide. At $t=0$ , the rod is aligned with the $x$ -axis and the bead is one foot from the origin and has no radial velocity $(dR/dt=0)$ .

(a)

Find a differential equation for $R(t)$ . Answer: $\ddot{R}(t)-\omega_{o}^{2}R(t)=0$ .
(b)

Turn this equation into a differential equation for $R(\theta)$ . Answer: $\frac{d^{2}R(\theta)}{d\theta^{2}}-R(\theta)=0$ .
(c)

How far will the bead have moved after one revolution of the rod? How far after two? Answer: $R(\theta=2\pi)=267.7\,\mathrm{ft}$ , $R(\theta=4\pi)=1.43\times 10^{5}\,\mathrm{ft}$ .
(d)

What is the speed of the bead after one revolution of the rod (use $\omega_{0}=2\pi\,\mathrm{rad}/\mathrm{s}=1\,{{\rm rev}/{\rm sec}}$ )? Answer: $v=1682.3\,{\mathrm{ft/s}}$
(e)

How much kinetic energy does the bead have after one revolution and where did it come from? Answer: ${E_{\mathrm{K}}}=(2.83\times 10^{6})\times(mass)(\,{\mathrm{ft/s}})^{2}$ .

Filename:sfig8-7-2

Figure 19.33

19.1.10 The new gun gets old and rusty.

Reconsider the bead on a rod in problem 19.1. This time, friction cannot be neglected. The friction coefficient is $\mu$ . At the instant of interest the bead with mass $m$ has radius $R_{0}$ with rate of change $\dot{R}_{0}$ . The angle $\theta$ is zero and $\omega$ is a constant. Neglect gravity.

(a)

What is $\ddot{R}$ at this instant? Give your answer in terms of any or all of $R_{0}$ , $\dot{R}_{0}$ , $\omega$ , $m$ , $\mu$ , $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ , $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ , $\hat{\boldsymbol{\imath}}$ , and $\hat{\boldsymbol{\jmath}}$ . Answer: $\ddot{R}=R_{0}\omega^{2}-2\mu\dot{R}_{0}\omega$ .
(b)

After a very long time it is observed that the angle $\phi$ between the path of the bead and the rod/trough is nearly constant. What, in terms of $\mu$ , is this $\phi$ ? Answer: $\phi=\sin^{-1}\left[\frac{1}{\sqrt{(\sqrt{1+\mu^{2}}-\mu)^{2}+1}}\right]$ .

19.1.11 A newer kind of gun. As an attempt to make an improvement on the ‘new gun’ demonstrated in problem 19.1, a person adds a length $\ell$ to the shaft on which the bead slides. Assume there is no friction between the bead (mass m) and the wire. Assume the bead starts at $s=0$ with $\dot{s}=v_{0}$ . The rigid rod, on which the bead slides, rotates at a constant rate $\omega=\omega_{0}$ . Find $s(t)$ in terms of $\ell,m,v_{0}$ , and $\omega_{0}$ .

Filename:sfig8-7-2a

Figure 19.34

19.1.12 Slippery bead on straight rotating stick. A long stick with mass $m_{s}$ rotates at a constant rate and a bead, modeled as a point mass, slides on the stick. The stick rotates at constant rate ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ . Neglect gravity. The bead has mass $m_{b}$ . Initially the mass is at a distance $R_{0}$ from the hinge point on the stick and has no radial velocity ( $\dot{R}=0$ ). The initial angle of the stick is $\theta=0$ , measured counterclockwise from the positive $x$ axis.

(a)

What is the torque, as a function of the net angle the stick has rotated, $\theta$ , required in order to keep the stick rotating at a constant rate ?
(b)

What is the path of the bead in the $x y$ -plane? (Draw an accurate picture showing about one half of one revolution.)
(c)

How long should the stick be if the bead is to fly in the negative $x$ -direction when it gets to the end of the stick?
(d)

Add friction. How does the speed of the bead over one revolution depend on $\mu$ , the coefficient of friction between the bead and the wire? Make a plot of $|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\hbox{one revolution}}|$ versus $\mu$ . $[$ Note, you have been working with $\mu=0$ in the problems above. $]$

19.1.13 Mass on a lightly greased slotted turntable or spinning uniform rod. Assume that the rod/turntable in the figure is massless and also free to rotate. Assume that at $t=0$ , the angular velocity of the rod/turntable is $1\,\mathrm{rad}/\mathrm{s}$ , that the radius of the bead is one meter, and that the radial velocity of the bead, $dR/dt$ , is zero. The bead is free to slide on the rod. Where is the bead at $t=5\sec$ ?

Filename:sfig8-7-2again

Figure 19.35

19.1.14 A bead slides in a frictionless slot in a turntable. The turntable spins a constant rate $\omega$ . The slot is straight and goes through the center of the turntable. If the bead is at radius $R_{o}$ with $\dot{R}=0$ at $t=0$ , what are the components of the acceleration vector in the directions normal and tangential to the path of the bead after one revolution? Neglect gravity.

19.1.15 A bead of mass $m=1\,$ gm is constrained to slide in a straight frictionless slot in a disk which is spinning counterclockwise at constant rate $\omega=3\,\mathrm{rad}/\mathrm{s}$ . At time $t=0$ , the slot is parallel to the $x$ -axis and the bead is in the center of the disk moving out (in the plus $x$ direction) at a rate $\dot{R}_{o}=.5\,\mathrm{m}/\mathrm{s}$ . After a net rotation $\theta$ of one and one eighth ( $1.125$ ) revolutions, what is the force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ of the disk on the bead? Express this answer in terms of $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ . Make the unreasonable assumption that the slot is long enough to contain the bead for this motion.

Filename:sfig8-7-2disks

Figure 19.36

19.1.16 Bead on springy leash in a slot on a turntable. The bead in the figure is held by a spring that is relaxed when the bead is at the origin. The constant of the spring is $k$ . The turntable speed is controlled by a strong stiff motor.

(a)

Assume $\omega=0$ for all time. What are possible motions of the bead?
(b)

Assume $\omega=\omega_{0}$ is a constant. What are possible motions of the bead? Notice there are two cases depending on the value of $\omega_{0}$ . What is going on here?

Filename:sfig8-4-4

Figure 19.37

Mass in a slot.

19.1.17 A small bead with mass $m$ slides without friction on a rigid rod which rotates about the $z$ axis with constant $\omega$ (maintained by a stiff motor not shown in the figure). The bead is also attached to a spring with constant $K$ , the other end of which is attached to the rod. The spring is relaxed when the bead is at the center position. Assume the bead is pulled to a distance $d$ from the center of the rod and then released with an initial $\dot{R}=0$ . If needed, you may assume that $K>m\omega^{2}$ .

(a)

Derive the equation of motion for the position of the bead $R(t)$ .
(b)

How is the motion affected by large versus small values of $K$ ?
(c)

What is the magnitude of the force of the rod on the bead as the bead passes through the center position? Neglect gravity. Answer in terms of $m$ , $\omega$ , $d$ , $K$ .
(d)

Write an expression for the Coriolis acceleration. Give an example of a situation in which this acceleration is important and explain why it arises.

Filename:sfig8-4-4a

Figure 19.38

19.1.18 A small ball of mass $m_{b}=500\ \,{\mathrm{grams}}$ may slide in a slender tube of length $l=1.2\,\mathrm{m}$ . and of mass $m_{t}=1.5\,\mathrm{kg}$ . The tube rotates freely about a vertical axis passing through its center C. (Hint: Treat the tube as a slender rod.)

(a)

If the angular velocity of the tube is $\omega=8\,\mathrm{rad}/\mathrm{s}$ as the ball passes through $C$ with speed relative to the tube, $v_{0}$ , calculate the angular velocity of the tube just before the ball leaves the tube. Answer: $\omega^{-}=4\,\mathrm{rad}/\mathrm{s}$ , where the minus sign ‘-’ means ‘just before leaving’.
(b)

Calculate the angular velocity of the tube just after the ball leaves the tube. Answer: $\omega^{+}=4\,\mathrm{rad}/\mathrm{s}$ , where the plus sign ‘+’ means ‘just after leaving’.
(c)

If the speed of the ball as it passes $C$ is $v=1.8\,\mathrm{m}/\mathrm{s}$ , determine the transverse and radial components of velocity of the ball as it leaves the tube. Answer: $v_{r}=3.841\,\mathrm{m}/\mathrm{s}$ , $v_{t}=2.4\,\mathrm{m}/\mathrm{s}$ .
(d)

After the ball leaves the tube, what constant torque must be applied to the tube (about its axis of rotation) to bring it to rest in $10\,{\rm s}$ ? Answer: torque= $0.072\,{\rm N}\mathord{\cdot}{\rm m}$

Filename:sfig8-4-4b

Figure 19.39

19.1.19 Toy train car on a turn-around. The $0.1\,\mathrm{kg}$ toy train’s speedometer reads a constant $1m/s$ when, heading west, it passes due north of point $O$ . The train is on a level straight track which is mounted on a spinning turntable whose center is ‘pinned’ to the ground. The turntable spins at the constant rate of $2\,\mathrm{rad}/\mathrm{s}$ . What is the force of the turntable on the train? (Don’t worry about the $z$ -component of the force.) Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=-0.6\,{\mbox{N}}\hat{\boldsymbol% {\jmath}}$ .

Filename:sfig8-4-4c

Figure 19.40

19.1.20 Due to forces not shown, the cart moves to the right with constant acceleration $a_{x}$ . The ball $B$ has mass $m_{B}$ . At time $t=0$ , the string $A B$ is cut. Find

(a)

the tension in string $B C$ before cutting, Answer: $T_{BC}=m\frac{\sqrt{34}}{13}(2a_{x}+g)$ .
(b)

the absolute acceleration of the mass at the instant of cutting, Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=\frac{1}{34}[(25a_{x}+15g)\hat{% \boldsymbol{\imath}}+(15a_{x}-25g)\hat{\boldsymbol{\jmath}}]$
(c)

the tension in string $B C$ at the instant of cutting. Answer: $T_{BC}=m\frac{1}{\sqrt{34}}(5a_{x}+3g)$ .

Filename:pfigure-s94h13p2

Figure 19.41

19.1.21 The forked arm mechanism pushes the bead of mass $1\,\mathrm{kg}$ along a frictionless hyperbolic spiral track given by $r=0.5\theta(\,\mathrm{m}/\,{\rm rad})$ . The arm rotates about its pivot point at $O$ with constant angular acceleration $\ddot{\theta}=1\,\mathrm{rad}/\mathrm{s}^{2}$ driven by a motor (not shown). The arm starts from rest at $\theta=0^{{}^{\circ}}$ .

(a)

Determine the radial and transverse components of the acceleration of the bead after $2\,{\rm s}$ have elapsed from the start of its motion.
(b)

Determine the magnitude of the net force on the mass at the same instant in time.

Filename:pfigure-f93f5

Figure 19.42

19.1.22 A rod is on the palm of your hand at point $A$ . Its length is $\ell$ . Its mass $m$ is assumed to be concentrated at its end at $C$ . Assume that you know $\theta$ and $\dot{\theta}$ at the instant of interest. Also assume that your hand is accelerating both vertically and horizontally with ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm hand}=a_{hx}\hat{% \boldsymbol{\imath}}+a_{hy}\hat{\boldsymbol{\jmath}}$ . Coordinates and directions are as marked in the figure.

(a)

Draw a Free-Body Diagram of the rod.
(b)

Assume that the hand is stationary. Solve for $\ddot{\theta}$ in terms of $g$ , $\ell$ , $m$ , $\theta$ , and $\dot{\theta}$ .
(c)

If the hand is not stationary but $\ddot{\theta}$ has been determined somehow, find the vertical force of the hand on the rod in terms of $\ddot{\theta}$ , $\dot{\theta}$ , $\theta$ , $g$ , $\ell$ , $m$ , $a_{hx}$ , and $a_{hy}$ .

Filename:pfigure-s94h13p3

Figure 19.43

Rod on a Moving Support(2-D).

19.1.23 Balancing a broom. Assume the hand is accelerating to the right with acceleration $\boldsymbol{a}=a\hat{\boldsymbol{\imath}}$ . What is the force of the hand on the broom in terms of $m,\ \ell,\ \theta,\ \dot{\theta},\ a,\ \hat{\boldsymbol{\imath}},\ \hat{% \boldsymbol{\jmath}}$ , and $g$ ? (You may not have any $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ or $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ in your answer.)

Filename:p-s96-p3-3

Figure 19.44

19.2 Mechanics of one-degree-of-freedom 2-D mechanisms

A one-degree-of-freedom mechanism is a collection of parts linked together so that they can move in only one way^†^†margin: No physical system can move in only one way. The idea that a machine has one degree of freedom is an idealization that takes the rigid-object description of the parts and the ideal nature of the connections literally. In fact, all parts deform at least a little, and no connections are so precise as to be exact geometric constraints. It may be reasonable to respect the standard rigid-object idealizations and consider a machine as one-degree-of-freedom for basic analysis, as we do in this section. When, for example, trying to figure out why a machine vibrates in an undesirable way it may also be reasonable to relax some of the assumptions we make here and consider more degrees of freedom. .

The word “freedom” has to be taken lightly here because in practice even the one “freedom” is often controlled or restricted. Frankly, most machine designers don’t trust the laws of mechanics to enforce motions that they want. Instead they choose kinematic restrictions that enforce the desired motions and then use a motor, a computer controlled actuator or big flywheel to keep that motion moving at a prescribed rate.

We consider here machines that can move in just one way, whether or not that one motion is free. So in this sense, “one”-degree-of-freedom machines include machines with no freedom at all, just so long as they move in only one way.

Some familiar examples of one-degree-of-freedom mechanisms are a 1-D spring and mass, a pendulum, a slider-crank, a grounded 4-bar linkage, and a gear train.

Most ideal constraints are workless constraints

A fruitful equation for studying one-degree-of-freedom mechanisms is power balance, or for conservative systems, energy balance. The reason these equations are so useful is because most ideal connections are workless. That is:

the net work of the interaction forces (and moments) of a pair of parts that are connected with the standard ideal connections is zero. This includes welds, frictionless hinges, frictionless sliding contact, rolling contact, or parts connected by a massless inextensible link.

Example: A frictionless hinge is a workless constraint

Filename:bikefork1-ang-mom

Figure 19.45: A frictionless hinge is a workless constraint. The net work of the interaction force on the two contacting bodies is zero.

Body $\cal A$ is connected to body $\cal B$ by a frictionless hinge at C (see fig. 19.45). The force on body $\cal B$ at C from $\cal A$ is ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}$ and the force on $\cal A$ from body $\cal B$ at C is $-{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}$ . The power of the interaction force on body $\cal B$ is $P_{\cal A{\rm on}\cal B}={\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}% \cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$ . This power contributes to the increase in the kinetic energy of $\cal B$ . The power of the interaction force on $\cal A$ is $P_{\cal B{\rm on}\cal A}=-{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}% \cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}=-P_{\cal A{\rm on}\cal B}$ . So the contribution to the increase in the kinetic energy of body $\cal A$ is minus the contribution to body $\cal B$ and the net power on the system of two bodies is zero. Writing this out,

$\displaystyle\overbrace{P_{\mbox{total}}}^{\text{The net power of the pair of % interaction forces on the pair of bodies}}$	$\displaystyle=$	$\displaystyle P_{\cal B{\rm on}\cal A}+P_{\cal A{\rm on}\cal B}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}\cdot{\accentset% {\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}+(-{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{C})\cdot{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}$
	$\displaystyle=$	$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}-{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{C})\cdot{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}$
	$\displaystyle=$	$\displaystyle 0.$

Basically the same situation holds for all the standard ideal connections as explained in the box on page 19.2.

If one of two interacting bodies is known to be stationary, like the ground, then the work of the constraint forces is zero on both of the bodies. Thus the work of the hinge force on a pendulum, and the ground reaction forces on a frictionlessly sliding body or the ground force on a perfectly rolling body is zero. But be careful with the words “workless constraint forces”, however.

The workless constraint connecting moving bodies $\cal A$ and $\cal B$ is likely to do positive work on one of the bodies and negative work on the other.

It is just the net work on the two bodies which is zero.

Box 19.2 Ideal constraints and workless constraints

All of the ideal constraints we consider are interactions between two bodies $\cal A$ and $\cal B$ . One of these could be the ground. Let’s take the interaction force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ and moment ${\accentset{\rightharpoonup}{\boldsymbol{M}}}$ to be the force and moment of $\cal A$ on $\cal B$ . The point of interaction is A on $\cal A$ and B on $\cal B$ . By the principle of action and reaction, the net power of the interaction force on the two bodies is

	$\displaystyle P$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}+{\accentset{% \rightharpoonup}{\boldsymbol{M}}}\cdot{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\cal B}+(-{\accentset{\rightharpoonup}{\boldsymbol{F}}% })\cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A}}+% (-{\accentset{\rightharpoonup}{\boldsymbol{M}}})\cdot{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}_{\cal A}$
		$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}+{\accentset{% \rightharpoonup}{\boldsymbol{M}}}\cdot{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\cal B/A}.$

All of our ideal constraints are designed to exactly make these dot products zero. The ideal hinge is considered in the text. Another example is perfect rolling. In that case the interaction moment is assumed to be zero. The no-slip condition means that ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}=% \mathbf{0}$ . On the other hand for frictionless sliding there ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}$ can have a component tangent to the surfaces. But that is exactly the direction where the friction force is assumed to be zero.

And so it is for all of the ideal “workless” constraints.

Examples of non-workless constraints, that is, interactions that contribute to the energy equations are: sliding with non-zero friction, joints with non-zero friction torques, joints with motors, or interactions mediated by springs, dampers or actuators.

Energy method: single degree of freedom systems

Although linear and angular momentum balance apply to a single degree of freedom system and all of its parts, often one finds what one wants with a single scalar equation, namely energy or power balance.

Imagine a complex machine that only has one degree of freedom, meaning the position of the whole machine is determined by a single configuration variable, call it $q$ . Further assume that the machine has no motion when $\dot{q}=0$ . The variable $q$ could be, for example, the angle of one of the linked-together machine parts. Also, assume that the machine has no dissipative parts: no friction, no collisions, no inelastic deformation. Because $q$ characterizes the position of all of the parts of the system we can, in principle, calculate the potential energy of the system as a function of $q$ ,

{E_{\mathrm{P}}}={E_{\mathrm{P}}}(q).

We find this function by adding up the potential energies of all the springs in the machine and the gravitational potential energies of the parts.

Similarly we can write the system’s kinetic energy in terms of $q$ and its rate of change $\dot{q}$ . Because at any configuration the velocity of every point in the system is proportional to $\dot{q}$ we can write the kinetic energy as:

{E_{\mathrm{K}}}=M(q)\dot{q}^{2}/2

where $M(q)$ is a function that one can determine by calculating the machine’s total kinetic energy in terms of $q$ and $\dot{q}$ and then factoring $\dot{q}^{2}$ out of the resulting expression.

Now, if we accept the equation of mechanical energy conservation we have

$\displaystyle{\rm constant}$	$\displaystyle=E_{\mathrm{T}}\quad\text{ conservation of energy,}$
$\displaystyle\Rightarrow\ \ 0$	$\displaystyle={d\over dt}E_{\mathrm{T}}\quad\text{ taking one time derivative,}$
	$\displaystyle={d\over dt}[{E_{\mathrm{P}}}+{E_{\mathrm{K}}}]\quad\text{ total % energy is potential plus kinetic}$
	$\displaystyle={d\over dt}[{E_{\mathrm{P}}}(q)+\frac{1}{2}M(q)\dot{q}^{2}]\quad% \text{ substituting from paragraphs above}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle={d\over dq}[{E_{\mathrm{P}}}(q)]\,\dot{q}+\frac{1}{2}{d\over dq}% [M(q)]\dot{q}\dot{q}^{2}+M(q)\dot{q}\ddot{q}$
$\displaystyle 0$	$\displaystyle={d\over dq}[{E_{\mathrm{P}}}(q)]+\left(\frac{1}{2}{d\over dq}[M(% q)]\right)\,\dot{q}^{2}+M(q)\,\ddot{q}\quad\text{ cancelling $\dot{q}$}$
$\displaystyle 0$	$\displaystyle=f_{1}(q)+f_{2}(q)\,\dot{q}^{2}+f_{3}(q)\,\ddot{q}$	(19.20)
	$\displaystyle\quad\mbox{with}\ f_{1}(q)\equiv{d\over dq}[{E_{\mathrm{P}}}(q)],$
	$\displaystyle\quad\phantom{\mbox{with}\ }f_{2}(q)\equiv\frac{1}{2}{d\over dq}[% M(q)],\hbox to36.135pt{\hfil and\hfil}$
	$\displaystyle\quad\phantom{\text{ with}\ }f_{3}(q)\equiv M(q).\par\par$

The cancellation of $\dot{q}$ above lacks mathematical rigor, but doesn’t cause problems^†^†margin: The cancellation of the factor $\dot{q}$ from equation 19.20 depends on $\dot{q}$ being other than zero. While moving, $\dot{q}$ is not zero. Strictly we cannot cancel the $\dot{q}$ term from the equation at the instants when $\dot{q}=0$ . However, to say that a differential equation is true except for certain instants in time is, in practice to say that it is always true, at least if we make reasonable assumptions about the smoothness of the motions. .The equation of motion is complicated because when we take the time derivative of a function of $M(q)$ and ${E_{\mathrm{P}}}(q)$ we have to use the chain rule. Also, because we have products of terms, we had to use the product rule.

Eqn. 19.20 is the general equation of motion of a conservative one-degree-of-freedom system. It is really just a special case of the equation of motion for one-degree-of-freedom systems found from power balance. Rather than memorizing eqn. (19.20) it is probably best to look at its derivation as an algorithm to be reproduced on a problem by problem basis.

Example: Spring and mass

Filename:bikefork-ang-mom

Figure 19.46: The familiar one degree of freedom spring and mass system.

