Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap9 Hydrostatics 11 Vibrations

Chapter 10 Dynamics in 1D

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10.0Introductionstudy PDF
10.1Force and motion in 1Dstudy PDF + solutions
10.2Energy methods in 1Dstudy PDF + solutions
10.3A mass and spring: the harmonic oscillatorstudy PDF + solutions
10.4Coupled motions in 1Dstudy PDF + solutions
10.5Collisions in 1Dstudy PDF + solutions

The scalar equation $F=ma$ introduces the concepts of motion and time derivatives to mechanics. In particular the equations of dynamics are seen to reduce to ordinary differential equations, the simplest of which have memorable analytic solutions. The harder differential equations need be solved on a computer. We explore various concepts and applications involving momentum, power, work, kinetic and potential energies, oscillations, collisions and multi-particle systems.

We now progress from statics to dynamics. As the names imply, statics generally concerns things that don’t move, or at least don’t accelerate much, whereas dynamics concerns things whose motion is of central interest. In statics, we neglected inertial (terms involving acceleration of mass). So, in statics the linear and angular momentum balance equations were reduced to force and moment balance. In dynamics the inertial terms in the momentum balance equations are important. In statics all the forces and moments cancel each other. In dynamics the forces and moments add up to cause the acceleration of mass.

Once you have mastered free-body diagrams and statics, the hard part of dynamics is learning how to keep track of motion. Keeping track of motion, without worrying about forces, is called kinematics. Kinematics is geometry in motion. The study of kinematics together with forces is dynamics, in the mechanics sense of the word

^†^†margin: Outside of mechanics the word dynamics means the study of anything which changes in time. For example, “relationship dynamics” concerns how people’s interpersonal feelings change over time, and the “dynamics of the market place” concerns fluctuations of prices due to supply, demand and so on. Electro-dynamics is the study of how voltage varies in time and dynamics of the hormonal system concerns how hormone levels go up and down.

Systems with this more general dynamics are sometimes called “dynamical systems”. The classic dynamical systems equation is

{d\over dt}z=f(z)

where

z

is a list of numbers describing the state of the system and

f(z)

is the set of rules that dictate how the system changes due to its present state. One sub-class of dynamical systems are the mechanical systems in this book where a set of equations, each of the form

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

are re-written in the form

\dot{z}=f(z),

as you will see.

Dynamics is called kinetics. We will develop our understanding of dynamics (kinetics) by considering progressively more complex kinematics.

This first dynamics chapter is limited to the unconstrained dynamics of one or more particles moving in one spatial dimension (1D). Each particle moves along a straight line (and not on a planar or spatial curve) and all forces are along that line. What is a particle?

A particle is a system idealized as being totally characterized by its position (as a function of time) and its (fixed) mass (read more on page 5.1).

Unconstrained motion. Finally, in this chapter we only consider cases where the applied forces are either given as a function of time or can be determined from the positions and velocities of the particles. The time-varying thrust from an engine might be thought of as a force given as a function of time. Gravity and springs cause forces which are functions of position. And the drag on a particle as it moves through air or water can be modeled as a force depending on velocity. We postpone until the next chapter forces caused by geometric constraints, for example the forces between particles connected by strings or rods. Such constraint forces need to be solved-for using dynamics. In contrast, the forces in this chapter can be found a priori from position, velocity or time.

Kinematics, acceleration and calculus. The main new concept here, which stays with us until the end of the book, is that things change with time. We keep track of that change using calculus. In particular, the equation $F=ma$ is a differential equation because

a\equiv\frac{d^{2}}{dt^{2}}x=\ddot{x}\ .

Any equation that has terms which are derivatives of functions is a differential equation. Because acceleration $a$ is defined in terms of derivatives any equation involving $a$ is a differential equation.

The organization of this chapter

The first three sections are a review and deepening of material from freshman physics: $F=ma$ , energy methods, and the harmonic oscillator. The last three sections concern multi-particle systems, collisions and collections of masses connected with springs and dashpots.

Before going on please get a better lay of the land by reading the introduction to mechanics in Chapter 1 and looking over the summary of mechanics on the inside cover.

10.1 Force and motion in 1D

Now we focus on special problems in which one particle moves on a straight line. With motion in only one direction, the kinematics is simple. It’s essentially a rehash of freshman calculus. Even in 1D, vectors can be useful because of their help with signs. But vectors are not really needed and we will not be zealous in their use (in this one chapter).

Filename:tfigure8-rel-ang-vel

Figure 10.1: One-dimensional position, velocity, and acceleration. Positive values for

x,\dot{x}

and

\ddot{x}

describe positions, velocities and accelerations in the positive

x

direction.

Position, velocity, and acceleration in one dimension

We can call the direction of motion the $\hat{\boldsymbol{\imath}}$ direction and the position of the particle, the distance of a particle in the $+\hat{\boldsymbol{\imath}}$ direction from a reference point, $x$ (see fig. 10.1). The particle might have some spacial extent, so to be precise we can define $x$ as the $x$ coordinate of the particle’s center-of-mass. We can write the position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ , velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ as

Filename:tfigure8-ang-vel-ex

Figure 10.2: Graphs of

x(t)

and

v(t)=\frac{dx}{dt}

versus time. The slope of the position curve

dx/dt

t^{*}

v(t^{*})

. And the slope of the velocity curve

dv/dt

t^{*}

a(t^{*})

position	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=x\hat{\boldsymbol{\imath}}$		$\displaystyle=x\hat{\boldsymbol{\imath}},$
velocity	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=v\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle{dx\over dt}\hat{\boldsymbol{\imath}}$		$\displaystyle=\dot{x}\hat{\boldsymbol{\imath}},$
andacceleration	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=a\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle{dv\over dt}\hat{\boldsymbol{\imath}}$	$\displaystyle={d^{2}x\over dt^{2}}\hat{\boldsymbol{\imath}}$	$\displaystyle=\ddot{x}\hat{\boldsymbol{\imath}}.$	(10.1)

Figure 10.2 shows example graphs of $x(t)$ and $v(t)$ versus time.

Signs. Vectors help with signs. When not using vectors we will take $v$ and $a$ to be positive if they have the same direction as increasing $x$ (or $y$ or whatever coordinate describes position). Even though we pedantically declare that ‘velocity is a vector’ and ‘acceleration is a vector’, we will loosely use the words ‘velocity’ and ‘acceleration’ to stand for scalars using these sign conventions.