Although the motion of a spring and mass system can be found easily enough from linear momentum balance, it is also a good example for energy balance (see fig. 19.46). Using conservation of energy for the spring and mass system:

$\displaystyle E_{\mathrm{T}}$	$\displaystyle=$	$\displaystyle constant$
$\displaystyle 0$	$\displaystyle=$	$\displaystyle{d\over dt}E_{\mathrm{T}}$
	$\displaystyle=$	$\displaystyle\dot{E_{\mathrm{K}}}+\dot{E_{\mathrm{P}}}$
	$\displaystyle=$	$\displaystyle{d\over dt}{(mv^{2}/2)}+{d\over dt}{(kx^{2}/2)}$
	$\displaystyle=$	$\displaystyle mv\dot{v}+kx\dot{x}$
$\displaystyle v=\dot{x}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle m\ddot{x}+kx.$

Similarly power balance could have been used to get the same result, looking at just the mass

$\displaystyle P$	$\displaystyle=$	$\displaystyle{d\over dt}{E_{\mathrm{K}}}$
$\displaystyle(-kx)(\dot{x})$	$\displaystyle=$	$\displaystyle{d\over dt}{(mv^{2}/2)}=mv\dot{v}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle kx+m\ddot{x}$

as before.

Example: Pendulum

Filename:summer95f-5-a

Figure 19.47: A rigid body suspended from a frictionless hinge is an energy conserving one-degree-of-freedom mechanism.

Consider a rigid body with mass $m$ and moment of inertia $I^{o}$ about a hinge which is a distance $\ell$ from the center-of-mass (see fig. 19.47). The familiar simple pendulum is another single degree of freedom system for which the equation of motion can be found from conservation of energy.

$\displaystyle E_{\mathrm{T}}$	$\displaystyle=$	$\displaystyle constant$
$\displaystyle\rightarrow 0$	$\displaystyle=$	$\displaystyle{d\over dt}E_{\mathrm{T}}=\dot{E_{\mathrm{K}}}+\dot{E_{\mathrm{P}}}$
	$\displaystyle=$	$\displaystyle{d\over dt}{(I^{o}\omega^{2}/2)}+{d\over dt}{(-mg\ell\cos\theta)}$
	$\displaystyle=$	$\displaystyle I^{o}\omega\dot{\omega}+-mg\ell(-\sin\theta)\dot{\theta}$
$\displaystyle\omega=\dot{\theta}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle\ddot{\theta}+{mg\ell\over I^{o}}\sin\theta.$

the pendulum equation that we have derived before by this and other means (angular momentum balance about point o).

The above examples are old friends which are handled easily with other techniques. Here is a problem which is much more difficult without the energy method.

Example: Three bars act like a simple pendulum

Filename:pfigure4-2-rp10

Figure 19.48: Three identical uniform bars are pinned with frictionless hinges and swing, obviously, something like a simple pendulum.

Assume all three bars in the structure shown in fig. 19.48are of equal length $\ell$ and have mass $m$ uniformly distributed along their length. It is intuitively obvious that this device swings back and forth something like a simple pendulum. But how can we get the laws of mechanics to tell us this? One approach, which will work in the end, is to draw free-body diagrams of all the parts, write linear and angular momentum balance for each, and then add and subtract equations to eliminate the unknown constraint forces at the various hinges.

The more direct approach is to write the energy equation, adding up the potential and kinetic energies of the parts, all evaluated in terms of the single configuration variable $\theta$ . Taking the potential energy to be zero at $\theta=\pi/2$ (when all centers of mass are at hinge height) we have

$\displaystyle E_{\mathrm{T}}$	$\displaystyle=$	$\displaystyle constant$
$\displaystyle 0$	$\displaystyle=$	$\displaystyle{d\over dt}E_{\mathrm{T}}=\dot{E_{\mathrm{P}}}+\dot{E_{\mathrm{K}}}$
	$\displaystyle=$	$\displaystyle\phantom{+}{d\over dt}\bigm{(}(I^{o}\omega^{2}/2)+(I^{o}\omega^{2% }/2)+(m(\ell\omega)^{2}/2)\bigm{)}$
		$\displaystyle+{d\over dt}\bigm{(}(-gm(\ell/2)\cos\theta$
		$\displaystyle-gm(\ell/2)\cos\theta)-gm\ell\cos\theta\bigm{)}$
$\displaystyle I^{o}=m\ell^{2}/3\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle{d\over dt}\bigm{(}5m\ell^{2}\omega^{2}/6\bigm{)}+{d\over dt}% \bigm{(}-2gm\ell\cos\theta\bigm{)}$
	$\displaystyle=$	$\displaystyle(5m\ell^{2}\omega\dot{\omega}/3+2gm\ell\omega\sin\theta\bigm{)}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle 5\ell\dot{\omega}/3+2g\sin\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle\ddot{\theta}+{6g\over 5\ell}\sin\theta$

which is the same governing equation as for a point-mass pendulum with length $5\ell/6$ . This is just half way between the following two cases. If the side links had no mass the equation would have been the same as for a point mass pendulum with length $\ell$

0=\ddot{\theta}+{g\over\ell}\sin\theta

and if the bottom link had no mass the equation would be the same as a stick hanging from one end which goes back and forth like a point mass pendulum with length $2\ell/3$ according to

0=\ddot{\theta}+{3g\over 2\ell}\sin\theta.

Filename:pfigure-blue-125-2

Figure 19.49: A person rides a bike. The pedaling leg is idealized as a pair of equal and opposite forces acting on the seat and pedal.

Example: One on the rim is like two on the frame.

A bicycle transmission is such that the speed of the bike relative to the ground is $n$ times the speed of the pedal relative to the frame :

v_{\rm bike}=nv_{\rm pedal/bike}

Assume the kinetic energy of the relative motion of a rider’s legs can be neglected, as can be the weight of the rider’s leg. At the moment in question the velocity of the pedal is parallel to the direction from the seat to the leg. Thus the free-body diagram of the bike/person system, leaving out the pedaling leg is as shown in fig. 19.49. Let’s assume the bike and rider have mass $M$ and that the wheels have mass $m_{r}$ and $m_{f}$ concentrated on the rim (the hubs are considered part of the frame and the spokes are neglected). Neglecting air resistance etc.the power balance equation is:

\displaystyle P

\displaystyle=

\displaystyle\dot{E_{\mathrm{K}}}

(19.21)

Let’s do some side calculations for evaluating the terms in eqn. (19.21). First, the only forces that do work on the system as drawn are the force on the pedal and the force on the seat.

$\displaystyle P$	$\displaystyle=$	$\displaystyle-{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{P}}% \cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{seat}}% +{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{P}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{pedal}}$
	$\displaystyle=$	$\displaystyle-{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{P}}% \cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{seat}}% +{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{P}}\cdot\bigm{(}{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{bike}}+{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{pedal/bike}}% \bigm{)}$
	$\displaystyle=$	$\displaystyle\underbrace{-{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{P}}\cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{seat}}+{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{P}}% \cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{bike}}% }_{0}+\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{P% }}\cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{% pedal/bike}}}_{F_{P}v_{\rm pedal/bike}}$
	$\displaystyle=$	$\displaystyle F_{P}v_{\rm bike}/n$

The net power of the leg is expressed by the compression it carries times its extension rate. The kinetic energy of the wheel comes from both its rotation and its translation. The moment of inertia of a hoop about its center is $I=mr^{2}$ . For rolling contact $|\omega R|=v$ so, for one wheel:

$\displaystyle{E_{\mathrm{K}}}_{\rm wheel}$	$\displaystyle=$	$\displaystyle mv_{\rm bike}^{2}/2+I\omega^{2}/2$
	$\displaystyle=$	$\displaystyle mv_{\rm bike}^{2}/2+(mR^{2})(v/R)^{2}/2$
	$\displaystyle=$	$\displaystyle mv_{\rm bike}^{2}.$

The kinetic energy of a rolling hoop is twice that of a point mass moving at the same speed. Putting these results back in to eqn. (19.21) we have

$\displaystyle P$	$\displaystyle=$	$\displaystyle\dot{E_{\mathrm{K}}}$	(19.22)
$\displaystyle F_{P}v_{\rm bike}/n$	$\displaystyle=$	$\displaystyle{d\over dt}\bigm{(}Mv_{\rm bike}^{2}/2+(m_{r}+m_{f})v_{\rm bike}^% {2}\bigm{)}$	(19.23)
	$\displaystyle=$	$\displaystyle\bigm{(}M\dot{v}_{\rm bike}+2(m_{r}+m_{f})\dot{v}_{\rm bike}\bigm% {)}v_{\rm bike}$	(19.24)
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F_{P}/n$	$\displaystyle=$	$\displaystyle(M+2(m_{r}+m_{f}))\dot{v}_{\rm bike}$	(19.25)
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{v}_{\rm bike}$	$\displaystyle=$	$\displaystyle{F_{P}\over n(M+2(m_{r}+m_{f}))}.$	(19.26)

The bigger the pedal force, the bigger the acceleration, obviously. The higher the gear ratio, the less the acceleration; the faster gears let you pedal slower for a given bike speed, but demand more pedal force for a given acceleration. A heavier bike accelerates less. But the contribution to slowing a bike is twice as much for mass added to the rim as for mass added to the frame or body.

Some comments. $n$ typically ranges from about 1.7 to 8 for a new 21 speed bike and is about 5 for an adult European, Indian or Chinese 1-speed. For a given speed of bicycle riding your feet go $n$ times slower relative to your body than for walking or running at that speed. This calculation is for accelerating a bike on level ground with no wind and rolling resistance. The net speed of a bike in a bike race is not so dependent on weight, because the main enemy is wind resistance. To the extent that weight is a problem it is for steady uphill travel. In this case the mass on the rim makes the same contribution as mass on the frame.

Vibrations

The preponderance of systems where vibrations occur is not due to the fact that so many systems look like a spring connected to a mass, a simple pendulum, or a torsional oscillator. Instead there is a general class of systems which can be expected to vibrate sinusoidally near some equilibrium position. These systems are one-degree -of-freedom (one DOF) near an energy minimum.

In detail why this works out is explained in Box 19.2.

Box 19.3 One degree of freedom systems near a potential energy minimum are harmonic oscillators

In order to specialize to the case of oscillations, we want to look at a one degree of freedom system near a stable equilibrium point, a potential energy minimum.

At a potential energy minimum we have, as you will recall from ‘max-min’ problems in calculus, that $d{E_{\mathrm{P}}}(q)/dq=0$ . To keep our notation simple, let’s assume that we have defined $q$ so that $q=0$ at this minimum. Physically this means that $q$ measures how far the system is from its equilibrium position. That means that if we take a Taylor series approximation of the potential energy the expression for potential energy can be expressed as follows:

	$\displaystyle{E_{\mathrm{P}}}$	$\displaystyle\approx$	$\displaystyle{\rm const}+\underbrace{d{E_{\mathrm{P}}}\over dq}_{0}\cdot q+% \frac{1}{2}\underbrace{d^{2}{E_{\mathrm{P}}}\over dq^{2}}_{K_{\mbox{equiv}}}% \cdot q^{2}+\dots$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad{d{E_{\mathrm{P}}}% \over dq}$	$\displaystyle\approx$	$\displaystyle K_{\mbox{equiv}}\cdot q$		(19.27)

Applying this result to equation( 19.20) we get:

0=K_{\mbox{equiv}}q+\frac{1}{2}{d\over dq}[M(q)]\dot{q}^{2}+M(q)\ddot{q}.

(19.28)

We now write $M(q)$ in terms of its Taylor series. We have

M(q)=M(0)+\left.dM/dq\right|_{0}\cdot q+\dots

(19.29)

and substitute this result into equation 19.28. We have not finished using our assumption that we are only going to look at motions that are close to the equilibrium position $q=0$ where $q$ is small. The nature of motion close to an equilibrium is that when the deflections are small, the rates and accelerations are also small. Thus, to be consistent in our approximation we should neglect any terms that involve products of $q,\dot{q},\ \hbox{or}\ \ddot{q}$ . Thus the middle term involving $\dot{q}^{2}$ is negligibly smaller than other terms. Similarly, using the Taylor series for $M(q)$ , the last term is well approximated by $M(0)\ddot{q}$ , where $M(0)$ is a constant which we will call $M_{{\rm equiv}}$ . Now we have for the equation of motion:

0={d\over dt}E_{\mathrm{T}}\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad 0% =K_{\mbox{equiv}}q+M_{\mbox{equiv}}\ddot{q},

(19.30)

which you should recognize as the harmonic oscillator equation. So we have found that for any energy conserving one degree of freedom system near a position of stable equilibrium, the equation governing small motions is the harmonic oscillator equation. The effective stiffness is found from the potential energy by $K_{\mbox{equiv}}=d^{2}{E_{\mathrm{P}}}/dq^{2}$ and the effective mass is the coefficient of $\dot{q}^{2}/2$ in the expansion for the kinetic energy ${E_{\mathrm{K}}}$ . The displacement of any part of the system from equilibrium will thus be given by

A\sin(\lambda\,t)+B\cos(\lambda\,t)

(19.31)

with $\lambda^{2}=K_{\mbox{equiv}}/M_{\mbox{equiv}}$ , and $A$ and $B$ determined by the initial conditions. So we have found that all stable non-dissipative one-degree-of-freedom systems oscillate when disturbed slightly from equilibrium and we have found how to calculate the frequency of vibration.

Examples of 1 DOF harmonic oscillators

In the previous section, we have shown that any non-dissipative one-degree-of-freedom system that is near a potential energy minimum can be expected to have simple harmonic motion. Besides the three examples we have given so far, namely,

•

a spring and mass,
•

a simple pendulum, and
•

a rigid body and a torsional spring,

there are examples that are somewhat more complex, such as

•

a cylinder rolling near the bottom of a valley,
•

a cart rolling near the bottom of a valley, and a
•

a four bar linkage swinging freely near its energy minimum.

The restriction of this theory to systems with only one-degree-of-freedom is not so bad as it seems at first sight. First of all, it turns out that simple harmonic motion is important for systems with multiple-degrees-of-freedom. We will discuss this generalization in more detail later with regard to normal modes. Secondly, one can also get a good understanding of a vibrating system with multiple-degrees-of-freedom by modeling it as if it has only one-degree-of-freedom.

Example: Cylinder rolling in a valley

Consider the uniform cylinder with radius $r$ rolling without slip in an cylindrical ‘ideal’ valley of radius $R$ .

For this problem we can calculate ${E_{\mathrm{K}}}$ and ${E_{\mathrm{P}}}$ in terms of $\theta$ . Briefly,

$\displaystyle{E_{\mathrm{P}}}$	$\displaystyle=$	$\displaystyle-mg(R-r)\cos\theta$
$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle\frac{1}{2}\left(\frac{3}{2}mr^{2}\right)\left(\frac{\dot{\theta}% (R-r)}{r}\right)^{2}$
	$\displaystyle=$	$\displaystyle\frac{3}{4}m(R-r)^{2}\dot{\theta}^{2}$

So we can derive the equation of motion using the fact of constant total energy.

Filename:pfigure-blue-68-1

Figure 19.50: Cylinder rolling without slip in a cylinder. This gives the pendulum equation, but the effective pendulum length does not tend to

R

r\rightarrow 0

(see text).

$\displaystyle 0$	$\displaystyle=$	$\displaystyle\frac{d}{dt}(E_{T})=\frac{d}{dt}({E_{\mathrm{K}}}+{E_{\mathrm{P}}})$
	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left(\underbrace{-mg(R-r)\cos\theta}_{{E_{\mathrm{P}% }}}+\underbrace{\frac{3}{4}m(R-r)^{2}\dot{\theta}^{2}}_{{E_{\mathrm{K}}}}\right)$
	$\displaystyle=$	$\displaystyle(mg(R-r)\sin\theta)\dot{\theta}+\frac{3}{2}(R-r)^{2}\dot{\theta}% \ddot{\theta}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle mg(R-r)\sin\theta+\frac{3}{2}(R-r)^{2}m\ddot{\theta}$

Now, assuming small angles, so $\theta\approx\sin\theta$ , we get

	$\displaystyle g(R-r)\theta+\frac{3}{2}(R-r)^{2}\ddot{\theta}$	$\displaystyle=$	$\displaystyle 0$		(19.32)
	$\displaystyle\ddot{\theta}+\underbrace{\left(\frac{2}{3}\frac{g}{(R-r)}\right)% }_{\lambda^{2}}\theta$	$\displaystyle=$	$\displaystyle 0$		(19.33)

This is, naturally, our old friend the harmonic oscillator equation. The period is a funny combination of terms. If $r\ll R$ it looks like a point mass pendulum with length $3R/2$ , more than $R$ . That is, the rolling effect doesn’t go away and make the roller act like a point mass even when the radius goes to zero. See page 17.4 for the angular momentum approach to this problem.

Although, in some abstract way the energy approach always works, practically speaking it has limitations for systems where the configuration is not easily found from one configuration variable.

Example: A four bar linkage analyzed with energy methods

Although probably not usually the best approach, energy methods can be used to find the motions of a 4-bar linkage. Take a four-bar linkage with one bar grounded. Assume the bars all have different lengths. This is a one-DOF system which can use the angle $\theta$ of one of the links as a configuration variable. But finding the potential energy as a single formula in terms of all of the links in terms of $\theta$ is more trigonometry than most of us like. And then finding the kinetic energy in terms of $\theta$ and $\dot{\theta}$ is close enough to impossible that people don’t do it.

So, though it is true that there are functions ${E_{\mathrm{P}}}(\theta)$ and ${E_{\mathrm{K}}}(\theta,\dot{\theta})$ and that the equations of motion could be written in terms of them, it is really not practical to do so.

How do you find the motions of a 4-bar linkage in practice? It’s more tricky. One approach is to solve the kinematics by integrating kinematic differential equations, as in Sample 18.5 on page 18.5. Then you set up and solve the balance equations of the separate parts as in Sample 19.95 on page 19.95

SAMPLE 19.11

Filename:pfigure-blue-58-1

Figure 19.51:

A plate pendulum. A $2a\times 2b$ rectangular plate of mass $m$ hangs from two parallel, massless links EA and FD of length $\ell$ each. The links are hinged at both ends so that when the plate swings, its edges AD and BC remain horizontal at all times. The only driving force present is gravity. Find the equation of motion of the plate.

Solution

Filename:pfigure-blue-157-1

Figure 19.52:

The given system is a single DOF system. So, we need just one configuration variable and the equation of motion will be just one scalar equation in this variable. Let us take angle $\theta$ (fig. 19.52) as our configuration variable.

Filename:summer95f-5

Figure 19.53:

The free-body diagram of the plate is shown in fig. 19.53. Note that the link forces ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{2}$ act along the links because massless links are two-force bodies. Let $(x,y)$ be the coordinates of the center-of-mass. Then the linear momentum balance for the plate gives

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1}+{\accentset{\rightharpoonup}% {\boldsymbol{F}}}_{2}-mg\hat{\boldsymbol{\jmath}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}=m(\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{y}\hat{% \boldsymbol{\jmath}}).

Now, we can eliminate both the unknown link forces from this equation by dotting the equation with $\hat{\boldsymbol{n}}=\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}}$ , a unit vector normal to the links. Then, we have

	$\displaystyle-mg(\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{n}})$	$\displaystyle=$	$\displaystyle m\bigl{[}\ddot{x}(\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol% {n}})+\ddot{y}(\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{n}})\bigr{]}$
	$\displaystyle-g\sin\theta$	$\displaystyle=$	$\displaystyle\ddot{x}\cos\theta+\ddot{y}\sin\theta.$		(19.34)

Now we need to find a relationship between $x$ and $\theta$ , and $y$ and $\theta$ , so that we can write $\ddot{x}$ and $\ddot{y}$ in terms of our configuration variable $\theta$ and its derivatives. From fig. 19.52, we have

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G}}$	$\displaystyle=$	$\displaystyle\overbrace{(\ell\sin\theta+a)}^{x}\hat{\boldsymbol{\imath}}+% \overbrace{(-\ell\cos\theta-b)}^{y}\hat{\boldsymbol{\jmath}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{x}$	$\displaystyle=$	$\displaystyle\ell(\cos\theta\cdot\ddot{\theta}-\sin\theta\cdot\dot{\theta}^{2})$
$\displaystyle\ddot{y}$	$\displaystyle=$	$\displaystyle\ell(\sin\theta\cdot\ddot{\theta}+\cos\theta\cdot\dot{\theta}^{2})$

Substituting these expressions for $\ddot{x}$ and $\ddot{y}$ in eqn. (19.34) , we get

	$\displaystyle-g\sin\theta$	$\displaystyle=$	$\displaystyle\ell\ddot{\theta}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{\theta}+\frac{g}{% \ell}\sin\theta$	$\displaystyle=$	$\displaystyle 0.$

Answer: $\ddot{\theta}+\frac{g}{\ell}\sin\theta=0$

This is the equation of a simple pendulum! Well, the plate does behave just like a simple pendulum in the given mechanism. From the expressions for the $x$ and $y$ coordinates of the center-of-mass, we have

$\displaystyle x-a$	$\displaystyle=$	$\displaystyle\ell\sin\theta$
$\displaystyle y+b$	$\displaystyle=$	$\displaystyle-\ell\cos\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}(x-a)^{2}$	$\displaystyle+$	$\displaystyle(y+b)^{2}=\ell^{2}$

that is, the center-of-mass follows a circle of radius $\ell$ centered at $(-a,b)$ . Since the orientation of the plate never changes (AD and BC always remain horizontal), the plate has no angular velocity. Thus the motion of the plate is equivalent to the motion of a particle of mass $m$ going in a circle centered at $(-a,b)$ and driven by gravity. That is the simple pendulum.

SAMPLE 19.12

Filename:pfigure-blue-127-2

Figure 19.54:

Equation of motion from power balance. A slider crank mechanism is shown in fig. 19.54 where the crank is a uniform wheel of mass $m$ and radius $R$ anchored at the center and the connecting rod AB is a massless rod of length $\ell$ . The rod is driven by a piston at B with a known force $F(t)=F_{0}\cos\Omega t$ . There is no gravity. Find the equation of motion of the wheel.

Solution The given mechanism is a one DOF system. So, let us choose a single configuration variable, $\theta$ for specifying the configuration of the system and derive an equation that determines $\theta$ . Since the applied force is given and the point of application of the force has a simple motion (vertical), it will be easy to calculate power of this force. Also, the connecting rod is massless, so it does not enter into dynamic calculations. The wheel rotates about its center and, therefore, it is easy to calculate its kinetic energy. So, we use the power balance, ${\dot{E}_{\mathrm{K}}}=P$ , here to find the equation of motion of the wheel. Since ${E_{\mathrm{K}}}=\frac{1}{2}I^{\mathrm{cm}}_{zz}\dot{\theta}^{2}$ and $P={\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}$ , we have

I^{\mathrm{cm}}_{zz}\ddot{\theta}\,\dot{\theta}=F(t)\hat{\boldsymbol{\jmath}}% \cdot v_{B}\hat{\boldsymbol{\jmath}}=F(t)v_{B}

(19.35)

Filename:pfigure-s94h13p4

Figure 19.55:

Now, we need to find $v_{B}$ and express it using the configuration variable $\theta$ and its derivatives. There are several ways we could find $v_{B}$ . Vectorially, we could write, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}\equiv v% _{B}\hat{\boldsymbol{\jmath}}={\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{A}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB% }\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}$ where ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=\dot{\phi}\hat{% \boldsymbol{k}}$ . Dotting both sides of this equation with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ we can find $\phi$ in terms of $\theta$ and $v_{B}$ in terms of $\theta$ and $\dot{\theta}$ . But, for a change, let us use geometry here.

See fig. 19.55. From triangle ABO, we have

$\displaystyle\frac{R}{\sin(90^{o}-\phi)}$	$\displaystyle=$	$\displaystyle\frac{\ell}{\sin(90^{o}+\theta)}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ell\cos\phi$	$\displaystyle=$	$\displaystyle R\cos\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-\ell\sin\phi\cdot\dot{\phi}$	$\displaystyle=$	$\displaystyle-R\sin\theta\cdot\dot{\theta}$
$\displaystyle\dot{\phi}$	$\displaystyle=$	$\displaystyle\frac{R\sin\theta}{\ell\sin\phi}\dot{\theta}$
Now,
$\displaystyle y_{B}$	$\displaystyle=$	$\displaystyle R\sin\theta-\ell\sin\phi$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v_{B}$	$\displaystyle\equiv$	$\displaystyle\dot{y}_{B}=R\cos\theta\cdot\dot{\theta}-\ell\cos\phi\cdot\dot{\phi}$
	$\displaystyle=$	$\displaystyle R\dot{\theta}\cos\theta-R\cos\theta\cdot\frac{R\sin\theta}{\ell% \sin\phi}\dot{\theta}$
	$\displaystyle=$	$\displaystyle R\dot{\theta}\left(\cos\theta-\frac{R\sin\theta\cos\theta}{\sqrt% {\ell^{2}-R^{2}\cos^{2}\theta}}\right)$

Substituting this expression for $v_{B}$ in power balance eqn. (19.35), we get

	$\displaystyle I^{\mathrm{cm}}_{zz}\ddot{\theta}\,\cancel{\dot{\theta}}$	$\displaystyle=$	$\displaystyle F(t)\cdot R\cancel{\dot{\theta}}\left(\cos\theta-\frac{\sin 2% \theta}{2\sqrt{(\ell/R)^{2}-\cos^{2}\theta}}\right)$
	$\displaystyle\ddot{\theta}$	$\displaystyle=$	$\displaystyle\frac{RF_{0}\cos\Omega t\cdot}{\frac{1}{2}mR^{2}}\left(\cos\theta% -\frac{\sin 2\theta}{2\sqrt{(\ell/R)^{2}-\cos^{2}\theta}}\right).$

This is the required equation of motion. As is evident, it is a nonlinear ODE which requires numerical solution on a computer if we would like to plot $\theta(t)$ .

Answer: $\ddot{\theta}=\frac{2F_{0}\cos\Omega t\cdot}{mR}\left(\cos\theta-\frac{\sin 2% \theta}{2\sqrt{(\ell/R)^{2}-\cos^{2}\theta}}\right)$

SAMPLE 19.13

Filename:pfigure-blue-90-2

Figure 19.56: End A of bar AB is free to slide on the frictionless horizontal surface while end B is going in circles with a disk rotating at a constant rate.

Instantaneous dynamics of slider crank. A uniform rigid rod AB of mass $m$ and length $\ell=4R$ has one of its ends pinned to the rim of a disk of radius $R$ . The other end of the bar is free to slide on a frictionless horizontal surface. A motor, connected to the center of the disk at O, keeps the disk rotating at a constant angular speed $\omega_{D}$ . At the instant shown, end B of the rod is directly above the center of the disk making $\theta$ to be $30^{o}$ .

1.

Find all the forces acting on the rod.
2.

Is there a value of $\omega_{D}$ which makes end A of the rod lift off the horizontal surface when $\theta=30^{o}$ ?

Solution The disk is rotating at constant speed. Since end B of the rod is pinned to the disk, end B is going in circles at constant rate. The motion of end B of the rod is completely prescribed. Since end A can only move horizontally (assuming it has not lifted off yet), the orientation (and hence the position of each point) of the rod is completely determined at any instant during the motion. Therefore, the rod represents a zero degree of freedom system.

Forces on the rod: The free-body diagram of the rod is shown in Fig. 19.57. The pin at B exerts two forces $B_{x}$ and $B_{y}$ while the surface in contact at A exerts only a normal force $N$ because there is no friction. Now, we can write the momentum balance equations for the rod. The linear momentum balance ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ ) for the rod gives

Filename:s92f1p7

Figure 19.57: Free-body diagram of the bar.

B_{x}\hat{\boldsymbol{\imath}}+(B_{y}+N-mg)\hat{\boldsymbol{\jmath}}=m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}.

(19.36)

The angular momentum balance about the center-of-mass G ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/G}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{/G}$ ) of the rod gives

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A/G}\times N\hat{\boldsymbol{% \jmath}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/G}\times(B_{x}\hat{% \boldsymbol{\imath}}+B_{y}\hat{\boldsymbol{\jmath}})=I_{zz/G}\ \alpha_{rod}% \hat{\boldsymbol{k}}.

(19.37)

From these two vector equations we can get three scalar equations (the Angular Momentum Balance gives only one scalar equation in 2-D since the quantities on both sides of the equation are only in the $\hat{\boldsymbol{k}}$ direction), but we have six unknowns — $B_{x},\,B_{y},\,N,\,{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$ (counts as two unknowns), and $\alpha_{rod}$ . Therefore, we need more equations. We have already used the momentum balance equations, hence, the extra equations have to come from kinematics.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}+\overbrace{{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{rod}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{A/B}}^{{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{A/B}}$
$\displaystyle\hbox to36.135pt{or\hfil}v_{A}\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle\omega_{D}R\hat{\boldsymbol{\imath}}+\omega_{rod}\hat{\boldsymbol% {k}}\times\ell(-\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{\boldsymbol% {\jmath}})$
	$\displaystyle=$	$\displaystyle(\omega_{D}R+\omega_{rod}\ell\sin\theta)\hat{\boldsymbol{\imath}}% -\omega_{rod}\ell\cos\theta\hat{\boldsymbol{\jmath}}$

Dotting both sides of the equation with $\hat{\boldsymbol{\jmath}}$ we get

0=\omega_{rod}\ell\cos\theta\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}% \quad\omega_{rod}=0.

Also,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}+\overbrace{\dot% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{A/B}+\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{rod}}_{{\accentset{\rightharpoonup}{% \boldsymbol{0}}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% rod}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A/B})}^{{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{A/B}}$
$\displaystyle\hbox to36.135pt{or\hfil}a_{A}\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle-\omega_{D}^{2}R\hat{\boldsymbol{\jmath}}+\dot{\omega}_{rod}\hat{% \boldsymbol{k}}\times\ell(-\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{% \boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-(\omega^{2}_{D}R+\dot{\omega}_{rod}\ell\cos\theta)\hat{% \boldsymbol{\jmath}}+\dot{\omega}_{rod}\ell\sin\theta\hat{\boldsymbol{\imath}}.$

Dotting both sides of this equation by $\hat{\boldsymbol{\jmath}}$ we get

\dot{\omega}_{rod}=-\frac{\omega_{D}^{2}R}{\ell\cos\theta}.

(19.38)

Now, we can find the acceleration of the center-of-mass:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}+\overbrace{\dot% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{G/B}+\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{rod}}_{{\accentset{\rightharpoonup}{% \boldsymbol{0}}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% rod}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/B})}^{{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{G/B}}$
	$\displaystyle=$	$\displaystyle-\omega_{D}^{2}R\hat{\boldsymbol{\jmath}}+\dot{\omega}_{rod}\hat{% \boldsymbol{k}}\times\frac{1}{2}\ell(-\cos\theta\hat{\boldsymbol{\imath}}-\sin% \theta\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-(\omega^{2}_{D}R+\frac{1}{2}\dot{\omega}_{rod}\ell\cos\theta)% \hat{\boldsymbol{\jmath}}+\frac{1}{2}\dot{\omega}_{rod}\ell\sin\theta\hat{% \boldsymbol{\imath}}.$

Substituting for $\dot{\omega}_{rod}$ from eqn. (19.38) and $30^{o}$ for $\theta$ above, we obtain

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}=-\frac{1}{2}\omega_{D}^{2}R(% \frac{1}{\sqrt{3}}\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}).

Substituting this expression for ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$ in eqn. (19.36) and dotting both sides by $\hat{\boldsymbol{\imath}}$ and then by $\hat{\boldsymbol{\jmath}}$ we get

	$\displaystyle B_{x}$	$\displaystyle=$	$\displaystyle-\frac{1}{2\sqrt{3}}m\omega_{D}^{2}R,$
	$\displaystyle B_{y}+N$	$\displaystyle=$	$\displaystyle-\frac{1}{2}m\omega_{D}^{2}R+mg$		(19.39)

From eqn. (19.37)

$\displaystyle\frac{1}{2}\ell[(B_{y}-N)\cos\theta-B_{x}\sin\theta]\hat{% \boldsymbol{k}}$	$\displaystyle=$	$\displaystyle\frac{1}{12}m\ell^{2}(-\frac{\omega_{D}^{2}R}{\ell\cos\theta})% \hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{or\hfil}B_{y}-N$	$\displaystyle=$	$\displaystyle-\frac{1}{6}\frac{m\omega_{D}^{2}R}{\cos^{2}\theta}+B_{x}\tan\theta$	(19.40)
	$\displaystyle=$	$\displaystyle-\frac{2}{9}m\omega_{D}^{2}R-\frac{1}{6}m\omega_{D}^{2}R$
	$\displaystyle=$	$\displaystyle-\frac{7}{18}m\omega_{D}^{2}R$

From eqns. (19.39) and (19.40)

B_{y}=\frac{1}{2}\left(mg-\frac{8}{9}m\omega_{D}^{2}R\right)

and

N=\frac{1}{2}\left(mg-\frac{1}{9}m\omega_{D}^{2}R\right).

2.

Lift off of end A: End A of the rod loses contact with the ground when normal force $N$ becomes zero. From the expression for $N$ from above, this condition is satisfied when

$\displaystyle\frac{2}{9}m\omega_{D}^{2}R$ $\displaystyle=$ $\displaystyle mg$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega_{D}$ $\displaystyle=$ $\displaystyle 3\sqrt{\frac{g}{R}}.$

SAMPLE 19.14

Filename:twodisks-ang-mom

Figure 19.58:

A bar sliding on a sliding wedge. A bar AB of mass $m$ and length $\ell$ is hinged at end A and rests on a wedge of mass $M$ at the other end B. The contact at B is frictionless. The wedge is free to slide horizontally without any friction. The motion of the system is driven only by gravity. Find the equation of motion of the bar using

1.

momentum balance
2.

energy (or power) balance.

Solution

Filename:pfigure-blue-90-1

Figure 19.59:

(a) Momentum Balance: The bar and the wedge make up a single DOF system. To derive the equations of motion of the bar, let us choose $\theta$ as the configuration variable. The free-body diagram of the bar and the wedge are shown in fig. 19.59. Note that the normal reaction at B is normal to the wedge surface, i.e., ${\accentset{\rightharpoonup}{\boldsymbol{N}}}=N\hat{\boldsymbol{n}}$ . Now, the angular momentum balance for the bar about point A gives,

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{A}}$	$\displaystyle=$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}$
$\displaystyle I^{A}_{zz}\ddot{\theta}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle\frac{\ell}{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times(% -mg\hat{\boldsymbol{\jmath}})+\ell\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}% \times N\hat{\boldsymbol{n}}$	(19.41)
	$\displaystyle=$	$\displaystyle-\frac{1}{2}mg\ell\cos\theta\hat{\boldsymbol{k}}+N\ell\cos(\alpha% -\theta)\hat{\boldsymbol{k}}$	(19.41)

where the last line follows from the fact that $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=\cos\theta\hat{\boldsymbol{\imath% }}+\sin\theta\hat{\boldsymbol{\jmath}}$ , the unit normal $\hat{\boldsymbol{n}}=-\sin\alpha\hat{\boldsymbol{\imath}}+\cos\alpha\hat{% \boldsymbol{\jmath}}$ , and so, $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times\hat{\boldsymbol{\jmath}}=% \cos\theta\hat{\boldsymbol{k}}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times\hat{\boldsymbol{n}}=\cos(% \alpha-\theta)\hat{\boldsymbol{k}}$ . We now need to eliminate the unknown normal reaction $N$ from the above equation. Since the wedge is constrained to move only horizontally, we can write the linear momentum balance for the wedge as

N\hat{\boldsymbol{n}}\cdot\hat{\boldsymbol{\imath}}=M\ddot{x}\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}N=\frac{M\,\ddot{x}}{\hat{\boldsymbol{n}}\cdot\hat{% \boldsymbol{\imath}}}=\frac{M\,\ddot{x}}{\sin\alpha}

(19.42)

Filename:pfigure-blue-110-1

Figure 19.60:

Thus, we have found $N$ in terms of $\ddot{x}$ that we can use in eqn. (19.41) to get rid of $N$ . But, we now need to express $\ddot{x}$ in terms of our configuration variable $\theta$ and its derivatives. Consider the triangle ABC formed by the bar and the slanted edge of the wedge. Let $x=$ AC denote the horizontal position of the wedge. Then, from the law of sines, we have $\frac{x}{\sin(\alpha-\theta)}=\frac{\ell}{\sin\alpha}$ , so that

$\displaystyle x$	$\displaystyle=$	$\displaystyle\frac{\ell}{\sin\alpha}\sin(\alpha-\theta)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{x}$	$\displaystyle=$	$\displaystyle\frac{\ell}{\sin\alpha}\cos(\alpha-\theta)\cdot(-\dot{\theta})$	(19.43)
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{x}$	$\displaystyle=$	$\displaystyle\frac{\ell}{\sin\alpha}\bigl{[}\ddot{\theta}\cos(\alpha-\theta)+% \dot{\theta}^{2}\sin(\alpha-\theta)\bigr{]}.$	(19.44)

Now, substituting for $N$ in eqn. (19.41) from eqn. (19.42), using the expression for $\ddot{x}$ from above, and dotting the resulting equation with $\hat{\boldsymbol{k}}$ , we get

$\displaystyle\frac{1}{3}m\ell^{2}\ddot{\theta}$	$\displaystyle=$	$\displaystyle-\frac{mg\ell}{2}\cos\theta+\frac{M\ell^{2}\cos(\alpha-\theta)}{% \sin^{2}\alpha}\bigl{[}\ddot{\theta}\cos(\alpha-\theta)+\dot{\theta}^{2}\sin(% \alpha-\theta)\bigr{]}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{\theta}$	$\displaystyle=$	$\displaystyle-\frac{3mg\sin^{2}\alpha\cos\theta+3M\ell\dot{\theta}^{2}\sin 2(% \alpha-\theta)}{2m\ell\sin^{2}\alpha+6M\ell\cos^{2}(\alpha-\theta)}$	(19.45)
	$\displaystyle=$	$\displaystyle-\frac{3\bigl{[}(g/\ell)\sin^{2}\alpha\cos\theta+(M/m)\dot{\theta% }^{2}\sin 2(\alpha-\theta)\bigr{]}}{2\big{[}\sin^{2}\alpha+3(M/m)\cos^{2}(% \alpha-\theta)\bigr{]}}$	(19.45)

Answer: $\ddot{\theta}=-\frac{3\bigl{[}(g/\ell)\sin^{2}\alpha\cos\theta+(M/m)\dot{% \theta}^{2}\sin 2(\alpha-\theta)\bigr{]}}{2\big{[}\sin^{2}\alpha+3(M/m)\cos^{2% }(\alpha-\theta)\bigr{]}}$

(b) Power balance: Now we derive the equation of motion for the bar using power balance ${\dot{E}_{\mathrm{K}}}=P$ . For power balance, we have to consider both the bar and the wedge. The bar rotates about the fixed point A, therefore, its kinetic energy is $(1/2)I^{A}_{z}z\dot{\theta}^{2}$ . The wedge moves with horizontal speed $\dot{x}$ , therefore, its kinetic energy is $(1/2)M\dot{x}^{2}$ . Thus, ${E_{\mathrm{K}}}=1/2)I^{A}_{z}z\dot{\theta}^{2}+(1/2)M\dot{x}^{2}$ . The only force that contributes to power is the force of gravity on the rod because ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{G}}\cdot(-% mg\hat{\boldsymbol{\jmath}})$ is non-zero. The sliding contact at B is frictionless and hence the net power due to the contact force there is zero. Now, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{G}}=\frac{1% }{2}\ell\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ . So, $P=-\frac{1}{2}mg\ell\dot{\theta}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!% \theta}\cdot\hat{\boldsymbol{\jmath}})=-\frac{1}{2}mg\ell\dot{\theta}\cos\theta$ . Thus the power balance for the system gives

I^{A}_{zz}\ddot{\theta}\dot{\theta}+M\ddot{x}\dot{x}=-\frac{1}{2}mg\ell\dot{% \theta}\cos\theta.

(19.46)

We can simplify this equation further. Note that, $\ddot{\theta}\dot{\theta}=\frac{d}{dt}(\frac{1}{2}\dot{\theta}^{2})$ and $\ddot{x}\dot{x}=\frac{d}{dt}(\frac{1}{2}\dot{x}^{2})$ . So that we can write the above equation as

$\displaystyle\frac{d}{dt}\left(\frac{1}{2}I^{A}_{zz}\dot{\theta}^{2}\right)+% \frac{d}{dt}\left(\frac{1}{2}M\dot{x}^{2}\right)$	$\displaystyle=$	$\displaystyle-\frac{1}{2}mg\ell\dot{\theta}\cos\theta$
$\displaystyle\hbox to36.135pt{or\hfil}\int d\bigl{(}I^{A}_{zz}\dot{\theta}^{2}% +M\dot{x}^{2}\bigr{)}$	$\displaystyle=$	$\displaystyle-mg\ell\int\cos\theta\,d\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}I^{A}_{zz}\dot{\theta}^{% 2}+M\dot{x}^{2}$	$\displaystyle=$	$\displaystyle C-mg\ell\sin\theta$

where $C$ is a constant of integration to be determined from initial conditions. For example, when the bar begins to slide from rest, we have, at $t=0$ , $\dot{x}=0$ and $\dot{\theta}=0$ . At that instant, if $\theta(0)=\theta_{0}$ , then $C=mg\ell\sin\theta_{0}$ . So, we can write

I^{A}_{zz}\dot{\theta}^{2}+M\dot{x}^{2}=mg\ell(\sin\theta_{0}-\sin\theta).

Now, replacing $\dot{x}$ in this equation with the expression we obtained in eqn. (19.43), we have

$\displaystyle I^{A}_{zz}\dot{\theta}^{2}+M\ell^{2}\dot{\theta}^{2}\frac{\cos^{% 2}(\alpha-\theta)}{\sin^{2}\alpha}$	$\displaystyle=$	$\displaystyle mg\ell(\sin\theta_{0}-\sin\theta)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\theta}^{2}$	$\displaystyle=$	$\displaystyle\frac{mg\ell(\sin\theta_{0}-\sin\theta)}{I^{A}_{zz}+M\ell^{2}% \frac{\cos^{2}(\alpha-\theta)}{\sin^{2}\alpha}}$
	$\displaystyle=$	$\displaystyle\frac{mg\ell\sin^{2}\alpha(\sin\theta_{0}-\sin\theta)}{(1/3)m\ell% ^{2}\sin^{2}\alpha+M\ell^{2}\cos^{2}(\alpha-\theta)}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\theta}$	$\displaystyle=$	$\displaystyle\sqrt{\frac{3g}{\ell}}\sin\alpha\sqrt{\frac{\sin\theta_{0}-\sin% \theta}{\sin^{2}\alpha+3(M/m)\cos^{2}(\alpha-\theta)}}.$

This is a first order, nonlinear, ODE compared to the second order equation we got in eqn. (19.45). However, again, we need to resort to numerical solution if we wish to solve for $\theta(t)$ , in which case, this reduction to the first order equation does not save much work.

Answer: $\dot{\theta}=\sqrt{\frac{3g}{\ell}}\sin\alpha\sqrt{\frac{\sin\theta_{0}-\sin% \theta}{\sin^{2}\alpha+3(M/m)\cos^{2}(\alpha-\theta)}}$

Problems for 19.2 One-degree-of-freedom 2-D mechanisms

19.2.1 A conservative vibratory system has the following equation of conservation of energy.

m(\ell\dot{\theta})^{2}-mg\ell(1-\cos\theta)+K(a\theta)^{2}=E_{0}

where $E_{0}$ is a constant.