Example: Position, velocity, and acceleration in one dimension

If position is given as
                and $x(t)=3e^{4t/\,\mathrm{s}}\,\mathrm{m}$
                then $v(t)=dx/dt=12e^{4t/\,\mathrm{s}}\,\mathrm{m}/\mathrm{s}$
                and $a(t)=dv/dt=48e^{4t/\,\mathrm{s}}\,\mathrm{m}/\mathrm{s}^{2}$ .
So at, say, time $t=2\,\mathrm{s}$ the acceleration is

a|_{t=2\,\mathrm{s}}=48e^{4\cdot 2\,\mathrm{s}/\,\mathrm{s}}(\,\mathrm{m}/% \mathrm{s}^{2})=48\cdot e^{8}\,\mathrm{m}/\mathrm{s}^{2}\approx 1.43\cdot 10^{% 5}\,\mathrm{m}/\mathrm{s}^{2}.

Fussing with units. In this example above we used 1/s as part of the argument of the exponential function. Thus when the exponential $e^{4t/\,\mathrm{s}}$ is differentiated with respect to time $t$ the 1/s is grouped with the 4 as the coefficient of $t$ in the exponential. So the factor 4/s comes out front according to the chain rule of differentiation. Treating units as quantities, manipulated like all others makes the answer come out with the right units also. Note in the last line that, when the dimensional quantity ( $2\,\mathrm{s}$ ) is substituted in for the variable $t$ , the units cancel. For more on units see appendix 4.

Filename:tfigure8-ang-vel

Figure 10.3: One-dimensional kinematics of a particle: (a) is a graph of the acceleration of a particle

a(t)

; (b) is a graph of the particle velocity

v(t)

and the integral of

a(t)

from

t_{0}=0

t^{*}

, the shaded area under the acceleration curve; (c) is the position of the particle

x(t)

and the integral of

v(t)

from

t_{0}=0

t^{*}

, the shaded area under the velocity curve.

1D kinematics $\Leftrightarrow$ calculus.

One-dimensional kinematics problems can include almost all of the skills in elementary calculus. In kinematics you are often given position, velocity or acceleration as function of time and you have to differentiate or integrate to find one of the other quantities. For example, if you are given the velocity $v(t)$ as a function of time and are asked to find the acceleration $a(t)$ , you have to differentiate. If instead you were asked to find the position $x(t)$ , you would calculate an integral (see fig. 10.3). Using the fundamental theorem of calculus, we get the integral versions of the relations between position, velocity, and acceleration (see fig. 10.3).

	$\displaystyle x(t)$	$\displaystyle=$	$\displaystyle x_{0}+\int_{t_{o}}^{t}v(\tau)\,d\tau\qquad\mbox{with}\qquad x_{0% }=x(t_{0}),\mbox{\ and}$
	$\displaystyle v(t)$	$\displaystyle=$	$\displaystyle v_{0}+\int_{t_{o}}^{t}a(\tau)\,d\tau\qquad\mbox{with}\qquad v_{0% }=v(t_{0}).$

With indefinite notation, these equations can also be written as:

	$\displaystyle x$	$\displaystyle=\int v\,dt$
	$\displaystyle v$	$\displaystyle=\int a\,dt.$

If acceleration is given as a function of time, then position is found by integrating twice.

1D kinematics, bicycles and calculus. To put it another way, almost every calculus question could be phrased as a question about a bicycle speedometer

^†^†margin: To cover the range of calculus problems you need to be a very good rider, however, and be able to ride frontwards, backwards, at zero speed and infinitely fast.

With a bicycle speedometer (which includes a distance-measuring odometer) you can read your speed and distance travelled as functions of time. And given one of those two functions you could find the other using calculus. Acceleration is also of interest, but few bicycle speedometers also measure acceleration.

Differential equations

A differential equation is an equation that involves derivatives. Thus the equation relating position to velocity is

{dx\over dt}=v\qquad\mbox{or, more explicitly}\qquad{dx(t)\over dt}=v(t),

is a differential equation. An ordinary differential equation (ODE) is an equation that contains some terms that are ordinary derivatives (as opposed to partial derivatives and partial differential equations which we don’t use in this book).

Example: Calculating a derivative solves an ODE

Given that the height of an elevator as a function of time on its 5 seconds long 3 meter trip from the first to second floor is

y(t)=(3\,\mathrm{m}){\left(1-\cos\left({\pi t\over 5\,\mathrm{s}}\right)\right% )\over 2}

we can solve the differential equation $v={dy\over dt}$ by differentiating to get

v={dy\over dt}={d\over dt}\left[(3\,\mathrm{m}){\left(1-\cos\left({\pi t\over 5% \,\mathrm{s}}\right)\right)\over 2}\right]={3\pi\over 10}\sin\left({\pi t\over 5% \,\mathrm{s}}\right){\,\mathrm{m}/\mathrm{s}}

Note: this would be a harsh elevator because of the jump in the acceleration (not calculated above) at the start and stop.

A little less trivial is the case when you want to find a function when you are given the derivative.

Example: Integration solves a simple ODE

Assume that you start at home ( $x=0$ ) and, over about 30 seconds, you accelerate towards a steady-state speed of $4\,\mathrm{m}/\mathrm{s}$ according to (see fig. 10.4)

Filename:tfigure8-ang-accel

Figure 10.4: Velocity (speed) vs time for a 1000 s bike trip. The rider increases speed to exponentially approach 4 m/s.

v(t)=4(1-e^{-t/(30\,\mathrm{s})})\,\mathrm{m}/\mathrm{s}.

Your ride lasts 1000 seconds. We can find how far you go by solving

\dot{x}=v(t)\quad\mbox{with the initial condition}\quad x(0)=0.

This is simply solved by integration. Say, after 1000 seconds

\begin{array}[]{rclccc}x(t=1000\,\mathrm{s})&=&\int_{0}^{1000\,\mathrm{s}}v(t)% \ dt=\int_{0}^{1000\,\mathrm{s}}4(1-e^{-t/(30\,\mathrm{s})})(\,\mathrm{m}/% \mathrm{s})\ dt\\ \\ &=&\left.\left(4t+(120\,\mathrm{s})e^{-t/(30\,\mathrm{s})}\right)\right|^{1000% \,\mathrm{s}}_{0}\,\mathrm{m}/\mathrm{s}\\ \\ &=&\left((4\cdot 1000\,\mathrm{s}+(120\,\mathrm{s})e^{-100/3})-(0+(120\,% \mathrm{s})e^{0})\right)\,\mathrm{m}/\mathrm{s}\\ \\ &=&\left(4000-120+120e^{-100/3})\par\right)\,\mathrm{m}\\ \\ &\approx&3880\,\mathrm{m}\hfil\mbox{(to within an angstrom or so)}\\ \end{array}

This is only $120\,\mathrm{m}$ less than if the whole trip was travelled at a steady $4\,\mathrm{m}/\mathrm{s}$ (then $x=4\,\mathrm{m}/\mathrm{s}\times 1000\,\mathrm{s}=4000\,\mathrm{m}$ ).