(a)

Obtain the differential equation of motion of this system by differentiating this energy equation with respect to $t$ .
(b)

Determine the circular frequency of small oscillations of the system in part (a). HINT: (Let $\sin\cong\theta\cong$ and $\cos\theta\cong 1-\theta^{2}/2)$ .

19.2.2 A motor at O turns at rate $\omega_{o}$ whose rate of change is $\dot{\omega}_{o}$ . At the end of a stick connected to this motor is a frictionless hinge attached to a second massless stick. Both sticks have length $L$ . At the end of the second stick is a mass $m$ . For the configuration shown, what is $\ddot{\theta}$ ? Answer in terms of $\omega_{o}$ , $\dot{\omega}_{o}$ , $L$ , $m$ , $\theta$ , and $\dot{\theta}$ . Ignore gravity.

Filename:pfigure-blue-107-1

Figure 19.61

19.2.3

In problem 18.3, find

(a)

the angular momentum about point $O$ , Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=\left[mL^{2}\omega_{1}+\frac{1}{2}mr^{2}(% \omega_{1}+\omega_{2})\right]\hat{\boldsymbol{k}}$ .
(b)

the rate of change of angular momentum of the disk about point $O$ , Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}={\accentset{\rightharpoonup}{\boldsymbol{0% }}}$ .
(c)

the angular momentum about point $C$ , Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/C}$}}}=\frac{1}{2}mr^{2}(\omega_{1}+\omega_{2})\hat% {\boldsymbol{k}}$ .
(d)

the rate of change of angular momentum of the disk about point $C$ . Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}={\accentset{\rightharpoonup}{\boldsymbol{0% }}}$ .

Assume the rod is massless and the disk has mass $m$ .

Filename:pfigure-s94h14p4

Figure 19.62

19.2.4 Robot arm, 2-D . The robot arm AB is rotating about point A with $\omega_{AB}=5\,\mathrm{rad}/\mathrm{s}$ and $\dot{\omega}_{AB}=2\,\mathrm{rad}/\mathrm{s}^{2}$ . Meanwhile the forearm BC is rotating at a constant angular speed with respect to AB of $\omega_{BC/AB}=3\,\mathrm{rad}/\mathrm{s}$ . Gravity cannot be neglected. At the instant shown, find the net force acting on the object P which has mass $m=1\,\mathrm{kg}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}=-109.3\,{\mbox{N}}\hat{\boldsymbol{\imath}}-19.54\,{\mbox{N}}% \hat{\boldsymbol{\jmath}}$ .

Filename:pfigure-blue-112-1

Figure 19.63

19.2.5 A crude model for a column, shown in the figure, consists of two identical rods of mass $m$ and length $\ell$ with hinge connections, a linear torsional spring of stiffness $K$ attached to the center hinge (the spring is relaxed when $\theta=0$ ), and a load $P$ applied at the top end.

(a)

Obtain the exact nonlinear equation of motion.
(b)

Obtain the squared natural frequency for small motions $\theta$ .
(c)

Check the stability of the straight equilibrium state $\theta=0$ for all $P\geq 0$ via minimum potential energy. How do these results compare with those from part(b)?

Filename:tfigure8-alt-app2c

Figure 19.64

19.2.6 A thin uniform rod of mass $m$ rests against a frictionless wall and on a frictionless floor. There is gravity.

(a)

Draw a free-body diagram of the rod.
(b)

The rod is released from rest at $\theta=\theta_{0}\neq 0$ . Write the equation of motion of the rod. Answer: $\ddot{\theta}=-\frac{3g}{2L}\cos\theta$ .
(c)

Using the equation of motion, find the initial angular acceleration, $\dot{\boldsymbol{\omega}}_{AB}$ , and the acceleration of the center of mass, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$ , of the rod. Answer: From the diagram, we see $\dot{\boldsymbol{\omega}}_{AB}=-\ddot{\theta}(t=0)\hat{\boldsymbol{k}}=\frac{3% g}{2L}\cos\theta_{0}$ ,

${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}=\frac{3}{4}g\cos\theta_{0}[% \sin\theta_{0}\hat{\boldsymbol{\imath}}-\cos\theta_{0}\hat{\boldsymbol{\jmath}}]$ .
(d)

Find the reactions on the rod at points $A$ and $B$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{A}=\frac{3}{4}mg\sin\theta_{0}% \cos\theta_{0}\hat{\boldsymbol{\imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{B}=mg[1-\frac{3}{4}g\cos^{2}% \theta_{0}]\hat{\boldsymbol{\jmath}}$ .
(e)

Find the acceleration of point $B$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=\frac{3}{2}g\sin\theta_{0}% \cos\theta_{0}\hat{\boldsymbol{\imath}}$ .
(f)

When $\theta=\frac{\theta_{0}}{2}$ , find ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}$ and the acceleration of point $A$ . Answer: At $\theta=\frac{\theta_{0}}{2}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{AB}=\sqrt{\frac{3g}{L}(% \sin\theta_{0}-\sin\frac{\theta_{0}}{2})}\hat{\boldsymbol{k}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=-3g\left[\frac{1}{2}\cos^{2}% \frac{\theta_{0}}{2}+\sin\frac{\theta_{0}}{2}(\sin\theta_{0}-\sin\frac{\theta_% {0}}{2})\right]\hat{\boldsymbol{\jmath}}$ .

Filename:sfig4-6-4a

Figure 19.65

19.2.7 Bar leans on a crooked wall. A uniform $3\,{\mathrm{lbm}}$ bar leans on a wall and floor and is let go from rest. Gravity pulls it down.

(a)

Draw a Free-Body Diagram of the bar.
(b)

Kinematics: find the velocity and acceleration of point B in terms of the velocity and acceleration of point A
(c)

Using equations of motion, find the acceleration of point A.
(d)

Do the reaction forces at A and B add up to the weight of the bar? Why or why not? (You do not need to solve for the reaction forces in order to answer this part.)

Filename:sfig4-6-3

Figure 19.66

19.2.8 Two blocks, each with mass $m$ , slide without friction on the wall and floor shown. They are attached with a rigid massless rod of length $\ell$ that is pinned at both ends. The system is released from rest when the rod makes an angle of 45^∘. What is the acceleration of the block B immediately after the system is released?

Filename:sfig4-6-3a

Figure 19.67

19.2.9 Slider Crank. 2-D . No gravity. Refer to the figure in problem 18.78. What is the tension in the massless rod $A B$ (with length $L$ ) when the slider crank is in the position with $\theta=0$ (piston is at maximum extent)? Assume the crankshaft has constant angular velocity $\omega$ , that the connecting rod $A B$ is massless, that the cylinder walls are frictionless, that there is no gas pressure in the cylinder, that the piston has mass $M$ and the crank has radius $R$ .

Filename:pfigure4-4-rp12

Figure 19.68

19.2.10 In problem 18.84, find the force on the wheel at point $D$ due to the rod.

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Figure 19.69

19.2.11 In problem 18.4, find the force on the rod at point $P$ .

Filename:pfigure-blue-49-2

Figure 19.70

19.2.12 Slider-Crank mechanism. The slider crank mechanism shown is used to push a $2\,{\mathrm{lbm}}$ block P. Arm AB and BC are each $0.5\,\mathrm{ft}$ long. Given that arm AB rotates counterclockwise at a constant 2 revolutions per second, what is the force on P at the instant shown? Answer: There are many solution methods. $-8\pi^{2}$ poundals to the left. (Note: $1$ poundal $=1\,\mathrm{ft}\cdot\,{\mathrm{lbm}}\cdot\,\mathrm{s}^{-2}$ .)

Filename:pfigure4-4-rp13

Figure 19.71

19.2.13 The large masses $m$ at $C$ and $A$ were supported by the light triangular plate ABC, whose corners follow the guide. B enters the curved guide with velocity $V$ . Neglecting gravity, find the vertical reactions (in the $y$ direction) at A and B. (Hint: for the rigid planar body $A B C$ , find ${\ddot{y}}_{A}$ and ${\ddot{y}}_{B}$ in terms of $a_{y}^{cm}$ and $\ddot{\theta}$ , assuming $\dot{\theta}=0$ initially.)

Filename:f92h7p1

Figure 19.72

19.2.14 An idealized model for a car comprises a rigid chassis of mass $M_{c}$ and four identical rigid disks (wheels) of mass $M_{w}$ and radius $R$ , as shown in the figure. Initially in motion with speed $v_{0}$ , the car is momentarily brought to rest by compressing an initially uncompressed spring of stiffness $k$ . Assume no frictional losses.

(a)

Assuming no wheel slip, determine the compression $\Delta$ of the spring required to stop the car. Answer: $\Delta=\sqrt{\frac{(M_{c}+6M_{w})v_{0}^{2}}{k}}$ .
(b)

While the car is in contact with the spring but still moving forward, in which direction is the tangential force on any one of the wheels (due to contact with the ground) pointing? Why? (Illustrate with a sketch). Answer: Tangential forces point toward the wall!
(c)

Repeat part (a) assuming now that the ground is perfectly frictionless from points A to B shown in the figure. Answer: $\Delta=\sqrt{\frac{(M_{c}+4M_{w})v_{0}^{2}}{k}}$ .

Filename:pfigure-blue-52-2

Figure 19.73

19.2.15 Assume Greg Lemond’s riding “tuck” was so good that you can neglect air resistance when you think about him and his bike. Further, you can regard his and the bike’s combined mass (all 70 $\,\mathrm{kg})$ ) as concentrated at a point in his stomach somewhere. Greg’s left foot has just fallen off the pedal so he is only pedaling with his right foot, which at the moment in question is at its lowest point in the motion. You note that, relative to the ground the right foot is only going 3/4 as fast as the bike (since it is going backwards relative to the bike), though you can’t make out all the radii of his frictionless gears and rigid round wheels. Greg, ever in touch with his body, tells you he is pushing back on the pedal with a force of 70 N. You would like to know Greg’s acceleration.

(a)

In your first misconceived experiment you set up a 70 $\,\mathrm{kg}$ bicycle in your laboratory and balance it with strings that cause no fore or aft forces. You tie a string to the vertically down right pedal and pull back with a force of 70 N. What acceleration do you measure?
(b)

What is Greg’s actual acceleration? [Hint: Greg’s massless leg is pushing forward on his body with a force of 70 N] You can neglect the mass of the wheels and other transmission parts (chain, crank, etc).

Filename:pfigure4-4-rp16

Figure 19.74

19.2.16 Which way does the bike accelerate? A bicycle with all frictionless bearings is standing still on level ground. A horizontal force $F$ is applied on one of the pedals as shown. There is no slip between the wheels and the ground. The bicycle is gently balanced from falling over sideways. It is heavy enough so that both wheels stay on the ground. Does the bicycle accelerate forward, backward, or not at all? Make any reasonable assumptions about the dimensions and mass. Justify your answer as clearly as you can, clearly enough to convince a person similar to yourself but who has not seen the experiment performed.

Filename:pfigure4-4-rp14

Figure 19.75

19.3 Dynamics of multi-degree-of-freedom 2-D mechanisms

A typical machine has many parts. They may work in concert as a one degree-of-freedom system or they may be designed to move independently.

One degree-of-freedom machines. Some mechanisms are designed so the parts work together to do something, one thing, well. A car going straight down a road has dozens of moving parts: cylinders, connecting rods, crank shaft, cam shaft, transmission gears, drive shaft, differential, wheel axles, wheels and the car itself. All of these parts move, each in its own different way, approximately like a rigid object. But in the simplest description

^†^†margin: A car has one degree-of-freedom? For studying the acceleration of a car going straight one might ignore the flexibility of the drive shaft and the bounce of the suspension, etc.thus leading to a 1 DOF model. But if you want to see how acceleration causes the car to tip back, or how the car oscillates when the engine lugs, then a 2, or more, DOF model would be appropriate.

they move as a one-degree-of freedom mechanism towards the end of fighting friction and moving the car down the road. If the crankshaft rotates a small amount, each part moves a corresponding amount; the amount of rotation of the crank shaft determines the position of all of the parts. And all of the velocities and accelerations of all of the bits of mass in the car can be determined by the rotation $\theta_{\text{crank}}$ and its derivatives ( $dot\theta_{\text{crank}}$ and $ddot\theta_{\text{crank}}$ ). Thus a car, even with lots of pieces moving this way and that, might be well-modeled for some purposes as a one degree-of-freedom mechanism. For one-degree-of-freedom mechanisms the methods of Section 19.2 may be appropriate.

Multi DOF systems In some modeling of machines it is important to keep track of multiple degrees of freedom (See fig. 18.65 on page 18.65 for examples of simple mechanisms). There are various reasons that a multi-DOF analysis is relevant:

•
Some machines intentionally have various ways of moving that are controlled by various motors. Classic examples include:
- –
  
  Robots,
- –
  
  Animal bodies.
•

Some systems intentionally have various degrees of freedom so as to respond smoothly to disturbances, for example the suspensions on the bottoms of cars and washing machines.
•

Some systems have undesirable degrees of freedom due to the parts which are nominally rigid actually being elastic. Thus a machine which is intended to have one degree of freedom might have various undesirable vibration modes.

Mechanical analysis of multi-DOF systems. To solve problems with multiple degrees of freedom the basic strategy is, as described at the start of the chapter:

1.

Draw FBDs of each object,
2.

Pick configuration variables,
3.

Write linear and angular momentum balance equations
4.

Solve the equations for variables of interest (usually forces and second derivatives of the configuration variables).
5.

Set up and solve the resulting differential equations (if you are trying to find the motion).

There are two basic approaches to these multi-object problems which, for lack of better language we call “brute-force” and “clever”.

A.

In the brute force approach you write three times as many scalar balance equations as you have objects (limiting attention to 2D where each object has 3 DOFs and 3 independent momentum balance equations). That is, for example, for each free-body diagram you write linear momentum balance and angular momentum balance about the center of mass. Then you take this set of $3n$ equations and add and subtract them to solve for variables of interest.

This approach is quite suitable for computers so most commercial general purpose dynamic simulators use a variant of this approach. For individual use with packaged software, the brute-force approach is generally both more reliable and more time consuming.
B.

In the clever approach you write as many scalar momentum balance equations as you have unknowns. For example, if you have 2 degrees of freedom and you are concerned with motions and not reaction forces, you write 2 equations. You do this by finding momentum balance equations that do not include the variables you are not interested in. Usually this involves using angular momentum balance about hinge points, or linear momentum balance orthogonal to sliding contacts.

The clever approach does not always work; the four-bar linkage is the classic problem case. However the desire to find such minimal sets of equations of motion it is historically important

^†^†margin: Attempts to automate the clever approach, that is to quickly find minimal equations of motion, led to Lagrange equations which led to Hamilton’s equations which led to quantum mechanics (but we won’t be that clever here).

.

At this point in the subject there are no quick simple problems. All problems are involved, especially if taken from start all the way to plotting solutions to the differential equations. The examples that follow emphasize getting to the equations of motion. The skills for numerically solving the differential equations and plotting the solutions are the same as from the start of dynamics so are not discussed until the sample problems which show all the work from setup to solution.

Example: Block sliding on sliding block: clever approach

Block 1 with mass $m_{1}$ rolls without friction on ideal massless wheels at A and B (see fig. 19.76). Block 2 with mass $m_{2}$ rolls down the tipped top of block 1 on ideal massless rollers at C and D. The locations of G₁ relative to A and B, and of G₂ relative to C and D are known. How do blocks 1 and 2 move?

First look at the free-body diagram of the system and note that there are no unknown forces in the $\hat{\boldsymbol{\imath}}$ direction. So, for the system

$\displaystyle\Bigm{\{}\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{i}}$	$\displaystyle=$	$\displaystyle\dot{{\accentset{\rightharpoonup}{\boldsymbol{L}}}}\Bigm{\}}\cdot% \hat{\boldsymbol{\imath}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle m_{1}\ddot{x}+m_{2}(\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{y}^{% \prime}\hat{\boldsymbol{\jmath}}^{\prime})\cdot\hat{\boldsymbol{\imath}}$	(19.47)
	$\displaystyle=$	$\displaystyle(m_{1}+m_{2})\ddot{x}-\ddot{y}^{\prime}m_{2}\cos\theta.$	(19.47)

Looking at the free-body diagram of mass 2 note that there are no unknown forces in the $\hat{\boldsymbol{\jmath}}^{\prime}$ direction, so

Filename:pfigure4-4-rp15

Figure 19.76: Block 2 rolls on block 1 which rolls on the ground. All rollers are ideal (frictionless and massless). The moving

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

frame moves with the lower block and is oriented with the slope.

$\displaystyle\Bigm{\{}\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{i}}$	$\displaystyle=$	$\displaystyle\dot{{\accentset{\rightharpoonup}{\boldsymbol{L}}}}\Bigm{\}}\cdot% \hat{\boldsymbol{\jmath}}^{\prime}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m_{2}g\sin\theta$	$\displaystyle=$	$\displaystyle m_{2}(\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{y}^{\prime}\hat{% \boldsymbol{\jmath}}^{\prime})\cdot\hat{\boldsymbol{\jmath}}^{\prime}$
$\displaystyle m_{2}g\sin\theta$	$\displaystyle=$	$\displaystyle m_{2}(-\cos\theta\,\ddot{x}+\ddot{y}^{\prime}).$	(19.48)

Eqns. 19.47 and 19.48 are a system of two equations in the two unknowns $\ddot{x}$ and $\ddot{y}^{\prime}$

\begin{array}[]{rrcl}(m_{1}+m_{2})\,\ddot{x}&-m_{2}\cos\theta\,\ddot{y}^{% \prime}&=&0\\ -m_{2}\cos\theta\,\ddot{x}&+m_{2}\,\ddot{y}^{\prime}&=&m_{2}g\sin\theta\\ \end{array}

which can be solved for $\ddot{x}$ and $\ddot{y}^{\prime}$ by hand or on the computer. Finding $x(t)$ and $y^{\prime}(t)$ is then easy because both $\ddot{x}$ and $\ddot{y}^{\prime}$ are constants.

Now we look at the same example, but proceed in a more naive manner.

Example: Block sliding on sliding block: brute force approach

Now we look at the free-body diagrams of the two separate blocks. We will use 3 balance equations from each free-body diagram, taking account of the kinematic constraints.

For the lower block we have

$\displaystyle{\rm AMB}_{G_{1}}$	$\Rightarrow$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/G_{1}% }}=\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/G_{1}}}$
		$\displaystyle\mbox{where}\qquad\sum{\accentset{\rightharpoonup}{\boldsymbol{M}% }}_{\mathrm{/G_{1}}}=\phantom{+}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {\!\mathrm{D/G_{1}}}\times(-F_{D}\hat{\boldsymbol{\imath}}^{\prime})$
		$\displaystyle\phantom{\mbox{where}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathrm{/G_{1}}}=}+{\accentset{\rightharpoonup}{\boldsymbol{% r}}}_{\!\mathrm{C/G_{1}}}\times(-F_{C}\hat{\boldsymbol{\imath}}^{\prime})$
		$\displaystyle\phantom{\mbox{where}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathrm{/G_{1}}}=}+{\accentset{\rightharpoonup}{\boldsymbol{% r}}}_{\!\mathrm{B/G_{1}}}\times F_{B}\hat{\boldsymbol{\jmath}}$
		$\displaystyle\phantom{\mbox{where}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathrm{/G_{1}}}=}+{\accentset{\rightharpoonup}{\boldsymbol{% r}}}_{\!\mathrm{A/G_{1}}}\times F_{A}\hat{\boldsymbol{\jmath}}$
		$\displaystyle\mbox{and}\qquad\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}% _{\mathrm{/G_{1}}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\mbox{\qquad( % ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{1}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$)}$
$\displaystyle{\rm and\ LMB}$	$\Rightarrow$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}% }=\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$
		$\displaystyle\mbox{where}\qquad\sum{\accentset{\rightharpoonup}{\boldsymbol{F}% }}_{\!\!\mathrm{i}}=-F_{D}\hat{\boldsymbol{\imath}}^{\prime}-F_{C}\hat{% \boldsymbol{\imath}}^{\prime}+F_{B}\hat{\boldsymbol{\jmath}}+F_{A}\hat{% \boldsymbol{\jmath}}-m_{1}g\hat{\boldsymbol{\jmath}}$
		$\displaystyle\mbox{and}\qquad\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}% =m_{1}\ddot{x}\hat{\boldsymbol{\imath}}$

Similarly for the upper block:

$\displaystyle{\rm AMB}_{G_{2}}$	$\Rightarrow$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/G_{2}% }}=\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/G_{2}}}$
		$\displaystyle\mbox{where}\qquad\sum{\accentset{\rightharpoonup}{\boldsymbol{M}% }}_{\mathrm{/G_{1}}}=\phantom{+}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {\!\mathrm{D/G_{2}}}\times F_{D}\hat{\boldsymbol{\imath}}^{\prime}$
		$\displaystyle\phantom{\mbox{where}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathrm{/G_{2}}}=}+{\accentset{\rightharpoonup}{\boldsymbol{% r}}}_{\!\mathrm{C/G_{2}}}\times F_{C}\hat{\boldsymbol{\imath}}^{\prime}$
		$\displaystyle\mbox{and}\qquad\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}% _{\mathrm{/G_{2}}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\mbox{\qquad( % ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{1}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$)}$
$\displaystyle{\rm and\ LMB}$	$\Rightarrow$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}% }=\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$
		$\displaystyle\mbox{where}\qquad\sum{\accentset{\rightharpoonup}{\boldsymbol{F}% }}_{\!\!\mathrm{i}}=F_{D}\hat{\boldsymbol{\imath}}^{\prime}+F_{C}\hat{% \boldsymbol{\imath}}^{\prime}-m_{2}g\hat{\boldsymbol{\jmath}}^{\prime}$
		$\displaystyle\mbox{and}\qquad\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}% =m_{2}(\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{y}^{\prime}\hat{\boldsymbol{% \jmath}}^{\prime})$

Eqns. 19.3-19.3 can be written as scalar equations by dotting the LMB equations with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ and the AMB equations with $\hat{\boldsymbol{k}}$ . All is known in these 6 equations but the six scalars: $F_{A},F_{B},F_{C},F_{D},\ddot{x},$ and $\ddot{y}^{\prime}$ . These could be set up as a matrix equation and solved on the computer, or you could try to find your way through by adding and subtracting equations. In any case you could solve for $\ddot{x}$ and $\ddot{y}^{\prime}$ and thus have differential equations to solve to find the motions.

One quick inference one can make is from looking at the equations: no term in the coefficients of the unknowns depends on $x,\dot{x},y,$ or $\dot{y}^{\prime}$ . So all of the reactions $F_{A},F_{B},F_{C},$ and $F_{D}$ as well as the accelerations $\ddot{x}$ and $\ddot{y}^{\prime}$ are constants in time (until the upper mass hits the ground).

Example: Block sliding on sliding block: even more brute force

This “multi-body” problem can be solved in an even more naive and more brute force manner. The method is the same as shown in Section 14.1 on page 14.1 for a one dimensional problem.

We would use 6 configuration variables: the $x$ and $y$ coordinates of $G_{1}$ and $G_{2}$ and the rotations of the two bodies:

x_{1},y_{1},\theta_{1},x_{2},y_{2},\hbox to36.135pt{\hfil and\hfil}\theta_{2}

That the two bodies don’t rotate would be expressed indirectly by noting that the accelerations of points A and B on the lower mass must be in the $\hat{\boldsymbol{\imath}}$ direction. These two equations would be added to the 6 linear and angular momentum balance equations. Similar constraint equations would be written for the interactions at C and D. Altogether there are now 6 configuration variables and 4 constraint forces. But there are 6 differential equations of motion and 4 constraint equations. Thus at one instant in time a set of 10 simultaneous equations needs to be solved. Then these are used to evaluate the right hand sides in the differential equations.

These 10 equations are an impractical mess for solving one simple problem like this. But these equations lend themselves to easy automation and this is closest to the approach used by general purpose dynamics simulators.

The example above is particularly simple because block 1 moves in a straight line without rotating and block 2 moves in a straight line without rotating relative to block 1. Even for a system of just two objects the situation could be much more complex if the first object had a complex motion and the second a complex motion relative to the first. But because of the preponderance of hinges in the world, circular motion, and motion relative to circular motion, is the most complex motion that need be considered by many engineers. Here is a version of the most common example of that class.

Example: A two link robot arm: finesse the finding of reactions

Filename:tfigure4-spherical-rotaxis

Figure 19.77: A two link robot arm. Free-body diagrams are shown of the whole system (including the motor torque at the shoulder

M_{s}

) and of the fore-arm (including the motor torque at the elbow

M_{e}

) .

A robot arm has two links. There are motors that apply known torque $M_{s}$ at the shoulder (reacted by the base and a torque $M_{e}$ at the elbow (reacted by the upper arm). Dimensions are as marked. This system has 2 degrees of freedom. So we need 2 configuration variables and 2 independent balance equations to find the motion.

The natural configuration variables are the angles of the upper arm relative to a fixed reference and the angle of the lower arm relative to a fixed reference. It would also be natural to instead use the angle of the lower arm relative to the upper arm. This leads to simpler equations in the end, but more work in set up.

Angular momentum balance for the system about the shoulder contains no unknown reaction forces, nor does angular momentum balance of the fore-arm about the hinge. So we base our work on these two equations:

	$\displaystyle\mbox{System AMB${}_{/O}$}\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/O}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/O}}$		(19.53)
	$\displaystyle\mbox{Forearm AMB${}_{/E}$}\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/E}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/E}}.$		(19.54)

The goal, equations of motion, is reached by evaluating the left and right sides of these equations in terms of known geometric and mass quantities as well as the configuration variables When we write $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/O}}$ and $\sum\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/O}}$ we implicitly mean for the whole system. And $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/E}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/E}}$ apply just to the forearm.

At each step in the calculations below imagine the results can be substituted into the later steps. We don’t do that here because the expressions grow in size. Further, if the angles and their rates of change are known, as they are when doing most dynamics problems, the intermediate calculations will result in numbers, and these do not become more numerous as the calculation proceeds.

$\displaystyle\hat{\boldsymbol{\lambda}}_{1}=\cos\theta_{1}\hat{\boldsymbol{% \imath}}+\sin\theta_{1}\hat{\boldsymbol{\jmath}}$	and	$\displaystyle\hat{\boldsymbol{\lambda}}_{2}=\cos\theta_{2}\hat{\boldsymbol{% \imath}}+\sin\theta_{2}\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{1}/O}% }=\ell_{1}\hat{\boldsymbol{\lambda}}_{1},\ {\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{E/O}}=\ell_{3}\hat{\boldsymbol{\lambda}}_{1}$	and	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{2}/E}% }=\ell_{2}\hat{\boldsymbol{\lambda}}_{2}$
	and	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{2}/O}% }={\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{E/O}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{2}/E}}.$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{1}/O}}=% -\dot{\theta}_{1}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{% G_{1}/O}}+\ddot{\theta}_{1}\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{1}/O}}$	,
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{E/O}}=-% \dot{\theta}_{1}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{E% /O}}+\ddot{\theta}_{1}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{E/O}}$	,
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{2}/E}}=% -\dot{\theta}_{2}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{% G_{2}/E}}+\ddot{\theta}_{2}\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{2}/E}}$	and	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{2}/O}}=% {\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{E/O}}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{2}/E}}.$

These terms are all we need to evaluate the 4 terms in Eqns. 19.53 and 19.54.

			$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/O}}={% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{1}/O}}\times(m_{1}g% \hat{\boldsymbol{\imath}})+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{G_{2}/O}}\times(-m_{2}g\hat{\boldsymbol{\jmath}})+M_{s}\hat{% \boldsymbol{k}},$
			$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/E}}={% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{2}/E}}\times(m_{2}g% \hat{\boldsymbol{\imath}})+M_{e}\hat{\boldsymbol{k}},$
			$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/O}}=m% _{1}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{1}/O}}\times{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{1}/O}}+\ddot{\theta}_% {1}I_{1}\hat{\boldsymbol{k}}+m_{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{\!\mathrm{G_{2}/O}}\times{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \mathrm{G_{2}/O}}+\ddot{\theta}_{2}I_{2}\hat{\boldsymbol{k}},$
		and	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/E}}=m% _{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{2}/E}}\times{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{2}/O}}+\ddot{\theta}_% {2}I_{2}\hat{\boldsymbol{k}}.$

Once these are substituted into Eqns. 19.53 and 19.54 one has 2 vector equations with only $\hat{\boldsymbol{k}}$ components. In other words we have two scalar equations in the two unknowns $\ddot{\theta}_{1}$ and $\ddot{\theta}_{2}$ . One can go through the algebra and solve for them explicitly, but the expressions are quite complex, even when simplified. At given values of $\theta_{1},\dot{\theta}_{1},\theta_{2},$ and $\dot{\theta}_{2}$ however these are just two linear equations in two unknowns.

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 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 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.

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. Why? Because finding all of the bar angles from one angle is a trigonometric mess. And differentiating that mess once (for velocities) and then once again (for accelerations) makes a huge mess.

SAMPLE 19.15

Filename:tfigure5-7

Figure 19.78:

Dynamics of sliding wedges. A wedge shaped body of mass $m_{2}$ sits on a frictionless ground. Another wedge shaped body of mass $m_{1}$ is gently placed on the inclined face of the stationary wedge. The top wedge starts to slide down. The coefficient of friction between the two wedges is $\mu$ . Find the sliding acceleration of the top wedge along the incline (i.e., the relative acceleration of $m_{1}$ with respect to $m_{2}$ ).

Solution

Filename:tfigure5-gen-rigid-body

Figure 19.79:

The free-body diagrams of the two wedges are shown in fig. 19.79. Note that the friction force is $\mu N$ since the wedges are sliding with respect to each other (if they were not sliding already then the friction force is an unknown force $F\leq\mu N$ ). Let the absolute acceleration of $m_{2}$ be ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{2}}=a_{2}\hat{% \boldsymbol{\imath}}$ . Then, the absolute acceleration of $m_{1}$ is ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{1}}={\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{2}}+{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{\mathrm{1/2}}=a_{2}\hat{\boldsymbol{\imath}}+a_{\rm rel}\hat% {\boldsymbol{\lambda}}$ . Now, we can write the linear momentum balance for $m_{1}$ and $m_{2}$ as follows.

	$\displaystyle N\hat{\boldsymbol{n}}-m_{1}g\hat{\boldsymbol{\jmath}}-\mu N\hat{% \boldsymbol{\lambda}}$	$\displaystyle=$	$\displaystyle m_{1}(a_{2}\hat{\boldsymbol{\imath}}+a_{\rm rel}\hat{\boldsymbol% {\lambda}})$		(19.55)
	$\displaystyle(R-m_{2}g)\hat{\boldsymbol{\jmath}}-N\hat{\boldsymbol{n}}+\mu N% \hat{\boldsymbol{\lambda}}$	$\displaystyle=$	$\displaystyle m_{2}a_{2}\hat{\boldsymbol{\imath}}$		(19.56)

where $\hat{\boldsymbol{\lambda}}=\cos\alpha\hat{\boldsymbol{\imath}}-\sin\alpha\hat{% \boldsymbol{\jmath}}$ and $\hat{\boldsymbol{n}}=\sin\alpha\hat{\boldsymbol{\imath}}+\cos\alpha\hat{% \boldsymbol{\jmath}}$ . Here, we have 4 independent scalar equations (from the two 2-D vector equations) in four unknowns $N,R,a_{2}$ , and $a_{\rm rel}$ . Thus, we can certainly solve for them. We are, however, only interested in $a_{\rm rel}$ . So, we should try to find the answer with fewer calculations. Dotting eqn. (19.55) with $\hat{\boldsymbol{\lambda}}$ , we have

m_{1}a_{\rm rel}=-m_{1}a_{2}\cos\alpha-\mu N+m_{1}g\sin\alpha

(19.57)

So, to find $a_{\rm rel}$ , we need $a_{2}$ and $N$ . Dotting eqn. (19.55) with $\hat{\boldsymbol{n}}$ , we have

m_{1}a_{2}\sin\alpha=N-m_{1}g\cos\alpha,

(19.58)

and dotting eqn. (19.56) with $\hat{\boldsymbol{\imath}}$ , we have

m_{2}a_{2}=-N\sin\alpha+\mu N\cos\alpha.

(19.59)

Solving eqn. (19.58) and (19.59) simultaneously, and using new variables (for convenience) $M=m_{1}/m_{2},\,C=\cos\alpha$ , and $S=\sin\alpha$ , we get

a_{2}=\frac{MC(\mu C-S)}{1-MS(\mu C-S)}\,g,\quad N=m_{1}g\,\frac{C}{1-MS(\mu C% -S)}

Substituting these expression in eqn. (19.57), we get

a_{\rm rel}=gS-\frac{gC\bigl{[}MC(\mu C-S)-\mu\bigr{]}}{1-MS(\mu C-S)}

Answer: $a_{\rm rel}=g\sin\alpha-\frac{g\cos\alpha\bigl{[}\frac{m_{1}}{m_{2}}\cos\alpha% (\mu\cos\alpha-\sin\alpha)-\mu\bigr{]}}{1-\frac{m_{1}}{m_{2}}\sin\alpha(\mu% \cos\alpha-\sin\alpha)}$

Note that when there is no friction $(\mu=0)$ , the expression for $a_{\rm rel}$ reduces to

a_{\rm rel}=gS+\frac{gMC^{2}S}{1+MS^{2}}

and if we let $m_{2}\rightarrow\infty$ (i.e., $m_{2}$ represents fixed ramp) so that $M\rightarrow 0$ , then $a_{\rm rel}=g\sin\alpha$ which is the acceleration of a point mass down a frictionless ramp of slope $\tan\alpha$ .

SAMPLE 19.16

Filename:tfigure5-term1-a

Figure 19.80:

Dynamics of a new gun. A new gun consists of a uniform rod AB of mass $m_{1}$ and a small collar C of mass $m_{2}$ that slides freely on the rod. A motor at A rotates the rod with constant torque $T$ .

1.

Find the equations of motion of the collar.
2.

Show that if $T=0$ then the equations of motion imply conservation of angular momentum about point A.

Solution

Filename:tfigure5-term1-b

Figure 19.81:

Let us denote the configuration of the collar with $R$ , the radial distance from the fixed point A along the rod, and $\theta$ , the angular displacement of the rod. We need to find differential equations that determine $R$ and $\theta$ as functions of time. The free-body diagram of the whole system (rod and collar together) and that of the collar is shown in fig. 19.81. We can write angular momentum balance for the whole system about point A so that the unknown reaction force F at A does not enter the equations. Noting that the acceleration of the collar is ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}=(\ddot{R}-R\dot{% \theta}^{2})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}% +R\ddot{\theta})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ , and letting $I_{1}\equiv I^{A}_{z}z$ be the moment of inertia of the rod about A, we have,

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{A}}$
$\displaystyle T\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle I_{1}\ddot{\theta}\hat{\boldsymbol{k}}+R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}\times m_{2}\bigl{[}(\ddot{R}-R\dot{\theta}^{2})\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}+(2\dot{R}\dot{\theta}+R\ddot{\theta})% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}\bigr{]}$
	$\displaystyle=$	$\displaystyle I_{1}\ddot{\theta}\hat{\boldsymbol{k}}+m_{2}R^{2}\ddot{\theta}% \hat{\boldsymbol{k}}+2m_{2}R\dot{R}\dot{\theta}\hat{\boldsymbol{k}}$

Dotting this equation with $\hat{\boldsymbol{k}}$ , we have

\ddot{\theta}=\frac{T}{I_{1}+m_{2}R^{2}}-\frac{2m_{2}R}{I_{1}+m_{2}R^{2}}\dot{% R}\dot{\theta}.

(19.60)

Thus we have obtained the equation of motion for $\theta$ . Now we consider the free-body diagram of the collar alone and write the linear momentum balance for it in the $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ direction, i.e., $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\cdot(\sum{\accentset{% \rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{\boldsymbol{a}% }})$ , so that we do not have to care about the unknown normal reaction ${\accentset{\rightharpoonup}{\boldsymbol{N}}}$ . So, we have,

$\displaystyle 0$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\cdot m_{2}\bigl{[}(% \ddot{R}-R\dot{\theta}^{2})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+(2\dot% {R}\dot{\theta}+R\ddot{\theta})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!% \theta}\bigr{]}$
	$\displaystyle=$	$\displaystyle m_{2}(\ddot{R}-R\dot{\theta}^{2})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{R}$	$\displaystyle=$	$\displaystyle R\dot{\theta}^{2}.$	(19.61)

Thus we have the required equations of motion. Note that eqn. (19.60) and (19.61) are coupled nonlinear differential equations. So, to find $\theta(t)$ and $R(t)$ we need to solve them numerically.

Answer: $\ddot{\theta}=\frac{T}{I_{1}+m_{2}R^{2}}-\frac{2m_{2}R}{I_{1}+m_{2}R^{2}}\dot{% R}\dot{\theta},\qquad\ddot{R}=R\dot{\theta}^{2}$

2.

Now we set $T=0$ in our equations of motion. Note that the equation for $R$ is independent of $T$ . The equation for $\theta$ becomes

$\ddot{\theta}=-\frac{2m_{2}R}{I_{1}+m_{2}R^{2}}\dot{R}\dot{\theta}.\hbox to36.% 135pt{\hfil$\Rightarrow$\hfil}(I_{1}+m_{2}R^{2})\ddot{\theta}+2m_{2}R\dot{R}% \dot{\theta}=0$

But the last expression is simply $\dot{H}_{A}$ for the system. Thus we have $\dot{H}_{A}=0$ which implies that $H_{A}=$ constant. That is conservation of angular momentum about point A.

SAMPLE 19.17 Numerical solutions of new gun equations. Consider Sample 19.79 again. Set up the equations of motion for numerical solution. Take $T=1\,{\rm N}\mathord{\cdot}{\rm m},\ell=1\,\mathrm{m},m_{2}=1\,\mathrm{kg}$ , and $m_{1}=m_{2}/3$ . Carry out numerical solutions for the following cases.

1.

Let the system start from rest at $\theta=0$ and $R=0.1\,\mathrm{m}$ . Find the solution from $t=0$ to $t=1\,\mathrm{s}$ . Plot $R(t),\,\theta(t)$ and $R(\theta)$ (in polar coordinates).
2.

Find the solution till the collar leaves the rod. What is the speed of the collar at this instant?
3.

Compute and plot the total energy of the system as a function of time. Also, plot the work done by the torque as a function of time and show that the work done is equal to the total energy of the system at each instant.
4.

Vary torque $T$ and carry out solutions for several values of $T$ . Find the terminal value of $\theta_{f}$ (when the collar leaves the rod) for each $T$ . Justify your observation about $\theta_{f}$ by plotting $\dot{R}/\dot{\theta}$ as a function of $T$ .

Solution

We first need to write the equations of motion, eqn. (19.60) and (19.61), as a set of first order ODEs. We can easily do so by introducing new variables $\omega\equiv\dot{\theta}$ and $v_{R}\equiv\dot{R}$ , so that we have,

\left(\begin{array}[]{c}\dot{\theta}\\ \dot{\omega}\\ \dot{R}\\ \dot{v}_{R}\end{array}\right)=\left(\begin{array}[]{c}\omega\\ \frac{T}{I_{1}+m_{2}R^{2}}-\frac{2m_{2}R}{I_{1}+m_{2}R^{2}}v_{R}\omega\\ v_{R}\\ R\omega^{2}\end{array}\right)

Given the values of all constants, we only need to specify the initial conditions for $\theta,\omega,R$ , and $v_{R}$ for solving these equations numerically.

We use the following pseudocode to carry out the numerical solution.

     Set T = 1, L = 1, m2 = 1, m1 = m2/3
     Let I1 = m1*L^2/3, I2 = m2*R^2,
     ODEs = {thetadot = w,
             wdot = (T-2*m2*R*vR*w)/(I1+I2),
             Rdot = vR,
             vRdot = R*w^2}
     IC = {theta = 0, w = 0, R = 0.1, vR = 0}ΨΨ
     Solve ODEs with IC for t=0 to t=1
     Plot t vs R, Plot t vs theta,
     Polarplot theta vs R

Filename:sfig1-2-12

Figure 19.82:

Filename:sfig4-6-5

Figure 19.83:

The $R(t)$ and $\theta(t)$ plots obtained from the numerical solution are shown in fig. 19.83 and the polar plot of $R(\theta)$ is shown in fig. 19.82.

2.

We do not know a priori the value of $t$ at which the collar leaves the rod. So, we have to carry out the solution for some assumed $t_{f}$ which gives us $R(t_{f})>\ell$ so that we know the collar has gone past the end of the rod. We then plot $R(t)$ , including the unreal value of $R(t_{f})>\ell$ , and find the time $t$ at which $R(t)=\ell$ , either by zooming into the graph or by interpolation (although, there are various sophisticated algorithms to find this $t$ ). Following the method of zooming into the graph

(see fig. 19.84)

we find the terminal value of $t$ to be 1.147 $\,\mathrm{s}$ . We carry out the numerical solution again from $t=0$ to $t_{f}=0.147\,\mathrm{s}$ and find that $R(t_{f})=1\,\mathrm{m},\,v_{R}(t_{f})=2.13\,\mathrm{m}/\mathrm{s}$ , and $\omega(t_{f})=1.03\,\mathrm{rad}/\mathrm{s}$ , so that $v_{f}=\sqrt{\dot{R}^{2}+(R\dot{\theta})^{2}}=2.37\,\mathrm{m}/\mathrm{s}.$ This is the terminal speed of the collar.

Filename:sfig4-6-5a

Figure 19.84: Finding the time $t$ at which the collar leaves the rod from the graph of $R(t)$ .

Filename:sfig4-6-5b

Figure 19.85: Angle $\theta$ at which the collar leaves the rod vs torque $T$ .
3.

The work done by the torque is $W=T\theta$ at any instant. The system only possesses kinetic energy. So the energy of the system at any instant is $E=E_{1}+E_{2}$ where $E_{1}=\frac{1}{2}I_{1}\omega^{2}$ and $E_{2}=\frac{1}{2}m_{2}(\dot{R}^{2}+R^{2}\omega^{2})$ . Computing these quantities for the solution obtained above, we plot $W$ and $E$ vs $t$ as shown in fig. 19.86. Clearly, $W=E$ .

Filename:sfig4-6-5c

Figure 19.86: Work done by the torque and the kinetic energy of the system.

Filename:sfig4-6-5d

Figure 19.87: The ratio of terminal $\dot{R}/\dot{\theta}$ vs torque $T$ .
4.

Now we take several values of $T$ (0.1, 0.5, 1, 1.5, 2, 2.5, and 3) and carry out the numerical solutions for each $T$ . We note the terminal values of $\theta,\omega(=\dot{\theta})$ , and $v_{R}(=\dot{R})$ . By plotting the terminal value of $\theta$ against $T$ (fig. 19.85), we see that the collar leaves the rod at exactly the same $\theta=2.86\,{\rm rad}$ for each $T$ ! But this is possible only if $\dot{R}$ and $\dot{\theta}$ both change in the same proportion for each $T$ . So, plot the ratio $\dot{R}/\dot{\theta}$ just for the terminal values against $T$ and find that the ratio is indeed constant (see fig. 19.87).

SAMPLE 19.18

Filename:sfig4-5-6

Figure 19.88:

Dynamics of a sliding-base pendulum. A cart of mass $M$ slides down a frictionless inclined plane as shown in the figure. A simple pendulum of mass $m$ and length $\ell$ hangs from the center-of-mass of the cart. Find the equation of motion of the pendulum.