Unlike the integral above, many integrals cannot be evaluated by hand (analytically).

Example: An ODE that leads to an intractable integral

Assume now that

v(t)={4t\over t+e^{-t/(30\,\mathrm{s})}\,\mathrm{s}}\,\mathrm{m}.

Again we have a bike trip where you start at zero speed and approach a steady speed of $4\,\mathrm{m}/\mathrm{s}$ . So your position as a function of time should be similar. Following the last example, we have

\dot{x}=v(t)\quad\mbox{with the initial condition}\quad x(0)=0

with the given $v(t)$ . The integral for position is then

\begin{array}[]{rclccc}x(t=1000\,\mathrm{s})&=&\int_{0}^{1000\,\mathrm{s}}v(t)% \ dt=\int_{0}^{1000\,\mathrm{s}}{4t\over t+e^{-t/(30\,\mathrm{s})}\,\mathrm{s}% }\,\mathrm{m}\ dt\\ \\ &=&\dots\par\end{array}

(10.2)

which is the kind of thing you have nightmares about seeing on an exam. You couldn’t solve this integral if your life depended on it. No one could. There is no formula for $x(t)$ that solves the differential equation, unless you regard eqn. (10.2) as a formula. In days of old they would say ‘the problem has been reduced to quadrature’ meaning that the remaining work was evaluating an integral

^†^†margin: Literally quadrature means finding a square (a ‘quad’) with area equal to the area under a given curve. The phrase ‘reduced to quadrature’ is used more generally to mean integration, even if the integration is of several variables.

, even if they didn’t know how to evaluate it exactly.

Just because a differential equation can’t be solved analytically with pencil and paper doesn’t mean it can’t be solved numerically. Most often the setup for numerical solution is not that difficult. Note that for numerical solution you either need dimensionless calculations, or at least need all variables to be in consistent units.

Example: Numerical solution of ODE

One of many ways to evaluate the integral of the above example numerically is by the following pseudo code.

     ODE = { xdot = 4 t / (t+e^(-t/30)) }
     IC  = { x(0) = 0  }
     solve ODE with IC and evaluate at t=1000

The result is $x\approx 3988\,\mathrm{m}$ which is also, as expected because of the similarity with the previous example, only slightly shy of the steady-speed approximation of $4000\,\mathrm{m}$ .

^†^†margin: Numerical error vs real difference. When you notice such small differences (12m out of 4000m) based on computer calculation you need to question whether the difference is something real in the problem or, rather, is due to numerical errors. Two ways to check are with so-called convergence tests and by using your canned package’s error estimate. We checked that the numerical integration is accurate to about

10^{-3}

m, less than the 1m resolution that we printed (no need to type lots of digits with little information). Thus the

\approx 12

m difference between the constant

v

solution and the solution where

v

approaches a constant, is a real difference. The startup is genuinely quicker in the second example (12 m behind constant

v

versus 120 m behind constant

v

), and not a numerical artifact.

More differential equations.

As mentioned, because dynamics equations contain derivatives they are all differential equations. A catalogue of the simplest differential equations and their solutions is given in box 3.1 on page 3.1.

The equations of dynamics

We want to understand kinetics (mechanics, dynamics), not just kinematics. The subject of mechanics is held up by the three pillars of material properties, geometry, and ‘Newton’s laws’ (see page 0.1). Here we begin to flesh out the ‘Newton’s laws’ pillar beyond statics (the first 8 chapters of this book), using kinematics (we just started with that above) to the third pillar, dynamics.

Linear momentum balance

For a particle moving in the $x$ direction the velocity and acceleration are ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=v\hat{\boldsymbol{\imath}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=a\hat{\boldsymbol{\imath}}$ .

Thus the linear momentum and its rate of change are

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{L}}}$	$\displaystyle\equiv\sum m_{i}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i}$	$\displaystyle=m{\accentset{\rightharpoonup}{\boldsymbol{v}}}=mv\hat{% \boldsymbol{\imath}},\mbox{ and}$
	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$	$\displaystyle\equiv\sum m_{i}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{i}$	$\displaystyle=m{\accentset{\rightharpoonup}{\boldsymbol{a}}}=\,\mathrm{m}a\hat% {\boldsymbol{\imath}}.$

Filename:tfigure8-ang-accel-ex

Figure 10.5: Three possible free-body diagrams for 1D particle motion. It is assumed that

{\accentset{\rightharpoonup}{\boldsymbol{F}}}

represents the sum of all forces acting on the particle. Note, even though the inertial term

m a

is of special interest, it does not show on the FBD. See the free-body diagram Chapter 3 for the prescription about this. See box 15.1 on page 15.1 for a further warning about d’Alembert ‘forces’.

Using any of the free-body diagrams in fig. 10.5, where ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=F\hat{\boldsymbol{\imath}}$ , the equation of linear momentum balance,

^†^†margin: Angular momentum? We skip thinking about angular momentum balance in this section. Why? If we pick an origin on the line of travel, all terms on both sides of all angular momentum balance equations are zero, and the equation

0=0

tells us nothing new.

eqn. I from the front inside cover, or equation 12.1 reduces to:

F\hat{\boldsymbol{\imath}}=ma\hat{\boldsymbol{\imath}}

(10.3)

which in scalar form is the central subject of this section.

F=ma

In scalar form, $F$ is the net force to the right and $a$ is the acceleration to the right. For the equation $F=ma$ to have content each of the terms must have some meaning in other contexts. And, at least intuitively, each does (see box 10.1 on page 10.1).

Force. The force $F$ could come from a spring, or a fluid or from your hand pushing the thing to the right or left, or any combination of these things. The most general case we want to consider here is that the force is determined by the position and velocity of the particle as well as the present time. Thus

F=f(x,v,t).

(10.4)

What do we mean ‘determined by’? We mean that we have an independent way of knowing the force from its position, velocity and time, even without thinking yet about the linear momentum balance equation $F=ma$ . Special cases would be, say,

\begin{array}[]{rccll}F&=&f(t)&=F_{0}\sin(\beta t)&\mbox{an oscillating load, % }\\ F&=&mg&&\mbox{the force of earth's gravity}\\ F&=&f(v)&=-cv&\mbox{a linear viscous drag,}\\ F&=&f(x)&=-kx&\mbox{a linear spring, and}\\ F&=&f(x,v,t)&=-kx-cv+F_{0}\sin(\beta t)&\mbox{a combination of forces.}\\ \end{array}

All elementary 1D particle mechanics problems can be reduced to the solution of this pair of coupled first order differential equations,

\begin{array}[]{rclr}{dv\over dt}&=&\underbrace{f(x,v,t)/m}_{a(t)}&\hfill(a)\\ \\ {dx\over dt}&=&v(t)&(b)\end{array}

(10.5)

where the function $f(x,v,t)$ is given and $x(t)$ and $v(t)$ are to be found.