Solution

Filename:sfig4-5-6a

Figure 19.89:

Let us measure the angular displacement of the pendulum with respect to the cart with angle $\theta$ measured anticlockwise from the normal to the inclined plane. Let $s$ be the position of the cart along the inclined plane from some reference point. Then, the acceleration of the cart can be written as ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}=\ddot{s}\hat{% \boldsymbol{\lambda}}$ and the acceleration of the pendulum mass as ${\accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{\mathrm{C}}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \mathrm{rel}}=\ddot{s}\hat{\boldsymbol{\lambda}}+\ell\ddot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}-\ell\dot{\theta}^{2}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}$ .

The free-body diagram of the cart and the pendulum system is shown in fig. 19.89. Writing angular momentum balance of the system about point C, we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{C}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{C}}$
$\displaystyle\ell\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times(-mg\hat{% \boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle\ell\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times m(\ddot{s}% \hat{\boldsymbol{\lambda}}+\ell\ddot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}-\ell\dot{\theta}^{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R})$
$\displaystyle-mgl\sin(\theta-\alpha)\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle m\ddot{s}\ell\cos\theta\hat{\boldsymbol{k}}+m\ell^{2}\ddot{% \theta}\hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{\theta}$	$\displaystyle=$	$\displaystyle-\frac{g}{\ell}\sin(\theta-\alpha)-\frac{\ddot{s}}{\ell}\cos\theta.$	(19.62)

To find $\ddot{s}$ , we write the linear momentum balance for the whole system in the $\hat{\boldsymbol{\lambda}}$ (so that we do not involve the unknown normal reaction $N$ ) direction.

$\displaystyle\hat{\boldsymbol{\lambda}}\cdot(-Mg\hat{\boldsymbol{\jmath}}-mg% \hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\lambda}}\cdot\bigl{[}M\ddot{s}\hat{\boldsymbol{% \lambda}}+m(\ddot{s}\hat{\boldsymbol{\lambda}}+\ell\ddot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}-\ell\dot{\theta}^{2}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R})\bigr{]}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}(M+m)g\sin\sin\alpha$	$\displaystyle=$	$\displaystyle(M+m)\ddot{s}+m\ell\ddot{\theta}\cos\theta-m\ell\dot{\theta}^{2}\sin\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{s}$	$\displaystyle=$	$\displaystyle g\sin\alpha-\frac{m\ell}{M+m}(\ddot{\theta}\cos\theta+\dot{% \theta}^{2}\sin\theta)$

Substituting eqn. (LABEL:eq:cartpend.s) in eqn. (19.62) and rearranging terms, we get

\ddot{\theta}=-\frac{g}{\ell}\sin\theta\frac{(1+\frac{m}{M})\cos\alpha}{1+% \frac{m}{M}\sin^{2}\theta}+\frac{\frac{m}{M}\dot{\theta}^{2}\sin\theta\cos% \theta}{1+\frac{m}{M}\sin^{2}\theta}

Answer: $\ddot{\theta}=-\frac{g}{\ell}\sin\theta\frac{(1+\frac{m}{M})\cos\alpha}{1+% \frac{m}{M}\sin^{2}\theta}+\frac{\frac{m}{M}\dot{\theta}^{2}\sin\theta\cos% \theta}{1+\frac{m}{M}\sin^{2}\theta}$

Note that if we set $\alpha=0$ and let $M\rightarrow\infty$ so that the cart behaves like a fixed ground, then we recover the equation of simple pendulum, $\ddot{\theta}=-\frac{g}{\ell}\sin\theta$ , from the equation of motion above. It is a good practice to carry out such simple checks wherever possible.

Remarks: We could write the equations of motion, eqn. (19.62) and eqn. (LABEL:eq:cartpend.s) in the coupled form as

\left[\begin{array}[]{cc}\ell&\cos\theta\\ m\ell\cos\theta&M+m\end{array}\right]\left(\begin{array}[]{c}\ddot{\theta}\\ \ddot{s}\end{array}\right)=\left(\begin{array}[]{c}-\frac{g}{\ell}\sin(\theta-% \alpha)\\ (M+m)g\sin\alpha-m\ell\dot{\theta}^{2}\sin\theta\end{array}\right)

and leave it at that, since for most computational purposes, it is enough. It is not so hard to find expressions for $\ddot{\theta}$ and $\ddot{s}$ from here by solving the matrix equation, even by hand.

SAMPLE 19.19

Filename:sfig4-6-8

Figure 19.90:

Resonant capture. A slightly unbalanced motor mounted on an elastic machine part is modeled as a spring mass system with a simple pendulum of mass $m$ and length $\epsilon$ driven by a constant torque $T$ as shown in the figure. The spring has stiffness $k$ and the motor has mass $M$ . There is no friction between mass $M$ and the horizontal surface.

1.

Find the equation of motion of the system.
2.

Take $M=m=1\,\mathrm{kg},\,k=1\,{\rm N}/{\rm m},\,T=15\times 10^{-3}\,{\rm N}% \mathord{\cdot}{\rm m}$ . Solve (numerically) the equations of motion with zero initial conditions and plot $x(t)$ and $\theta(t)$ for $t=0$ to 100 $\,\mathrm{s}$ .

Solution

Filename:sfig4-6-8a

Figure 19.91:

The free-body diagram of the system is shown in fig. 19.91. Let the angular displacement of the eccentric mass $m$ at some instant $t$ be $\theta$ . At the same instant, let the displacement of the motor be $x$ from the relaxed state of the spring. Then we can write the acceleration of the motor as $\ddot{x}\hat{\boldsymbol{\imath}}$ and that of the eccentric mass as ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}=\ddot{x}\hat{% \boldsymbol{\imath}}+\epsilon\ddot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}-\epsilon\dot{\theta}^{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}$ . Now, we can write the angular momentum balance for the system about point C (fixed in the stationary frame of reference but instantly coincident with the center-of-mass of motor M) as

$\displaystyle T\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle\epsilon\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times m(% \ddot{x}\hat{\boldsymbol{\imath}}+\epsilon\ddot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}-\epsilon\dot{\theta}^{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R})$
	$\displaystyle=$	$\displaystyle m\epsilon\ddot{x}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}% \times\hat{\boldsymbol{\imath}})+m\epsilon^{2}\ddot{\theta}(\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!R}\times\hat{\boldsymbol{e}}_{\scriptscriptstyle\!% \theta})$
	$\displaystyle=$	$\displaystyle m\epsilon(-\ddot{x}\sin\theta+\epsilon\ddot{\theta})\hat{% \boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\epsilon\ddot{\theta}-% \sin\theta\,\ddot{x}$	$\displaystyle=$	$\displaystyle\frac{T}{m\epsilon}$	(19.64)

This is just one scalar equation in $\ddot{\theta}$ and $\ddot{x}$ . We need one more independent equation $\ddot{\theta}$ and $\ddot{x}$ without involving any other unknowns. So, we write the linear momentum balance for the system in the $x$ -direction:

Filename:sfig4-6-8b

Figure 19.92:

Filename:sfig5-5-2

Figure 19.93:

$\displaystyle-kx$	$\displaystyle=$	$\displaystyle M\ddot{x}+m(\ddot{x}\hat{\boldsymbol{\imath}}+\epsilon\ddot{% \theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}-\epsilon\dot{\theta}^% {2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R})\cdot\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle(M+m)\ddot{x}-m\epsilon\ddot{\theta}\sin\theta-m\epsilon\dot{% \theta}^{2}\cos\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\epsilon\sin\theta\,% \ddot{\theta}-\Bigl{(}\frac{M+m}{m}\Bigr{)}\ddot{x}$	$\displaystyle=$	$\displaystyle\frac{k}{m}x-\epsilon\dot{\theta}^{2}\cos\theta.$	(19.65)

Thus we have the required equations of motion. We can write eqn. (19.64) and (19.65) compactly as

\left[\begin{array}[]{cc}\epsilon&-\sin\theta\\ \epsilon\sin\theta&-\frac{m+M}{M}\end{array}\right]\left(\begin{array}[]{c}% \ddot{\theta}\\ \ddot{x}\end{array}\right)=\left(\begin{array}[]{c}\frac{T}{m\epsilon}\\ \frac{k}{m}x-\epsilon\dot{\theta}^{2}\cos\theta.\end{array}\right)

We use the following pseudocode to solve the equations of motion. Note that we first convert the two second order ODEs into four first order ODEs by introducing new variables $\omega=\dot{\theta}$ and $u=\dot{x}$ .

    Set T = 0.015, m = 1, M = 1, k = 1, e = 1
    A=[e -sin(theta); e*sin(theta) -(1+M/m)];
    b = [T/(m*e); k/m*x-e*omega^2*cos(theta)];
    solve A*acln = b for acln  % acln = accelerations
    ODEs = {omega = thetadot, u  = xdot,
            omegadot = acln(1), udot = acln(2)}
    IC   = {theta = 0, x = 0, omega = 0, u = 0}
    Solve ODEs with IC for t=0 to t=100

The plots of $x(t)$ and $\theta(t)$ obtained from the numerical solution are shown in fig. 19.92 and fig. 19.93, respectively. Note the resonance of $M$ for the given values of the system.

SAMPLE 19.20

Filename:sfig5-5-2a

Figure 19.94:

Dynamics using a rotating and translating coordinate system. Consider the rotating wheel of Sample 18.52 which is shown here again in Figure 19.94. At the instant shown in the figure find

1.

the linear momentum of the mass P and
2.

the net force on the mass P.

For calculations, use a frame $\mathcal{B}$ attached to the rod and a coordinate system in $\mathcal{B}$ with origin at point A of the rod OA.

Solution

Filename:sfig5-5-2b

Figure 19.95: The velocity of point

P^{\prime}

is the sum of two terms: the velocity of

O^{\prime}

and the velocity of

P^{\prime}

relative to

O^{\prime}

We attach a frame $\mathcal{B}$ to the rod. We choose a coordinate system $x^{\prime}y^{\prime}z^{\prime}$ in this frame with its origin O^′ at point A. We also choose the orientation of the primed coordinate system to be parallel to the fixed coordinate system $x y z$ (see Fig. 19.95), i.e., $\hat{\boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}},\quad\hat{% \boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}},\hbox to36.135pt{\hfil and% \hfil}\hat{\boldsymbol{k}}^{\prime}=\hat{\boldsymbol{k}}.$

Linear momentum of P: The linear momentum of the mass P is given by

{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{P}.

Clearly, we need to calculate the velocity of point P to find ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ . Now,

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

\displaystyle=

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

Note that O^′ and P^′ are two points on the same (imaginary) rigid body OAP^′. Therefore, we can find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}$ as follows:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}$	$\displaystyle=$	$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \mathcal{B}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{O^{\prime}/O}% }^{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{O^{\prime}}}+\overbrace{{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}/O^{\prime}}}^{{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}/O^{\prime}}}$
	$\displaystyle=$	$\displaystyle\omega_{1}\hat{\boldsymbol{k}}\times\ell(\cos\theta\hat{% \boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})+\omega_{1}\hat{% \boldsymbol{k}}\times r(\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{% \boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle\omega_{1}[(\ell+r)\cos\theta\hat{\boldsymbol{\jmath}}-(\ell-r)% \sin\theta\hat{\boldsymbol{\imath}}]$
	$\displaystyle=$	$\displaystyle 3\,\mathrm{rad}/\mathrm{s}\cdot[2.5\,\mathrm{m}\cdot\cos 30^{{}^% {\circ}}\hat{\boldsymbol{\jmath}}-1.5\,\mathrm{m}\cdot\sin 30^{{}^{\circ}}\hat% {\boldsymbol{\imath}}]$
	$\displaystyle=$	$\displaystyle(6.50\hat{\boldsymbol{\jmath}}-2.25\hat{\boldsymbol{\imath}})\,% \mathrm{m}/\mathrm{s}\quad\mbox{(same as in Sample~{}\ref{S8.2Dwheel}.)},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{rel}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P/\mathcal{B}}$
	$\displaystyle=$	$\displaystyle-\omega_{2}\hat{\boldsymbol{k}}^{\prime}\times r(\cos\theta\hat{% \boldsymbol{\imath}}^{\prime}-\sin\theta\hat{\boldsymbol{\jmath}}^{\prime})$
	$\displaystyle=$	$\displaystyle-\omega_{2}r(\cos\theta\hat{\boldsymbol{\jmath}}^{\prime}+\sin% \theta\hat{\boldsymbol{\imath}}^{\prime})$
	$\displaystyle=$	$\displaystyle-(2.16\hat{\boldsymbol{\jmath}}^{\prime}+1.25\hat{\boldsymbol{% \imath}}^{\prime})\,\mathrm{m}/\mathrm{s}$
	$\displaystyle=$	$\displaystyle-(2.16\hat{\boldsymbol{\jmath}}+1.25\hat{\boldsymbol{\imath}})\,% \mathrm{m}/\mathrm{s}.$

Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}+{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{rel}}$
	$\displaystyle=$	$\displaystyle(4.34\hat{\boldsymbol{\jmath}}-3.50\hat{\boldsymbol{\imath}})\,% \mathrm{m}/\mathrm{s}\qquad\mbox{and}\qquad\mbox{\hskip 72.26999pt}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{L}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$
	$\displaystyle=$	$\displaystyle 0.5\,\mathrm{kg}\cdot(4.34\hat{\boldsymbol{\jmath}}-3.50\hat{% \boldsymbol{\imath}})\,\mathrm{m}/\mathrm{s}$
	$\displaystyle=$	$\displaystyle(-1.75\hat{\boldsymbol{\imath}}+2.17\hat{\boldsymbol{\jmath}})\,{% \mathrm{kg}}\mathord{\cdot}{\mathrm{m}}/{\mathrm{s}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{L}}}=(-1.75\hat{\boldsymbol{\imath}}+% 2.17\hat{\boldsymbol{\jmath}})\,{\mathrm{kg}}\mathord{\cdot}{\mathrm{m}}/{% \mathrm{s}}$

Net force on P: From the

{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}}

for the mass P we get $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}_{P}$ . Thus to find the net force $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$ we need to find ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ .

The calculation of ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ is the same as in Sample 18.52 except that ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}$ is now calculated from

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}={\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{O^{\prime}}+{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{P^{\prime}/O^{\prime}}

where

Filename:sfig7-4-2

Figure 19.96:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{O^{\prime}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}% \times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{O^{\prime}/O})$
	$\displaystyle=$	$\displaystyle-\omega_{1}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{O^{% \prime}/O}$
	$\displaystyle=$	$\displaystyle-\omega_{1}^{2}\ell(\cos\theta\hat{\boldsymbol{\imath}}+\sin% \theta\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-(3\,\mathrm{rad}/\mathrm{s})^{2}\cdot 2\,\mathrm{m}\cdot(\cos 30% ^{{}^{\circ}}\hat{\boldsymbol{\imath}}+\sin 30^{{}^{\circ}}\hat{\boldsymbol{% \jmath}})$
	$\displaystyle=$	$\displaystyle-(15.59\hat{\boldsymbol{\imath}}+9.00\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}^{2},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}/O^{% \prime}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}% \times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}/O^{\prime}})$
	$\displaystyle=$	$\displaystyle-\omega_{1}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P^{% \prime}/O^{\prime}}$
	$\displaystyle=$	$\displaystyle-\omega_{1}^{2}r(\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta% \hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-(3\,\mathrm{rad}/\mathrm{s})^{2}\cdot 0.5\,\mathrm{m}\cdot(\cos 3% 0^{{}^{\circ}}\hat{\boldsymbol{\imath}}-\sin 30^{{}^{\circ}}\hat{\boldsymbol{% \jmath}})$
	$\displaystyle=$	$\displaystyle-(3.90\hat{\boldsymbol{\imath}}-2.25\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}^{2}.$

Thus,

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}=-(19.49\hat{% \boldsymbol{\imath}}+6.75\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}^{2}

which, of course, is the same as calculated in Sample 18.52. The other two terms, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cor}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{rel}}$ , are exactly the same as in Sample 18.52. Therefore, we get the same value for ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ by adding the three terms:

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=-(17.83\hat{\boldsymbol{% \imath}}+3.63\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}^{2}.

The net force on P is

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$
	$\displaystyle=$	$\displaystyle 0.5\,\mathrm{kg}\cdot(-17.83\hat{\boldsymbol{\imath}}-3.63\hat{% \boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}^{2}$
	$\displaystyle=$	$\displaystyle-(8.92\hat{\boldsymbol{\imath}}+1.81\hat{\boldsymbol{\jmath}})\,{% \mbox{N}}.$

Answer: $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=-(8.92\hat{\boldsymbol{% \imath}}+1.81\hat{\boldsymbol{\jmath}})\,{\mbox{N}}$

SAMPLE 19.21

Filename:pfigure-blue-99-1

Figure 19.97:

Inverse dynamics of a four bar mechanism. A four bar mechanism ABCD consists of three uniform bars AB, BC, and CD of length $\ell_{1},\,\ell_{2},\,\ell_{3}$ , and mass $m_{1},\,m_{2},\,m_{3}$ , respectively. The mechanism is driven by a torque $T$ applied at A such that bar AB rotates at constant angular speed. Write equations to find the torque $T$ at some instant $t$ .

Solution This is an inverse dynamics problem, that is, we are given the motion and we are supposed to find the forces (torque $T$ in this case) that cause that motion. We are given that rod AB rotates at constant angular speed, say $\dot{\theta}$ . From kinematics, we can find out angular velocities and angular accelerations of the other two bars as well as the accelerations of center-of-mass of each rod. Then we can write the momentum balance equations and compute the forces and moments required to generate this motion. So, in contrast to what we usually do, let us do the kinematics first. Please see Sample 18.5 on page 18.5. We found the angular velocities, $\dot{\beta}$ (eqn. (18.110)) and $\dot{\phi}$ (eqn. (18.109)), of rods BC and CD, respectively, in terms of $\dot{\theta}$ . We can rewrite those equations as

\left[\begin{array}[]{rr}-\ell_{2}\sin\beta&\ell_{3}\sin\phi\\ -\ell_{2}\cos\beta&\ell_{3}\cos\phi\end{array}\right]\left(\begin{array}[]{c}% \dot{\beta}\\ \dot{\phi}\end{array}\right)=\dot{\theta}\left(\begin{array}[]{c}\ell_{1}\sin% \theta\\ \ell_{1}\cos\theta\end{array}\right)

(19.66)

We wrote this equation in matrix form to make it easier for us to find the angular accelerations which we do by simply differentiating this equation once:

	$\displaystyle\left[\begin{array}[]{rr}-\ell_{2}\cos\beta\,\dot{\beta}&\ell_{3}% \cos\phi\,\dot{\phi}\\ \ell_{2}\sin\beta\,\dot{\beta}&-\ell_{3}\sin\phi\,\dot{\phi}\end{array}\right]% \left(\begin{array}[]{c}\dot{\beta}\\ \dot{\phi}\end{array}\right)$	$\displaystyle+$	$\displaystyle\left[\begin{array}[]{rr}-\ell_{2}\sin\beta&\ell_{3}\sin\phi\\ -\ell_{2}\cos\beta&\ell_{3}\cos\phi\end{array}\right]\left(\begin{array}[]{c}% \ddot{\beta}\\ \ddot{\phi}\end{array}\right)$
		$\displaystyle=$	$\displaystyle\ell_{1}\left(\begin{array}[]{c}\ddot{\theta}\sin\theta+\dot{% \theta}^{2}\cos\theta\\ \ddot{\theta}\cos\theta-\dot{\theta}^{2}\sin\theta\end{array}\right)$

Rearranging terms we get

	$\displaystyle\left[\begin{array}[]{rr}-\ell_{2}\sin\beta&\ell_{3}\sin\phi\\ -\ell_{2}\cos\beta&\ell_{3}\cos\phi\end{array}\right]\left(\begin{array}[]{c}% \ddot{\beta}\\ \ddot{\phi}\end{array}\right)$	$\displaystyle=$	$\displaystyle-\left[\begin{array}[]{rr}-\ell_{2}\cos\beta&\ell_{3}\cos\phi\\ \ell_{2}\sin\beta&\ell_{3}\sin\phi\end{array}\right]\left(\begin{array}[]{c}% \dot{\beta}^{2}\\ \dot{\phi}^{2}\end{array}\right)$		(19.82)
			$\displaystyle+\ell_{1}\left[\begin{array}[]{rr}\sin\theta&\cos\theta\\ \cos\theta&-\sin\theta\end{array}\right]\left(\begin{array}[]{c}\ddot{\theta}% \\ \dot{\theta}^{2}\end{array}\right)\qquad\qquad$		(19.82)

Thus, we can find the angular accelerations of BC and CD, $\ddot{\beta}$ and $\ddot{\phi}$ , because the quantities on the right hand side are known ( $\ddot{\theta}(=0)$ and $\dot{\theta}$ are given, and $\dot{\beta}$ and $\dot{\phi}$ are determined by eqn. (19.66)). Now, we can find the accelerations of center-of-mass of each rod as follows.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{1}}}$	$\displaystyle=$	$\displaystyle-\frac{\ell_{1}}{2}\dot{\theta}^{2}\hat{\boldsymbol{\lambda}}_{1}$	(19.83)
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{2}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B}}+{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{2}/B}}=-\ell_{1}\dot{% \theta}^{2}\hat{\boldsymbol{\lambda}}_{1}-\frac{\ell_{2}}{2}\dot{\beta}^{2}% \hat{\boldsymbol{\lambda}}_{2}+\ell_{2}\ddot{\beta}\hat{\boldsymbol{n}}_{2}$	(19.84)
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{3}}}$	$\displaystyle=$	$\displaystyle-\frac{\ell_{3}}{2}\dot{\phi}^{2}\hat{\boldsymbol{\lambda}}_{3}$	(19.85)

We are now ready to write momentum balance equations. Since we are only interested in finding the torque $T$ , we should try to write equations involving minimum number of unknown forces. So, we draw free-body diagrams of the whole mechanism, of part BCD, and of bar CD alone; and write angular momentum balance equations about appropriate points so that we involve only the unknown torque $T$ and the unknown reaction ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{D}$ at D. Thus, we will have only three scalar unknowns $T$ , $R_{D_{x}}$ and $R_{D_{y}}$ (since ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{D}=R_{D_{x}}\hat{\boldsymbol{% \imath}}+R_{D_{y}}\hat{\boldsymbol{\jmath}}$ ). So, we will need only three independent equations.

Consider the free-body diagram of the whole mechanism. We can write angular momentum balance about point A for the whole mechanism as

Filename:pfigure-blue-47-2

Figure 19.98:

T\hat{\boldsymbol{k}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm% {D/A}}\times{\accentset{\rightharpoonup}{\boldsymbol{R}}}_{D}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{A}}=\dot{\accentset{\rightharpoonup% }{\boldsymbol{H}}}_{\mathrm{1/A}}+\dot{\accentset{\rightharpoonup}{\boldsymbol% {H}}}_{\mathrm{2/A}}+\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{% \mathrm{3/A}}

(19.86)

where

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{1/A}}$	$\displaystyle=$	$\displaystyle I_{1}\ddot{\theta}\hat{\boldsymbol{k}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{1}}}\times m_{1}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{1}}}$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{2/A}}$	$\displaystyle=$	$\displaystyle I_{2}\ddot{\beta}\hat{\boldsymbol{k}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{2}}}\times m_{2}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{2}}}=I_{2}\ddot{\beta}\hat{% \boldsymbol{k}}+({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}+% {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{2}/B}})\times m_{2% }{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{2}}}$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{3/A}}$	$\displaystyle=$	$\displaystyle I_{3}\ddot{\phi}\hat{\boldsymbol{k}}+{\accentset{\rightharpoonup% }{\boldsymbol{r}}}_{\!\mathrm{G_{3}}}\times m_{3}{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{\mathrm{G_{3}}}=I_{3}\ddot{\phi}\hat{\boldsymbol{k}}+({% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{3}/D}})\times m_{3}{\accentset% {\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{3}}}.$

Filename:pfigure-blue-89-1

Figure 19.99:

Similarly, the angular momentum balance about point B for BCD gives

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{R}}}_{\mathrm{D}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{2/B}}+\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{3/B}}

(19.87)

where

	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{2/B}}$	$\displaystyle=$	$\displaystyle I_{2}\ddot{\beta}\hat{\boldsymbol{k}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{2}/B}}\times m_{2}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{2}}}$
	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{3/B}}$	$\displaystyle=$	$\displaystyle I_{3}\ddot{\phi}\hat{\boldsymbol{k}}+{\accentset{\rightharpoonup% }{\boldsymbol{r}}}_{\!\mathrm{G_{3}/B}}\times m_{3}{\accentset{\rightharpoonup% }{\boldsymbol{a}}}_{\mathrm{G_{3}}}$

and angular momentum balance of bar CD about point C gives

Filename:pfigure-blue-35-2

Figure 19.100:

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/C}}\times{% \accentset{\rightharpoonup}{\boldsymbol{R}}}_{\mathrm{D}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{3/C}}=I_{3}\ddot{\phi}\hat{% \boldsymbol{k}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{3}% /C}}\times m_{3}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G_{3}}}.

(19.88)

Note that we can easily write the position vectors in terms of $\ell_{1},\ell_{2},\ell_{3}$ and the unit vectors $(\hat{\boldsymbol{\lambda}}_{1},\hat{\boldsymbol{n}}_{1})$ , $(\hat{\boldsymbol{\lambda}}_{2},\hat{\boldsymbol{n}}_{2})$ and $(\hat{\boldsymbol{\lambda}}_{3},\hat{\boldsymbol{n}}_{3})$ where

	$\displaystyle\hat{\boldsymbol{\lambda}}_{1}=\cos\theta\hat{\boldsymbol{\imath}% }+\sin\theta\hat{\boldsymbol{\jmath}},$		$\displaystyle\hat{\boldsymbol{n}}_{1}=-\sin\theta\hat{\boldsymbol{\imath}}+% \cos\theta\hat{\boldsymbol{\jmath}}$
	$\displaystyle\hat{\boldsymbol{\lambda}}_{2}=\cos\beta\hat{\boldsymbol{\imath}}% +\sin\beta\hat{\boldsymbol{\jmath}},$		$\displaystyle\hat{\boldsymbol{n}}_{2}=-\sin\beta\hat{\boldsymbol{\imath}}+\cos% \beta\hat{\boldsymbol{\jmath}}$
	$\displaystyle\hat{\boldsymbol{\lambda}}_{1}=\cos\phi\hat{\boldsymbol{\imath}}+% \sin\phi\hat{\boldsymbol{\jmath}},$		$\displaystyle\hat{\boldsymbol{n}}_{1}=-\sin\phi\hat{\boldsymbol{\imath}}+\cos% \phi\hat{\boldsymbol{\jmath}}.$

We can put all the three angular momentum balance equations, (19.86), (19.87), and (19.88), in one matrix equation by dotting both sides of the equations with $\hat{\boldsymbol{k}}$ and assembling them as follows.

	$\displaystyle\left[\begin{array}[]{ccc}1&0&\ell_{4}\\ 0&\hat{\boldsymbol{k}}\cdot(\ell_{2}\hat{\boldsymbol{\lambda}}_{2}-\ell_{3}% \hat{\boldsymbol{\lambda}}_{3})\times\hat{\boldsymbol{\imath}}&\hat{% \boldsymbol{k}}\cdot(\ell_{2}\hat{\boldsymbol{\lambda}}_{2}-\ell_{3}\hat{% \boldsymbol{\lambda}}_{3})\times\hat{\boldsymbol{\jmath}}\\ 0&\hat{\boldsymbol{k}}\cdot(-\ell_{3}\hat{\boldsymbol{\lambda}}_{3}\times\hat{% \boldsymbol{\imath}})&\hat{\boldsymbol{k}}\cdot(-\ell_{3}\hat{\boldsymbol{% \lambda}}_{3}\times\hat{\boldsymbol{\jmath}})\end{array}\right]\left(\begin{% array}[]{c}T\\ R_{D_{x}}\\ R_{D_{y}}\end{array}\right)=\left(\begin{array}[]{c}\dot{H}_{123/A}\\ \dot{H}_{23/B}\\ \dot{H}_{3/C}\end{array}\right)$		(19.98)
			(19.99)

where $\dot{H}_{123/A}=\hat{\boldsymbol{k}}\cdot(\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\mathrm{1/A}}+\dot{\accentset{\rightharpoonup}{\boldsymbol{H% }}}_{\mathrm{2/A}}+\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{% 3/A}})$ , $\dot{H}_{23/B}=\hat{\boldsymbol{k}}\cdot(\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\mathrm{2/B}}+\dot{\accentset{\rightharpoonup}{\boldsymbol{H% }}}_{\mathrm{3/B}})$ , and $\dot{H}_{3/C}=\hat{\boldsymbol{k}}\cdot\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\mathrm{3/C}}$ .

Note that we know the $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ ’s on the right hand side and the matrix on the left side can be evaluated for any given $(\theta,\beta,\phi)$ . Thus we can solve for $T,R_{D_{x}}$ , and $R_{D_{y}}$ .

SAMPLE 19.22

Numerical solution of the inverse dynamics problem. Consider Sample 19.95 again. Using numerical solutions on a computer, find and plot torque $T$ against $\theta$ for one complete cycle of the drive arm AB. Take $m_{1}=m_{2}=m_{3}=1\,\mathrm{kg}$ and $\ell_{1}=400\,{\rm mm},\ell_{2}=400\sqrt{2}\,{\rm mm},\ell_{3}=800\sqrt{2}\,{% \rm mm}$ , and $\ell_{4}=1200\,{\rm mm}$ .

Solution

Since we have to plot $T$ against $\theta$ for one complete revolution, we need to find angular velocities, angular accelerations and center-of-mass accelerations for several values of $\theta$ and then solve for $T$ for each of those $\theta$ ’s. We can do this several ways. One way would be to first solve kinematic equations to find $\theta(t),\beta(t)$ , and $\phi(t)$ at discrete times over one complete cycle and then compute all other quantities at each $(\theta(t_{i}),\beta(t_{i}),\phi(t_{i}))$ where $t_{i}$ represents a discrete time. So, let us follow this method step by step with pseudocodes. Here, we assume that we have vector functions called dot and cross that compute the dot product and the cross product of two vectors that are given as input arguments.

Step-1: solve for angular positions.

Specify the given geometry

L1=0.4, L2=0.4*sqrt(2), L3=0.8*sqrt(2), L4=1.2

and use the pseudocode of Sample 18.5 to find $\theta(t_{i}),\beta(t_{i}),\phi(t_{i})$ for, say, 100 values of $t_{i}$ between 0 and 1 sec. Now, for each triad of $(\theta(t_{i}),\beta(t_{i}),\phi(t_{i}))$ , follow all the steps below.

Step-2: solve for angular velocities.

Since $\dot{\theta}=2\pi\rm rad/s$ is given, we only need to solve for $\dot{\beta}$ and $\dot{\phi}$ . We use eqn. (18.110) and eqn. (18.109) to compute $\dot{\beta}$ and $\dot{\phi}$ as follows (or modify the pseudocode of Sample 18.5 to save $\dot{\beta}$ and $\dot{\phi}$ along with the values for $\beta$ and $\phi$ .

  define thdot=thetadot, bdot=betadot, pdot=phidot
  thdot = 2*pi   % this is given
  set th = theta(ti), b = beta(ti), p = phi(ti)
  set unit vectors
    l1=[cos(th) sin(th) 0]’, n1=[-sin(th) cos(th) 0]’
    l2=[cos(b) sin(b) 0]’, n2=[-sin(b) cos(b) 0]’
    l3=[cos(p) sin(p) 0]’, n3=[-sin(p) cos(p) 0]’
  bdot = -(L1/L2)*(cross(n1,l3)/cross(n2,l3))*thdot
  pdot = (L1/L3)*(cross(n1,l2)/cross(n3,l2))*thdot

Step-3: solve for angular accelerations.

Now that we have $(\theta,\beta,\phi)$ and the corresponding values of $(\dot{\theta},\dot{\beta},\dot{\phi})$ , we can use eqn. (19.82) to calculate $\ddot{\beta}$ and $\ddot{\phi}$ (we are given $\ddot{\theta}=0$ ).

  define thddot=thetaddot, bddot=betaddot,
         pddot=phiddot
  thddot = 0   % this is given
  B = [-L2*sin(b) L3*sin(p); -L2*cos(b) L3*cos(p)]
  C = L1*[sin(th) cos(th); cos(th) -sin(th)]
  D = [-cos(b) cos(p); sin(b) -sin(p)]
  c = [thddot thdot^2]’,  d = [L2*bdot^2 L3*pdot^2]’
  assume w = [bddot pddot]’
  solve B*w = C*c + D*d for w

So, now we know $\ddot{\theta},\ddot{\beta},\ddot{\phi}$ also. We are now ready to compute $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ ’s required for dynamic calculations.

Step-4: set up equations and solve for unknown forces.

We need to set up and solve eqn. (19.99). Note that we need to compute several quantities for this equation but the vector computations are more or less straightforward.

   % set mass and inertia properties
   m1 = 1, m2 = 1, m3 = 1
   I1 = m1*L1^2/12, I2 = m2*L2^2/12, I3 = m3*L3^2/12
   % set fixed unit vectors
   i = [1 0 0]’, j = [0 1 0]’, k = [0 0 1]’
   % compute position vectors
   rA = [0;0;0],  rB = rA+L1*l1
   rC = rB+L2*l2, rD = L4*l4
   rG1 = L1/2*l1
   rG2 = rB+L2/2*l2
   rG3 = rD+L3/2*l3
   % compute center-of-mass accelerations
   aG1 = 0.5*L1*(tddot*n1-tdot^2*l1)
   aG2 = 2*aG1+0.5*L2*(bddot*n2-bdot^2*l2)    % aB = 2*aG1
   aG3 = 0.5*L3*(pddot*n3-pdot^2*l3)

Filename:pfigure4-3Dpend

Figure 19.101: Torque

T

as a function of

\theta

over one complete cycle of motion (

\theta(0)=\pi/2,\theta(1)=5\pi/2

   % compute Hdot_cms

   Hdot_cm1 = I1*tddot*uk
   Hdot_cm2 = I2*bddot*uk
   Hdot_cm3 = I3*pddot*uk
   % compute Hdots
   Hdot_123_A = Hdot_cm1 + cross(rG1, m1*aG1)
                 + Hdot_cm2 + cross(rG2, m2*aG2)
                   + Hdot_cm3 + cross(rG3, m3*aG3)
   Hdot_23_B = Hdot_cm2 + cross(rG2-rB, m2*aG2)
                 + Hdot_cm3 + cross(rG3-rB, m3*aG3)
   Hdot_3_C = Hdot_cm3 + cross(rG3-rC, m3*aG3)
   % set up the linear eqns for torque and RD
   b = [dot(Hdot_123_A,k) dot(Hdot_23_B,k) dot(Hdot_3_C,k)]
   A = [1  dot(k,cross(rD,i))     dot(k,cross(rD,j))
        0  dot(k,cross(rD-rB,i))  dot(k,cross(rD-rB,j))
Ψ        0  dot(k,cross(rD-rC,i))  dot(k,cross(rD-rC,j))]
   % let forces = [T  RDx  RDy]’
   solve A*forces = b for forces

Step-5, repeat calculations.

Now repeat Step-2 – Step-4 for each triad $(\theta,\beta,\phi)$ obtained in Step-1 and save the corresponding values of $T$ in a vector. Finally,

   plot T vs theta

Filename:pfigure-s94h6p3

Figure 19.102: Torque

T

as a function of time over one complete cycle of motion.

The plot thus obtained is shown in fig. 19.101. We can also plot $T$ vs time (as shown in fig. 19.102), and, of course, expect to see the same graph of $T$ since $\theta$ is just a linear function of $t$ . Note that the area under the graph of $T$ over one complete cycle must equal zero since the net impulse must be zero over one cycle.

Problems for 19.3 Multi-degree-of-freedom 2-D mechanisms

19.3.1 Particle on a springy leash. A particle with mass $m$ slides on a rigid horizontal frictionless plane. It is held by a string which is in turn connected to a linear elastic spring with constant $k$ . The string length is such that the spring is relaxed when the mass is on top of the hole in the plane. The position of the particle is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=x\hat{\boldsymbol{\imath}}+y\hat% {\boldsymbol{\jmath}}$ . For each of the statements below, state the circumstances in which the statement is true (assuming the particle stays on the plane). Justify your answer with convincing explanation and/or calculation.

(a)

The force of the plane on the particle is $mg\hat{\boldsymbol{k}}$ .
(b)

$\ddot{x}+{k\over m}x=0$
(c)

$\ddot{y}+{k\over m}y=0$
(d)

$\ddot{r}+{k\over m}r=0,\;\;\mbox{where }r=|{\accentset{\rightharpoonup}{% \boldsymbol{r}}}|$
(e)

$r=$ constant
(f)

$\dot{\theta}=$ constant
(g)

$r^{2}\dot{\theta}=$ constant.
(h)

$m(\dot{x}^{2}+\dot{y}^{2})+kr^{2}=$ constant
(i)

The trajectory is a straight line segment.
(j)

The trajectory is a circle.
(k)

The trajectory is not a closed curve.

Filename:pfigure-s94h6p4

Figure 19.103

Particle on a springy leash.

19.3.2 “Yo-yo” mechanism of satellite de-spinning. A satellite, modeled as a uniform disk, is “de-spun” by the following mechanism. Before launch two equal length long strings are attached to the satellite at diametrically opposite points and then wound around the satellite with the same sense of rotation. At the end of the strings are placed 2 equal masses. At the start of de-spinning the two masses are released from their position wrapped against the satellite. Find the motion of the satellite as a function of time for the two cases: (a) the strings are wrapped in the same direction as the initial spin, and (b) the strings are wrapped in the opposite direction as the initial spin.

19.3.3 A particle with mass $m$ is held by two long springs each with stiffness $k$ so that the springs are relaxed when the mass is at the origin. Assume the motion is planar. Assume that the particle displacement is much smaller than the lengths of the springs.

(a)

Write the equations of motion in cartesian components. Answer: $-k(x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}})=m(\ddot{x}\hat{% \boldsymbol{\imath}}+{\ddot{y}}\hat{\boldsymbol{\jmath}})$ .
(b)

Write the equations of motion in polar coordinates. Answer: $-kr\hat{\boldsymbol{e}}_{r}=m(\ddot{r}-r\dot{\theta}^{2})\hat{\boldsymbol{e}}_% {r}+m(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\boldsymbol{e}}_{t}$ .
(c)

Express the conservation of angular momentum in cartesian coordinates. Answer: $\frac{d}{dt}(x{\dot{y}}-y\dot{x})=0$ .
(d)

Express the conservation of angular momentum in polar coordinates. Answer: $\frac{d}{dt}(r^{2}\dot{\theta})=0$ .
(e)

Show that (a) implies (c) and (b) implies (d) even if you didn’t note them a-priori. Answer: dot equation in (a) with $(x\hat{\boldsymbol{\jmath}}-y\hat{\boldsymbol{\imath}})$ to get $(x{\ddot{y}}-y\ddot{x})=0$ which can be rewritten as (c); dot equation in (b) with $\hat{\boldsymbol{e}}_{t}$ to get $(r\ddot{\theta}+2\dot{r}\dot{\theta})=0$ which can be rewritten as (d).
(f)

Express the conservation of energy in cartesian coordinates. Answer: $\frac{1}{2}m(\dot{x}^{2}+{\dot{y}}^{2})+\frac{1}{2}k(x^{2}+y^{2})=const$ .
(g)

Express the conservation of energy in polar coordinates. Answer: $\frac{1}{2}m(\dot{r}^{2}+(r\dot{\theta})^{2})+\frac{1}{2}kr^{2}$ =const.
(h)

Show that (a) implies (f) and (b) implies (g) even if you didn’t note them a-priori. Answer: dot equation in (a) with ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{x}\hat{\boldsymbol{\imath}}% +{\dot{y}}\hat{\boldsymbol{\jmath}}$ to get $m(\ddot{x}\dot{x}+{\ddot{y}}{\dot{y}})+k(\dot{x}x+{\dot{y}}y)=0$ which can be rewritten as (f).
(i)

Find the general motion by solving the equations in (a). Describe all possible paths of the mass. Answer: $x=A_{1}\sin(\omega t+B_{1})$ , $y=A_{2}\sin(\omega t+B_{2})$ , where $\omega=\sqrt{\frac{k}{m}}$ . The general motion is an ellipse.
(j)

Can the mass move back and forth on a line which is not the $x$ or $y$ axis? Answer: Yes. Consider $B_{1}=B_{2}=0$ , $A_{1}=1$ , and $A_{2}=2$ .

Filename:pg84-3

Figure 19.104

19.3.4 Cart and pendulum A mass $m_{B}=6\,\mathrm{kg}$ hangs by two strings from a cart with mass $m_{C}=12\,\mathrm{kg}$ . Before string BC is cut string AB is horizontal. The length of string AB is $r=1\,\mathrm{m}$ . At time $t=0$ all masses are stationary and the string BC is cleanly and quietly cut. After some unknown time $t_{\rm{vert}}$ the string AB is vertical.

(a)

What is the net displacement of the cart $x_{C}$ at $t=t_{\rm{vert}}$ ? Answer: $0.33\,\mathrm{m}\hat{\boldsymbol{\imath}}$
(b)

What is the velocity of the cart $v_{C}$ at $t=t_{\rm vert}$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}=1.82\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\imath}}$ .
(c)

What is the tension in the string at $t=t_{\rm{vert}}$ ? Answer: $T\approx 240\,{\mbox{N}}$ .
(d)

(Optional) What is $t=t_{\rm{vert}}$ ? [You will either have to leave your answer in the form of an integral you cannot evaluate analytically or you will have to get part of your solution from a computer.]

Filename:pfigure-blue-36-1

Figure 19.105

19.3.5 A dumbbell slides on a floor. Two point masses $m$ at A and B are connected by a massless rigid rod with length $\ell$ . Mass B slides on a frictionless floor so that it only moves horizontally. Assume this dumbbell is released from rest in the configuration shown. [Hint: What is the acceleration of A relative to B?]

(a)

Find the acceleration of point B just after the dumbbell is released.
(b)

Find the velocity of point A just before it hits the floor

Filename:pfigure-s95q8

Figure 19.106

19.3.6 After a winning goal one second before the clock ran out a psychologically stunned hockey player (modeled as a uniform rod) stands nearly vertical, stationary and rigid. The player’s perfectly slippery skates start to pop out from under her as she falls. Her height is $\ell$ , her mass $m$ , her tip from the vertical $\theta$ , and the gravitational constant is $g$ .

(a)

What is the path of her center of mass as she falls? (Show clearly with equations, sketches or words.)
(b)

What is her angular velocity just before she hits the ice, a millisecond before she sticks out her hands and breaks her fall (first assume her skates remain in contact the whole time and then check the assumption)?
(c)

Find a differential equation that only involves $\theta$ , its time derivatives, $m$ , $g$ , and $\ell$ . (This equation could be solved to find $\theta$ as a function of time. It is a non-linear equation and you are not being asked to solve it numerically or otherwise.)

19.3.7 Falling hoop. A bicycle rim (no spokes, tube, tire, or hub) is idealized as a hoop with mass $m$ and radius $R$ . $G$ is at the center of the hoop. An inextensible string is wrapped around the hoop and attached to the ceiling. The hoop is released from rest at the position shown at $t=0$ .