Know a solution when you see one. How can you tell if candidate functions solve a differential equation? First you can tell that the initial conditions are satisfied by evaluating the expressions at $t=0$ . To check that the differential equations are satisfied, you plug the candidate solutions into the equation and see that an identity results. Differential equations are satisfied when the unknown functions therein are replaced with specific functions that make the equations correct.

Viscous drag.

If the only applied force is a viscous drag, $F=-cv$ (see fig. 10.6), then linear momentum balance ( $F=ma$ ) would be $-cv=ma$ and Eqns. 10.5 are

\begin{array}[]{rclr}{dv\over dt}&=&-cv/m\\ \\ {dx\over dt}&=&v\\ \end{array}

where $c$ and $m$ are constants and $x(t)$ and $v(t)$ are yet to be determined functions of time. Because the force only slows the particle there will be no motion, unless the particle was given some initial velocity. In general, you need to specify the initial position and velocity to find a solution. So we complete the problem statement by specifying initial conditions

Example: Slowing with viscous drag

Find $x(t)$ given that

Filename:tfigure8-ang-accel1

Figure 10.6: Three acceptable free-body diagrams showing the viscous drag on a particle moving to the right at speed

v

. The second FBD depends on being told on the side, say here, that

F=-cv

x(0)=x_{0}\hbox to36.135pt{\hfil and\hfil}v(0)=v_{0}

where $x_{0}$ and $v_{0}$ are given constants. Before worrying about how to solve such equations, you should know how to recognize a solution. The following two functions, assume for now that they fell from the sky, solve the differential equations.

\begin{array}[]{cccc}v(t)&=&v_{0}e^{-ct/m},\qquad\mbox{and}\\ x(t)&=&x_{0}+mv_{0}(1-e^{-ct/m})/c\\ \end{array}

Plugging this presumed solution for $v(t)$ into $\dot{v}=-cv/m$ gives, and this is what we want, $0=0$ . And similarly, when the presumed solution for $x(t)$ is plugged in to $\dot{x}=v$ you also get the ‘satisfying’ result that $0=0$ .

Replacing the unknown functions $v(t)$ and $x(t)$ with the given formulas gives an identity. Thus the given formulas satisfy (or solve) the differential equations.

Just like the case of integration (or equivalently the solution for $x$ of the ODE $\dot{x}=v(t)$ ), one often cannot find formulas for the solutions of differential equations.

Example: A dynamics problem with no pencil and paper solution

Consider the following case which models a particle in a sinusoidal force field with a second applied force that oscillates in time. Using the dimensional constants $c,d,F_{0},\beta$ , and $m$ ,

\begin{array}[]{rclr}{dv\over dt}&=&\left(c\sin(x/d)+F_{0}\sin(\beta t)\right)% /m\\ \\ {dx\over dt}&=&v\\ \\ {\mbox{with initial conditions $x(0)=0\hbox to36.135pt{\hfil and\hfil}v(0)=0$.% }}\\ \end{array}

There is no known formula for $x(t)$ that solves this ODE.

Just writing the ordinary differential equations and initial conditions is analogous to setting up an integral in freshman calculus. The solution is reduced to quadrature. Because numerical solution of sets of ordinary differential equations is a standard part of all modern computation packages you are in some sense done when you get this far. A computer can finish up for you.

Some special cases in 1D mechanics

There are various special cases of eqn. (10.5) which have simple solutions.

Example: The simplest dynamics problem.

1D, particle, no force. Formally working out the details,

\begin{array}[]{lcrcl}F=ma\ \textrm{with}\ F=0&\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}&0&=&ma\\ \textrm{definition of }a&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&\dot{v}&=&0% \\ \textrm{integrating}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&v&=&v_{0}\quad% \textrm{(= any constant)}\\ \textrm{integrating again}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&x&=&x_{0}% +v_{0}t\\ \end{array}

In a sense we have thus derived Newton’s first law, ‘an object in motion tends to stay in motion unless acted upon by a force’.

Constant force. Another simple case is constant force $F$ which leads to constant acceleration $a=F/m$ . Using calculus you should know well by now, you get the following formulas:

$\displaystyle a=const\Rightarrow$	$\displaystyle x$	$\displaystyle=x_{0}+v_{0}t+at^{2}/2$
$\displaystyle a=const\Rightarrow$	$\displaystyle v$	$\displaystyle=v_{0}+at$
$\displaystyle a=const\Rightarrow$	$\displaystyle v$	$\displaystyle=\pm\sqrt{v_{0}^{2}+2a(x-x_{0})}.$

These are much seen in high school physics because, by permuting what is given and what is unknown, one can make up 100 homework problems that can be solved with these formulas and without calculus.

Force given as a function of time. Say $F$ is given as $F=F(t)$ . This general case shows up when some kind of motor force is controlled by a human or computer to vary in time is some predetermined manner.

\begin{array}[]{lcrcl}F=ma\ \textrm{with}\ F=F(t)&\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}&F(t)&=&ma\\ \textrm{definition of}\ a&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&\dot{v}&=&% F(t)/m\\ \textrm{integrating}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&v&=&v_{0}+\frac% {1}{m}\int_{0}^{t}F(\tau)\,d\tau\end{array}

And we have to integrate once again to get position.

Example: Ramping up the acceleration at the start

If you get a car going by gradually depressing the ‘accelerator’ so that its acceleration increases linearly with time, we have

\begin{array}[]{cccccc}a&=&ct&&\mbox{(take $\,t=0\,$ at the start)}\\ \hbox to36.135pt{\hfil$\Rightarrow$\hfil}v(t)&=&\int^{t}_{0}ad\tau+v_{0}&=&% \int^{t}_{0}c\tau d\tau\quad=\quad ct^{2}/2\\ &&&&\mbox{(since $v_{0}=0$)}\\ \hbox to36.135pt{\hfil$\Rightarrow$\hfil}x(t)&=&\int^{t}_{0}vd\tau+x_{0}&=&% \int^{t}_{0}(c\tau^{2}/2)d\tau\quad=\quad ct^{3}/6\\ &&&&\mbox{(since $x_{0}=0$)}.\\ \end{array}

The distance the car travels is proportional to the cube of the time that has passed from dead stop.

The overall subject of ‘vibrations’ is in some sense about what happens when something is shaken. We can think of ‘shaking’ as applying a force which varies sinusoidally in time.

Example: Force varies sinusoidally in time.

Assume a 1 kg mass starts from rest and has a force of $F=2\cos(2\pi t/\,\mathrm{s})\,{\mbox{N}}$ applied. That’s a force that oscillates once per second with an amplitude of 2 N. What is the position at $t=10\,\mathrm{s}$ ?

\begin{array}[]{lcrcl}\par F=ma\ \textrm{with}\ F=F(t)&\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}&2\cos(2\pi t/\,\mathrm{s})\,{\mbox{N}}&=&m\dot{v}\\ \textrm{integrating}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&v&=&v_{0}+\frac% {1}{m}\int_{0}^{t}2\cos(2\pi t/\,\mathrm{s})\,{\mbox{N}}\,d\tau\\ \textrm{using freshman calculus}&\hbox to36.135pt{\hfil$\Rightarrow$\hfil}&v&=% &v_{0}+\frac{1}{\pi m}\sin(2\pi t/\,\mathrm{s})\,{\mbox{N}}\,\mathrm{s}\\ \textrm{IC: $v(0)=0\Rightarrow v_{0}=0$}&\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}&v&=&\frac{1}{\pi m}\sin(2\pi t/\,\mathrm{s})\,{\mbox{N}}\,\mathrm{s}\\ \textrm{integrate again, using $\dot{x}=v$}&\hbox to36.135pt{\hfil$\Rightarrow% $\hfil}&x&=&x_{0}-\frac{1}{2\pi^{2}m}\cos(2\pi t/\,\mathrm{s})\,{\mbox{N}}\,% \mathrm{s}^{2}\\ \textrm{IC: $x(0)=0\Rightarrow x_{0}=\frac{1}{2\pi^{2}m}$}&\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}&x&=&\frac{1}{2\pi^{2}m}\left(1-\cos(2\pi t/\,\mathrm{% s})\right)\,{\mbox{N}}\,\mathrm{s}^{2}\end{array}

Now we can substitute in $m=1\,\mathrm{kg}$ and $t=10\,\mathrm{s}$ to get $x=0.0\,\mathrm{m}$ .

The algebraic cancellation of units came about naturally from substituting in the definition of a Newton $1\,{\mbox{N}}=1\,\mathrm{kg}\,\mathrm{m}/\,\mathrm{s}^{2}$ . We carried the units through even though the final answer was 0.

Force depends on velocity. This case is encountered when, say, an object moves through a fluid and other forces, say gravity, are negligible. Here we have

\par F=ma\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F(v)=m\dot{v}.

This is solved by multiplying both sides by $d t$ and dividing both sides by $F(v)$ and integrating to get

\int dt=m\int\frac{dv}{F(v)}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\par t=m% \int_{v_{0}}^{v}\frac{dv^{\prime}}{F(v^{\prime})}\par

If we want to know position vs time we have to integrate once again.

Example: The slowing of a bullet.

The main force on a bullet after it leaves the gun and before it hits its mark is from air drag. This drag is roughly proportional to the speed squared, thus

F=ma\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-cv^{2}=m\dot{v}\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}-c\int dt=m\int_{v_{0}}^{v}\frac{dv^{\prime}}{v^{% \prime 2}}.

Carrying out the integrals ( $\int dt=t$ and $\int v^{-2}\,dv=-v^{-1}$ ) we get

ct=m\left(\frac{1}{v}-\frac{1}{v_{0}}\right)\hbox to36.135pt{\hfil$\Rightarrow% $\hfil}v=\frac{v_{0}}{cv_{0}t/m+1}

To get position we would integrate again to get:

x=\int_{0}^{t}v(t^{\prime})\,dt^{\prime}=\int_{0}^{t}\frac{v_{0}}{cv_{0}t^{% \prime}/m+1}\,dt^{\prime}=(m/c)\ln(1+cv_{0}t/m)

Interestingly, according to this equation (which becomes less and less accurate as the bullet slows and gravity and eventually viscous forces become important) the bullet goes an infinite distance before stopping.

Force varies with position. This case, where $F=F(x)$ will be treated in some detail in the next section on energy.

The simplest ODEs

The simplest and most common ODEs in dynamics, and in the rest of science and engineering, are

: Linear: e.g., no functions squared.
: First or second order: Have only first or second derivatives, respectively, and
: Constant coefficient: All multiples of the derivatives are constants, not functions of time.

The most essential of these are

\dot{x}=0,\dot{x}=C,\ddot{x}=0,\ddot{x}=C,\dot{x}=Cx,\dot{x}=-Cx,\ddot{x}=Cx\ % \&\ \ddot{x}=-Cx.

These are discussed in appendix 3.1 on page 3.1.

What do the terms in $F=ma$ mean?

Some say ‘ $F=ma$ is the definition of force’. This is a legitimate point of view. But, if we care about force for other reasons, which we do, it is not useful. For the equation $F=ma$ to be useful we need to be able to think about $F$ , $m$ and $a$ independently from each other.

Mass, $m$ . Now that we know about atoms (these centuries) and what they are made of (these decades) we can approximately (with about one percent accuracy) define the mass of a system by (in principle) by counting up the total number of protons and neutrons and multiplying by $1.67\cdot 10^{-27}\,\mathrm{kg}$ . That is, mass is a measure of the extent of matter. Given that we think of mass as the amount of matter, we could more accurately and more easily use a reference volume of a pure chemical substance as a reference. This way, with a good balance and some trouble, we could get an accuracy of parts per million. Officially,

mass is measured in comparison to a fancy piece of metal

locked in a box in some basement in some government building. That calibrated kilogram is accurate to about a part in 20,000,000 (See Appendix 4 starting on page 4). We can find the mass of a more complicated thing using that reference mass and a (very good) balance.

Acceleration, $a$ . Because this is a course in mechanics and not in philosophy of science, we will just accept the concepts of space ( $x$ ) and time ( $t$ ) as given and measurable (using rulers and timers). So acceleration is operationally well defined as

a=d^{2}x/dt^{2}

with no use of $F=ma$ . We have presupposed that we measure position relative to a Newtonian reference system. This definition is sensible to at least to about one part per billion for most engineering purposes (see page 0.2).

Force, $F$ . Here is where people argue. It’s easiest to define force using the deformation of solids. When one thing pushes on another, think of your little finger as caught in between. How much your finger is squeezed, as measured by how loud you yell, is a measure of force. More technically, we could look at the small amounts of deformation occurring where the bodies contact, and use the deformation as a measure of force. Or, more practically, we could interpose a calibrated material and measure its deformation. Such a chunk of material with deformation-measuring electronics is called a load cell. Load cells are sold by the millions (say, in bathroom scales). A load cell uses nothing about $F=ma$ to operate accurately.