(a)

Find $y_{G}$ at a later time $t$ in terms of any or all of $m$ , $R$ , $g$ , and $t$ .
(b)

Does $G$ move sideways as the hoop falls and unrolls?

Filename:pfigure4-rpi

Figure 19.107

19.3.8 A model for a yo-yo consists of a thin disk of mass $M$ and radius $R$ and a light drum of radius r, rigidly attached to the disk, around which a light inextensible cable is wound. Assuming that the cable unravels without slipping on the drum, determine the acceleration $a_{G}$ of the center of mass.

Filename:pfigure-blue-86-1

Figure 19.108

19.3.9 A uniform rod with mass $m_{R}$ pivots without friction about point $A$ in the $x y$ -plane. A collar with mass $m_{C}$ slides without friction on the rod after the string connecting it to point $A$ is cut. There is no gravity. Before the string is cut, the rod has angular velocity $\omega_{1}$ .

(a)

What is the speed of the collar after it flies off the end of the rod? Use the following values for the constants and initial conditions: $m_{R}=1\,\mathrm{kg}$ , $m_{C}=3\,\mathrm{kg}$ , $a=1\,\mathrm{m}$ , $\ell=3\,\mathrm{m}$ , and $\omega_{1}=1\,\mathrm{rad}/\mathrm{s}$
(b)

Consider the special case $m_{R}=0$ . Sketch (approximately) the path of the motion of the collar from the time the string is cut until some time after it leaves the end of the rod.

Filename:pfigure-blue-88-1

Figure 19.109

19.3.10 Assume the rod in the figure for problem 18.18 has polar moment of inertia $I o z z$ . Assume it is free to rotate. The bead is free to slide on the rod. Assume that at $t=0$ the angular velocity of the rod is $1\,\mathrm{rad}/\mathrm{s}$ , that the radius of the bead is one meter and that the radial velocity of the bead, $dR/dt$ , is zero.

(a)

Draw separate free-body diagrams of the bead and rod.
(b)

Write equations of motion for the system. Answer:

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

$\displaystyle(I_{zz}+mR^{2})\ddot{\theta}+2mR\dot{R}\dot{\theta}$ $\displaystyle=$ $\displaystyle 0\par$
(c)

Use the equations of motion to show that angular momentum is conserved. Answer: The second equation in part (b) can be rewritten in the form $\frac{d}{dt}\left[I_{zz}+mR^{2})\dot{\theta}\right]=0$ . The quantity inside the derivative is angular momentum; thus, it is conserved and equal to a constant, say, $(H_{/o})_{0}$ , which can be found in terms of the initial conditions.
(d)

Find one equation of motion for the system using: (1) the equations of motion for the bead and rod and (2) conservation of angular momentum.

Answer: $\ddot{R}-R\left[\frac{(H_{o})_{0}^{2}}{(I_{zz}^{O}+mR^{2})^{2}}\right]=0$ .
(e)

Write an expression for conservation of energy. Let the initial total energy of the system be, say, $E_{0}$ .

Answer: $E_{0}=\frac{1}{2}m\dot{R}^{2}+\frac{1}{2}\omega(H_{/o})_{0}$ .
(f)

As $t$ goes to infinity does the bead’s distance go to infinity? Its speed? The angular velocity of the turntable? The net angle of twist of the turntable? Answer: The bead’s distance goes to infinity and its speed approaches a constant. The turntable’s angular velocity goes to zero and its net angle of twist goes to a constant.

Filename:pfigure4-2-rp3

Figure 19.110

Coupled motion of bead and rod and turntable.

19.3.11 A primitive gun rides on a cart (mass $M$ ) and carries a cannon ball (of mass $m$ ) on a platform at a height $H$ above the ground. The cannon ball is dropped through a frictionless tube shaped like a quarter circle of radius $R$ .

(a)

If the system starts from rest, compute the horizontal speed (relative to the ground) that the cannon ball has as it leaves the bottom of the tube. Also find the cart’s speed at the same instant.
(b)

Compute the velocity of the cannon ball relative to the cart.
(c)

If two balls are dropped simultaneously through the tube, what speed does the cart have when the balls reach the bottom? Is the same final speed also achieved if one ball is allowed to depart the system entirely before the second ball is released? Why/why not?

Filename:pfigure-s94q6p1

Figure 19.111

19.3.12 Numerically simulate the coupled system in problem 19.3. Use the simulation to show that the net angle of the turntable or rod is finite.

(a)

Write the equations of motion for the system from part (b) in problem 19.3 as a set of first order differential equations. Answer:

$\displaystyle\dot{\theta}$ $\displaystyle=$ $\displaystyle\omega$

$\displaystyle\dot{\omega}$ $\displaystyle=$ $\displaystyle-\frac{2mRv\omega}{I_{zz}+mR^{2}}$

$\displaystyle\dot{R}$ $\displaystyle=$ $\displaystyle v$

$\displaystyle\dot{v}$ $\displaystyle=$ $\displaystyle\omega^{2}R$
(b)

Numerically integrate the equations of motion.

19.3.13 Two frictionless prisms of similar right triangular sections are placed on a frictionless horizontal plane. The top prism weighs $W$ and the lower one $n W$ . The prisms are held in the initial position shown and then released, so that the upper prism slides down along the lower one until it just touches the horizontal plane. The center of mass of a triangle is located at one-third of its height from the base. Compute the velocities of the two prisms at the moment just before the upper one reaches the bottom. Answer: $v_{2_{x}}=\sqrt{\frac{2g(a-b)\tan\phi}{(1+\frac{1}{n}+\left(\frac{n+1}{n}% \right)^{2}\tan^{2}\phi)}}$ , $v_{2_{y}}=v_{2_{x}}\frac{n+1}{n}\tan\phi$ , $v_{1_{x}}=-\frac{v_{2_{x}}}{n}$ , $v_{1_{y}}=0$ , where 2 refers to the top wedge and 1 refers to the lower wedge.

Filename:pfigure-spinningbrick

Figure 19.112

19.3.14 Mass slides on an accelerating cart. 2D. A cart is driven by a powerful motor to move along the $30^{\circ}$ sloped ramp according to the formula: $d=d_{0}+v_{o}t+a_{0}t^{2}/2$ where $d_{0}$ , $v_{0}$ , and $a_{0}$ are given constants. The cart is held from tipping over. The cart itself has a $30^{\circ}$ sloped upper surface on which rests a mass (given mass $m$ ). The surface on which the mass rests is frictionless. Initially the mass is at rest with regard to the cart.

(a)

What is the force of the cart on the mass? [in terms of $g,d_{0},v_{0},t,a_{0},m,g,\hat{\boldsymbol{\imath}}$ , and $\hat{\boldsymbol{\jmath}}$ .] Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m(g+a_{0})\left[\frac{\sqrt{3}}{% 4}\hat{\boldsymbol{\imath}}+\frac{3}{4}\hat{\boldsymbol{\jmath}}\right]$ .
(b)

For what values of $d_{0}$ , $v_{0}$ , $t$ and $a_{0}$ is the acceleration of the mass exactly vertical (i.e., in the $\hat{\boldsymbol{\jmath}}$ direction)? Answer: For $a_{0}=-g$ , the acceleration of the mass is exactly vertical; $d_{0}$ , $v_{0}$ , and $t$ could be anything.

Filename:pfigure-s94f1p2

Figure 19.113

19.3.15 A thin rod $A B$ of mass $W_{AB}=10\,{\mathrm{lbm}}$ and length $L_{AB}=2\,\mathrm{ft}$ is pinned to a cart $C$ of mass $W_{C}=10\,{\mathrm{lbm}}$ , the latter of which is free to move along a frictionless horizontal surface, as shown in the figure. The system is released from rest with the rod in the horizontal position.

(a)

Determine the angular speed of the rod as it passes through the vertical position (at some later time). Answer: angular speed = $8.8\,\mathrm{rad}/\mathrm{s}$ .
(b)

Determine the displacement $x$ of the cart at the same instant. Answer: displacement = $0.5\,\mathrm{ft}$ .
(c)

After the rod passes through vertical, it is momentarily horizontal but on the left side of the cart. How far has the cart moved when this configuration is reached?

Filename:s97f2

Figure 19.114

19.3.16 As shown in the figure, a block of mass $m$ rolls without friction on a rigid surface and is at position $x$ (measured from a fixed point). Attached to the block is a uniform rod of length $\ell$ which pivots about one end which is at the center of mass of the block. The rod and block have equal mass. The rod makes an angle $\theta$ with the vertical. Use the numbers below for the values of the constants and variables at the time of interest:

$\displaystyle\ell$	$\displaystyle=$	$\displaystyle 1\,\mathrm{m}$
$\displaystyle m$	$\displaystyle=$	$\displaystyle 2\,\mathrm{kg}$
$\displaystyle\theta$	$\displaystyle=$	$\displaystyle\pi/2$
$\displaystyle d\theta/dt$	$\displaystyle=$	$\displaystyle 1\,\mathrm{rad}/\mathrm{s}$
$\displaystyle d^{2}\theta/dt^{2}$	$\displaystyle=$	$\displaystyle 2\,\mathrm{rad}/\mathrm{s}^{2}$
$\displaystyle x$	$\displaystyle=$	$\displaystyle 1\,\mathrm{m}$
$\displaystyle dx/dt$	$\displaystyle=$	$\displaystyle 2\,\mathrm{m}/\mathrm{s}$
$\displaystyle d^{2}x/dt^{2}$	$\displaystyle=$	$\displaystyle 3\,\mathrm{m}/\mathrm{s}^{2}$

(a)

What is the kinetic energy of the system?
(b)

What is the linear momentum of the system (momentum is a vector)?

Filename:pg92-2

Figure 19.115

19.3.17 A pendulum of length $\ell$ hangs from a cart. The pendulum is massless except for a point mass of mass $m_{p}$ at the end. The cart rolls without friction and has mass $m_{c}$ . The cart is initially stationary and the pendulum is released from rest at an angle $\theta$ . What is the acceleration of the cart just after the mass is released? [Hints: ${\bf a}_{P}={\accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{P}/C$ . The answer is ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}=(g/3)\bf i$ in the special case when $m_{p}=m_{c}$ and $\theta=\pi/4$ . ]

Filename:summer95p2-3

Figure 19.116

19.3.18 Due to the application of some unknown force $F$ , the base of the pendulum A is accelerating with ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=a_{A}\boldsymbol{\imath}.$ There is a frictionless hinge at A. The angle of the pendulum $\theta$ with the $x$ axis and its rate of change $\dot{\theta}$ are assumed to be known. The length of the massless pendulum rod is $\ell$ . The mass of the pendulum bob $M$ . There is no gravity. What is $\ddot{\theta}$ ? (Answer in terms of $a_{A}$ , $\ell$ , $M$ , $\theta$ and $\dot{\theta}$ .)

Filename:p-f96-p3-3

Figure 19.117

19.3.19 Pumping a Swing Can a swing be pumped by moving the support point up and down?

For simplicity, neglect gravity and consider the problem of swinging a rock in circles on a string. Let the rock be mass $m$ attached to a string of fixed length $\ell$ . Can you speed it up by moving your hand up and down? How? Can you make a quantitative prediction? Let $x_{S}$ be a function of time that you can specify to try to make the mass swing progressively faster.

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Figure 19.118

19.3.20 Using free-body diagrams and appropriate momentum balance equations, find differential equations that govern the angle $\theta$ and the vertical deflection $y$ of the system shown. Be clear about your datum for $y$ . Your equations should be in terms of $\theta,y$ and their time-derivatives, as well as $M,m,\ell$ , and $g$ .

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Figure 19.119

19.3.21 The two blocks shown are released from rest at $t=0$ . There is no friction and the cable is initially taut. (a) What is the tension in the cable immediately after release? (Use any reasonable value for the gravitational constant). (b) What is the tension after 5 $\,{\rm s}$ ?

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Figure 19.120

19.3.22 Carts A and B are free to move along a frictionless horizontal surface, and bob C is connected to cart B by a massless, inextensible cord of length $\ell$ , as shown in the figure. Cart A moves to the left at a constant speed $v_{0}=1\,\mathrm{m}/\mathrm{s}$ and makes a perfectly plastic collision ( $e=0$ ) with cart B which, together with bob C, is stationary prior to impact. Find the maximum vertical position of bob C, $h_{max}$ , after impact. The masses of the carts and pendulum bob are $m_{A}=m_{B}=m_{C}=10\,\mathrm{kg}$ .

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Figure 19.121

19.3.23 A double pendulum is made of two uniform rigid rods, each of length $\ell$ . The first rod is massless. Find equations of motion for the second rod. Define any variables you use in your solution.

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Figure 19.122

19.3.24 A model of walking involves two straight legs. During the part of the motion when one foot is on the ground, the system looks like the picture in the figure, confined to motion in the $x-y$ plane. Write two equations from which one could find $\ddot{\theta}_{1},\hbox to36.135pt{\hfil and\hfil}\ddot{\theta}_{2}$ given $\theta_{1},\theta_{2},\dot{\theta}_{1},\dot{\theta}_{2}$ and all mass and length quantities.

\left[\begin{array}[]{rccl}\mbox{Hint: }\\ \sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/A}&=&\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{/A}&\mbox{whole system}\\ \sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/B}&=&\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{/B}&\mbox{for bar BC}\end{array}\right]

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Figure 19.123

Advanced problems in 2D motion

19.3.25 A pendulum is hanging from a moving support in the $x y$ -plane. The support moves in a known way given by ${\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)=r(t)\hat{\boldsymbol{\imath}}$ . For the following cases, find a differential equation whose solution would determine $\theta(t)$ , measured clockwise from vertical; then find an expression for $d^{2}\theta/dt^{2}$ in terms of $\theta$ and $r(t)$ :

(a)

with no gravity and assuming the pendulum rod is massless with a point mass of mass $m$ at the end,
(b)

as above but with gravity,
(c)

assuming the pendulum is a uniform rod of mass $m$ and length $\ell$ .

19.3.26 Robotics problem: balancing a broom stick by sideways motion. Try to balance a broom stick by moving your hand horizontally. Model your hand contact with the broom as a hinge. You can model the broom as a uniform stick or as a point mass at the end of a stick — your choice.

(a)

Equation of motion. Given the acceleration of your hand (horizontal only), the current tip, and the rate of tip of the broom, find the angular acceleration of the broom. Answer: For point mass: $\ddot{\phi}=(g/L)\sin\phi-(a_{hand}/L)\cos\phi$ .
(b)

Control? Can you find a hand acceleration in terms of the tip and the tip rate that will make the broom balance upright? Answer: There are many correct solutions. Test your solution with a computer simulation. Show your result with appropriate plots.

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Figure 19.124

19.3.27 Balancing the broom again: vertical shaking works too. You can balance a broom by holding it at the bottom and applying appropriate torques, as in problem 16.147 or by moving your hand back and forth in an appropriate manner, as in problem 19.124. In this problem, you will try to balance the broom differently. The lesson to be learned here is more subtle, and you should probably just wonder at it rather than try to understand it in detail.

In the previous balancing schemes, you used knowledge of the state of the broom ( $\theta$ and $\dot{\theta}$ ) to determine what corrective action to apply. Now balance the broom by moving it in a way that ignores what the broom is doing. In the language of robotics, what you have been doing before is ‘closed-loop feedback’ control. The new strategy, which is simultaneously more simple minded and more subtle, is an ‘open loop’ control. Imagine that your hand connection to the broom is a hinge.

(a)

Picture and model. Assume your hand oscillates sinusoidally up and down with some frequency and some amplitude. The broom is instantaneously at some angle from vertical. Draw a picture which defines all the variables you will use. Use any mass distribution that you like.
(b)

FBD. Draw a FBD of the broom.
(c)

Momentum balance. Write the equation of angular momentum balance about the point instantaneously coinciding with the hinge.
(d)

Kinematics. Use any geometry and kinematics that you need to evaluate the terms in the angular momentum balance equation in terms of the tip angle and its time derivatives and other known quantities (take the vertical motion of your hand as ‘given’). [Hint: There are many approaches to this problem.]
(e)

Equation of motion. Using the angular momentum balance equation, write a governing differential equation for the tip angle. Answer: $\ddot{\phi}-\left[\frac{{\ddot{y}}+g}{L}\right]\sin\phi=0$ , where $y(t)$ is the vertical displacement of your hand and $\phi(t)$ is the angle of the broom from the vertical.
(f)

Simulation. Taking the hand motion as given, simulate on the computer the system you have found.
(g)

Stability? Can you find an amplitude and frequency of shaking so that the broom stays upright if started from a near upright position? You probably cannot find linear equations to solve that will give you a control strategy. So, this problem might best be solved by guessing on the computer. Successful strategies require the hand acceleration to be quite a bit bigger than $g$ , the gravitational constant. The stability obtained is like the stability of an undamped uninverted pendulum — oscillations persist. You improve the stability a little by including a little friction in the hinge.
(h)

Dinner table experiment for nerdy eaters. If you put a table knife on a table and put your finger down on the tip of the blade, you can see that this experiment might work. Rapidly shake your hand back and forth, keeping the knife from sliding out from your finger but with the knife sliding rapidly on the table. Note that the knife aligns with the direction of shaking (use scratch-resistant surface). The knife is different from the broom in some important ways: there is no gravity trying to ‘unalign’ it and there is friction between the knife and table that is much more significant than the broom interaction with the air. Nonetheless, the experiment should convince you of the plausibility of the balancing mechanism. Because of the large accelerations required, you cannot do this experiment with a broom.

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Figure 19.125

The sketch of the knife on the table goes with part (h).

19.3.28 Double pendulum. The double pendulum shown is made up of two uniform bars, each of length $\ell$ and mass $m$ . The pendulum is released from rest at $\phi_{1}=0$ and $\phi_{2}=\pi/2$ . Just after release what are the values of $\ddot{\phi}_{1}$ and $\ddot{\phi}_{2}$ ? Answer in terms of other quantities. Answer: $\ddot{\phi}_{1}=0,\;\ddot{\phi}_{2}=-3g/2\ell$ .

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Figure 19.126

19.3.29 A rocker. A standing dummy is modeled as having massless rigid circular feet of radius $R$ rigidly attached to their uniform rigid body of length $L$ and mass $m$ . The feet do not slip on the floor.

(a)

Given the tip angle $\phi$ , the tip rate $\dot{\phi}$ and the values of the various parameters ( $m$ , $R$ , $L$ , $g$ ) find $\ddot{\phi}$ . [You may assume $\phi$ and $\dot{\phi}$ are small.]

Answer: $\ddot{\phi}+\left[3(R-\frac{L}{2})\frac{g}{L^{2}}\right]\phi=0$ .
(b)

Using the result of (a) or any other clear reasoning find the conditions on the parameters ( $m$ , $R$ , $L$ , $g$ ) that make vertical passive dynamic standing stable. [Stable means that if the person is slightly perturbed from vertically up that their resulting motion will be such that they remain nearly vertically up for all future time.]

Answer: This first-order approximation show that stable standing requires $R>\frac{L}{2}$ . $R=\frac{L}{2}$ is neutrally stable (like a wheel). (Someone questions this answer, so don’t trust it too much.)

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Figure 19.127

19.3.30 Consider a rigid spoked wheel with no rim. Assume that when it rolls a spoke hits the ground and doesn’t bounce. The body just swings around the contact point until the next spoke hits the ground. The uniform spokes have length $R$ . Assume that the mass of the wheel is $m$ , and that the polar moment of inertia about its center is $I$ (use $I=mR^{2}/2$ if you want to get a better sense of the solution). Assume that just before collision number $n$ , the angular velocity of the wheel is $\omega_{n-}$ , the kinetic energy is $T_{n-}$ , the potential energy (you must clearly define your datum) is $U_{n}-$ . Just after collision $n$ the angular velocity of the wheel is $\omega_{n+}$ . The Kinetic Energy is $T_{N+}$ , the potential energy (you must clearly define your datum) is $U_{n+}$ . The wheel has $k$ spokes (pick $k=4$ if you have trouble with abstraction). This problem is not easy. It can be answered at a variety of levels. The deeper you get into it the more you will learn.

(a)

What is the relation between $\omega_{n-}$ and $T_{n-}$ ?
(b)

What is the relation between $\omega_{n-}$ and $\omega_{n+}$ ?
(c)

Assume ‘rolling’ on level ground. What is the relation between $\omega_{n}^{+}$ and $\omega_{n+1}^{+}$ (the angular velocities just after two successive collisions?
(d)

Assume rolling down hill at slope $\theta$ . What is the relation between $\omega_{n+}$ and $\omega_{n+1}$ ?
(e)

Can it be true that $\omega_{n+}$ = $\omega_{(n+1)+}$ ? About how fast is the wheel going in this situation?
(f)

As the number of spokes $m$ goes to infinity, in what senses does this wheel become like an ordinary wheel?

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Figure 19.128

	$\displaystyle a_{n}$	$\displaystyle=$	$\displaystyle\frac{v^{2}}{\rho}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\rho$	$\displaystyle=$	$\displaystyle\frac{v^{2}}{a_{n}}=\frac{(0.9\,\mathrm{m}/\mathrm{s})^{2}}{1.2\,% \mathrm{m}/\mathrm{s}^{2}}=0.675\,\mathrm{m}.$

	$\displaystyle\frac{2}{9}m\omega_{D}^{2}R$	$\displaystyle=$	$\displaystyle mg$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega_{D}$	$\displaystyle=$	$\displaystyle 3\sqrt{\frac{g}{R}}.$

$\displaystyle\dot{\theta}$	$\displaystyle=$	$\displaystyle\omega$
$\displaystyle\dot{\omega}$	$\displaystyle=$	$\displaystyle-\frac{2mRv\omega}{I_{zz}+mR^{2}}$
$\displaystyle\dot{R}$	$\displaystyle=$	$\displaystyle v$
$\displaystyle\dot{v}$	$\displaystyle=$	$\displaystyle\omega^{2}R$