One reason it is nice to think of force as having a life away from $F=ma$ is that the whole coherent and useful subject of statics, useful for designing bridges and airplane landing gear, has little or no use for $F=ma$ . Alternatively, still without thinking about $F=ma$ , one could define force in terms of the net effect of earth’s gravitational pull on a calibrated mass at some officially-ordained location like Potsdam.

However you like to define force, the great result is that with any one of several possible independent definitions, things mostly work out. Miraculously, the same concept of force $F$ works no matter which of these three you take as defining:

$\displaystyle F$	$\displaystyle=mg,$	(10.6)
$\displaystyle F$	$\displaystyle=kx,\textrm{or}$	(10.7)
$\displaystyle F$	$\displaystyle=ma.$	(10.8)

Pick your favorite as fundamental and use the others with confidence.

Units of force. Most beginners prefer that forces be measured in Newtons. One Newton is 1 $\,\mathrm{kg}$ $\,\mathrm{m}/\mathrm{s}^{2}$ . But people not living in the SI world tend instead to use the kgf, also called a kilogram force or kp or kilopond. One kgf is the weight of a kilogram. It’s about 9.8 N. In the English system the force most analogous to the Newton is called the Poundal, it’s 1 $\,{\mathrm{lbm}}$ $\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ and is little used. More commonly used is the pound force, $\,{\rm lbf}$ . It’s the weight of a pound mass and is about 32.2 Poundals. Read more about units of force in box 4.1 on page 4.1.

SAMPLE 10.1 Time derivatives: The position of a particle varies with time as ${\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)=(C_{1}t+C_{2}t^{2})\hat{% \boldsymbol{\imath}}$ , where $C_{1}=4\,\mathrm{m}/\mathrm{s}$ and $C_{2}=2\,\mathrm{m}/\mathrm{s}^{2}$ .

1.

Find the velocity and acceleration of the particle as functions of time.
2.

Sketch the position, velocity, and acceleration of the particle against time from $t=0$ to $t=5\,\mathrm{s}$ .
3.

Find the position, velocity, and acceleration of the particle at $t=2\,\mathrm{s}$ .

Solution

We are given the position of the particle as a function of time. We need to find the velocity (time derivative of position) and the acceleration (time derivative of velocity).

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=$	$\displaystyle(C_{1}t+C_{2}t^{2})\hat{\boldsymbol{\imath}}=(4\,\mathrm{m}/% \mathrm{s}\,t+2\,\mathrm{m}/\mathrm{s}^{2}\,t^{2})\hat{\boldsymbol{\imath}}$	(10.9)
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle\equiv$	$\displaystyle\frac{d{\accentset{\rightharpoonup}{\boldsymbol{r}}}}{dt}=\frac{d% }{dt}(C_{1}t+C_{2}t^{2})\hat{\boldsymbol{\imath}}$	(10.10)
	$\displaystyle=$	$\displaystyle(C_{1}+2C_{2}t)\hat{\boldsymbol{\imath}}=(4\,\mathrm{m}/\mathrm{s% }+4\,\mathrm{m}/\mathrm{s}^{2}\;t)\hat{\boldsymbol{\imath}}$	(10.10)
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle\equiv$	$\displaystyle\frac{d{\accentset{\rightharpoonup}{\boldsymbol{v}}}}{dt}=\frac{d% }{dt}(C_{1}+2C_{2}t)\hat{\boldsymbol{\imath}}$	(10.11)
	$\displaystyle=$	$\displaystyle 2C_{2}\hat{\boldsymbol{\imath}}=(4\,\mathrm{m}/\mathrm{s}^{2})% \hat{\boldsymbol{\imath}}$	(10.11)

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=(4\,\mathrm{m}/\mathrm{s}+4\,% \mathrm{m}/\mathrm{s}^{2}\;t)\hat{\boldsymbol{\imath}},\;{\accentset{% \rightharpoonup}{\boldsymbol{a}}}=(4\,\mathrm{m}/\mathrm{s}^{2})\hat{% \boldsymbol{\imath}}.$

Thus, we find that the velocity is a linear function of time and the acceleration is time-independent (a constant).

Filename:sfig8-2-3

Figure 10.7:

2.

We plot eqns. (10.9, 10.10, and 10.11) against time by taking 100 points between $t=0$ and $t=5\,\mathrm{s}$ , and evaluating ${\accentset{\rightharpoonup}{\boldsymbol{r}}},{\accentset{\rightharpoonup}{% \boldsymbol{v}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ at those points. The plots are shown in fig. 10.7.

We can find the position, velocity, and acceleration at $t=2\,\mathrm{s}$ by evaluating their expressions at the given time instant:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}\|_{t=2\,\mathrm{s}}$	$\displaystyle=$	$\displaystyle[(4\,\mathrm{m}/\mathrm{s})\cdot(2\,\mathrm{s})+(2\,\mathrm{m}/% \mathrm{s}^{2})\cdot(2\,\mathrm{s})^{2}]\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle(16\,\mathrm{m})\hat{\boldsymbol{\imath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}\|_{t=2\,\mathrm{s}}$	$\displaystyle=$	$\displaystyle[(4\,\mathrm{m}/\mathrm{s})+(4\,\mathrm{m}/\mathrm{s}^{2})\cdot(2% \,\mathrm{s})]\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle(12\,\mathrm{m}/\mathrm{s})\hat{\boldsymbol{\imath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}\|_{t=2\,\mathrm{s}}$	$\displaystyle=$	$\displaystyle(4\,\mathrm{m}/\mathrm{s}^{2})\hat{\boldsymbol{\imath}}={% \accentset{\rightharpoonup}{\boldsymbol{a}}}(\mbox{for all t})$

Answer: $\mbox{At }t=2\,\mathrm{s},\;{\accentset{\rightharpoonup}{\boldsymbol{r}}}=(16% \,\mathrm{m})\hat{\boldsymbol{\imath}},\;{\accentset{\rightharpoonup}{% \boldsymbol{v}}}=(12\,\mathrm{m}/\mathrm{s})\hat{\boldsymbol{\imath}},\;{% \accentset{\rightharpoonup}{\boldsymbol{a}}}=(4\,\mathrm{m}/\mathrm{s}^{2})% \hat{\boldsymbol{\imath}}.$

SAMPLE 10.2 Math review: Solving simple differential equations. For the following differential equations, find the solution for the given initial conditions.

1.

$\frac{dv}{dt}=a,\;v(t=0)=v_{0},$ where $a$ is a constant.
2.

$\frac{d^{2}x}{dt^{2}}=a,\;x(t=0)=x_{0},\;\dot{x}(t=0)=\dot{x}_{0},$ where $a$ is a constant.

Solution

1.

$\displaystyle\frac{dv}{dt}$ $\displaystyle=$ $\displaystyle a\hbox to36.135pt{\hfil$\Rightarrow$\hfil}dv=a\,dt$

$\displaystyle\hbox to36.135pt{or\hfil}\int dv$ $\displaystyle=$ $\displaystyle\int a\,dt=a\int dt$

$\displaystyle\hbox to36.135pt{or\hfil}v$ $\displaystyle=$ $\displaystyle at+C,\quad\mbox{where C is a constant of integration}$

Now, substituting the initial condition into the solution,

$v(t=0)\equiv v_{0}=a\cdot 0+C\hbox to36.135pt{\hfil$\Rightarrow$\hfil}C=v_{0}.$

Therefore,

$v=at+v_{0}.$

Answer: $v=v_{0}+at$

Alternatively, we can use definite integrals:

$\int_{v_{0}}^{v}dv=\int_{0}^{t}a\,dt\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v% -v_{0}=at\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v=v_{0}+at.$

This is a second order differential equation in $x$ . We can solve this equation by first writing it as a first order differential equation in $v\equiv dx/dt$ , solving for $v$ by integration, and then solving again for $x$ in the same manner.

$\displaystyle\frac{d^{2}x}{dt^{2}}$	$\displaystyle=$	$\displaystyle a\hbox to36.135pt{\hfil or\hfil}\frac{dv}{dt}=a$
$\displaystyle\hbox to36.135pt{or\hfil}\int dv$	$\displaystyle=$	$\displaystyle\int a\,dt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v\equiv\dot{x}$	$\displaystyle=$	$\displaystyle at+C_{1}$	(10.12)
$\displaystyle\mbox{but, }v\equiv\frac{dx}{dt},\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}\int dx$	$\displaystyle=$	$\displaystyle\int at\,dt+\int C_{1}\,dt$
$\displaystyle\hbox to36.135pt{or\hfil}x$	$\displaystyle=$	$\displaystyle\frac{1}{2}at^{2}+C_{1}t+C_{2},$	(10.13)

where $C_{1}$ and $C_{2}$ are constants of integration. Substituting the initial condition for $\dot{x}$ in Eqn. (10.12), we get

\dot{x}(t=0)=\dot{x}_{0}=a\cdot 0+C_{1}\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}C_{1}=\dot{x}_{0}.

Similarly, substituting the initial condition for $x$ in Eqn. (10.13), we get

x(t=0)=x_{0}=\frac{1}{2}a\cdot 0+\dot{x}_{0}\cdot 0+C_{2}\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}C_{2}=x_{0}.

Therefore,

x(t)=x_{0}+\dot{x}_{0}t+\frac{1}{2}at^{2}.

Answer: $x(t)=x_{0}+\dot{x}_{0}t+\frac{1}{2}at^{2}$

SAMPLE 10.3 Constant speed motion: A ship cruises at a constant speed of 15 knots (15 nautical miles per hour) due Northeast. It passes a lighthouse at 8:30 am. The next lighthouse is approximately 35 nautical miles straight ahead. At what time does the ship pass the next lighthouse?

Solution We are given the distance $s$ and the speed of travel $v$ . We need to find how long it takes to travel the given distance.

	$\displaystyle s$	$\displaystyle=$	$\displaystyle vt$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}t$	$\displaystyle=$	$\displaystyle\frac{s}{v}=\frac{35\mbox{nautical miles}}{15\mbox{(nautical % miles)/hour}}=2.33\mbox{ hrs}.$

Now, the time at $t=0$ is 8:30 am. Therefore, the time after 2.33 hrs (2 hours 20 minutes) will be 10:50 am.

Answer: $10:50\mbox{ am}$

SAMPLE 10.4 Constant velocity motion: A particle travels with constant velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=5\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}}$ . The initial position of the particle is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{0}=2\,\mathrm{m}\hat{% \boldsymbol{\imath}}+3\,\mathrm{m}\hat{\boldsymbol{\jmath}}$ . Find the position of the particle at $t=3\,\mathrm{s}$ .

Solution Here, we are given the velocity, i.e., the time derivative of position:

{\accentset{\rightharpoonup}{\boldsymbol{v}}}\equiv\frac{d{\accentset{% \rightharpoonup}{\boldsymbol{r}}}}{dt}=v_{0}\hat{\boldsymbol{\imath}},\qquad% \mbox{where $v_{0}=5\,\mathrm{m}/\mathrm{s}$.}

We need to find ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ at $t=3\,\mathrm{s}$ , given that ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ at $t=0$ is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{0}$ .

Filename:sfig8-2-3a

Figure 10.8:

$\displaystyle d{\accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=$	$\displaystyle v_{0}\hat{\boldsymbol{\imath}}dt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\int_{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{0}}^{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}(t)}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=$	$\displaystyle\int_{0}^{t}v_{0}\hat{\boldsymbol{\imath}}dt=v_{0}\hat{% \boldsymbol{\imath}}\int_{0}^{t}dt$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{0}$	$\displaystyle=$	$\displaystyle v_{0}\hat{\boldsymbol{\imath}}t$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{0}+v_{0}t\hat{% \boldsymbol{\imath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}(3\,\mathrm{s})$	$\displaystyle=$	$\displaystyle(2\,\mathrm{m}\hat{\boldsymbol{\imath}}+3\,\mathrm{m}\hat{% \boldsymbol{\jmath}})+(5\,\mathrm{m}/\mathrm{s})\cdot(3\,\mathrm{s})\hat{% \boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle 17\,\mathrm{m}\hat{\boldsymbol{\imath}}+3\,\mathrm{m}\hat{% \boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=17\,\mathrm{m}\hat{\boldsymbol{% \imath}}+3\,\mathrm{m}\hat{\boldsymbol{\jmath}}$

Comments: We could solve this problem more compactly by working with scalars or components. It is given that the velocity is constant and is only in the $x$ -direction. Therefore, the $y$ -component of particle position will remain the same, i.e., $r_{y}={r_{0}}_{y}=3\,\mathrm{m}$ , and $r_{x}={r_{0}}_{x}+v_{x}t=2\,\mathrm{m}+(5\,\mathrm{m}/\mathrm{s})\cdot(3\,% \mathrm{s})=17\,\mathrm{m}$ . Thus, ${\accentset{\rightharpoonup}{\boldsymbol{r}}}(3\,\mathrm{s})=r_{x}\hat{% \boldsymbol{\imath}}+r_{y}\hat{\boldsymbol{\jmath}}=17\,\mathrm{m}\hat{% \boldsymbol{\imath}}+3\,\mathrm{m}\hat{\boldsymbol{\jmath}}$ .

SAMPLE 10.5 Constant acceleration: A $0.5\,\mathrm{kg}$ mass starts from rest and attains a speed of $20\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}$ in $4\,\mathrm{s}$ . Assuming that the mass accelerates at a constant rate, find the force acting on the mass.

Solution Here, we are given the initial velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}(0)={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ and the final velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ after $t=4\,\mathrm{s}$ . We have to find the force acting on the mass. The net force on a particle is given by ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ . Thus, we need to find the acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ of the mass to calculate the force acting on it.

Now, the velocity of a particle under constant acceleration is given by

{\accentset{\rightharpoonup}{\boldsymbol{v}}}(t)={\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{0}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}t.

Therefore, we can find the acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{v}}}(t)-{% \accentset{\rightharpoonup}{\boldsymbol{v}}}(0)}{t}$
	$\displaystyle=$	$\displaystyle\frac{20\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}-{% \accentset{\rightharpoonup}{\boldsymbol{0}}}}{4\,\mathrm{s}}$
	$\displaystyle=$	$\displaystyle 5\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{\imath}}.$

The force on the particle is

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}=(0.5\,\mathrm{kg})\cdot(5\,\mathrm{m}/\mathrm{s}^{2}\hat{% \boldsymbol{\imath}})=2.5\,{\mbox{N}}\hat{\boldsymbol{\imath}}.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=2.5\,{\mbox{N}}\hat{\boldsymbol{% \imath}}$

SAMPLE 10.6 Time of travel for a given distance: A ball of mass 200 gm falls freely under gravity from a height of 50 $\,\mathrm{m}$ . Find the time taken to fall through a distance of 30 $\,\mathrm{m}$ , given that the acceleration due to gravity $g=10\,\mathrm{m}/\mathrm{s}^{2}$ .

Solution The entire motion is in one dimension — the vertical direction. We can, therefore, use scalar equations for distance, velocity, and acceleration. Let $y$ denote the distance travelled by the ball. Let us measure $y$ vertically downwards, starting from the height at which the ball starts falling (see fig. 10.9).

Filename:sfig8-2-3b

Figure 10.9:

Under constant acceleration $g$ , we can write the distance travelled as

y(t)=y_{0}+v_{0}t+\frac{1}{2}gt^{2}.

Note that at $t=0$ , $y_{0}=0$ and $v_{0}=0$ . We are given that at some instant $t$ (that we need to find) $y=30\,\mathrm{m}$ . Thus,

	$\displaystyle y$	$\displaystyle=$	$\displaystyle\frac{1}{2}gt^{2}$
	$\displaystyle t$	$\displaystyle=$	$\displaystyle\sqrt{\frac{2y}{g}}=\sqrt{\frac{2\times 30\,\mathrm{m}}{10\,% \mathrm{m}/\mathrm{s}^{2}}}=2.45\,\mathrm{s}.$

Answer: $t=2.45\,\mathrm{s}$

SAMPLE 10.7

Time varying acceleration: A force $F(t)=F_{0}\sin\lambda t$ acts on an initially still cart of mass $m$ in a particular direction. Find the speed and the distance travelled by the cart as functions of time. Plot the acceleration, the speed and the displacement of the cart against time for $0\leq t\leq\pi\,\mathrm{s}$ , assuming $\lambda=1/\,\mathrm{s}$ . What are the speed and the displacement of the cart at $t=\pi\,\mathrm{s}$ if $F_{0}=1\,{\mbox{N}}$ and $m=1\,\mathrm{kg}$ ?

Solution We are given the applied force and the mass of the cart. Therefore, we know the acceleration ( $a=F/m$ ). Thus,

$\displaystyle a\equiv\frac{dv}{dt}$	$\displaystyle=$	$\displaystyle\frac{F_{0}}{m}\sin\lambda t$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}dv$	$\displaystyle=$	$\displaystyle a_{0}\sin\lambda tdt$
where $a_{0}=F_{0}/m$ . Hence,
$\displaystyle\int_{0}^{v(t)}dv$	$\displaystyle=$	$\displaystyle\int_{0}^{t}a_{0}\sin\lambda\tau d\tau$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v(t)$	$\displaystyle=$	$\displaystyle-\frac{a_{0}}{\lambda}(\cos\lambda t-1)$
	$\displaystyle=$	$\displaystyle\frac{a_{0}}{\lambda}(1-\cos\lambda t).$

Since the speed $v=\frac{dx}{dt}$ , we have,

$\displaystyle dx$	$\displaystyle=$	$\displaystyle\frac{a_{0}}{\lambda}(1-\cos\lambda t)dt$
$\displaystyle\int_{0}^{x(t)}dx$	$\displaystyle=$	$\displaystyle\int_{0}^{t}\frac{a_{0}}{\lambda}(1-\cos\lambda\tau)d\tau$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}x(t)$	$\displaystyle=$	$\displaystyle\frac{a_{0}}{\lambda}\left(t-\frac{1}{\lambda}\sin\lambda t\right).$

Answer: $v(t)=\frac{F_{0}/m}{\lambda}(1-\cos\lambda t),\quad x(t)=\frac{F_{0}/m}{% \lambda}\left(t-\frac{1}{\lambda}\sin\lambda t\right)$

Substituting $a_{0}=F_{0}/m=1\,{\mbox{N}}/1\,\mathrm{kg}=1\,\mathrm{m}/\mathrm{s}^{2}$ , $\lambda=1/\,\mathrm{s}$ and $t=\pi\,\mathrm{s}$ in the expressions for $v$ and $x$ above, we find the speed and the displacement (distance travelled by the cart) at $t=\pi$ seconds as follows.

$\displaystyle v(t=\pi\,\mathrm{s})$	$\displaystyle=$	$\displaystyle\frac{1\,\mathrm{m}/\mathrm{s}^{2}}{1/\,\mathrm{s}}(1-\cos\pi)$
	$\displaystyle=$	$\displaystyle 2\,\mathrm{m}/\mathrm{s},$
$\displaystyle x(t=\pi\,\mathrm{s})$	$\displaystyle=$	$\displaystyle 1\,\mathrm{m}/\mathrm{s}\left(\pi\,\mathrm{s}-\frac{1}{1/\,% \mathrm{s}}\sin(\frac{1}{\,\mathrm{s}}\cdot\pi\,\mathrm{s})\right)$
	$\displaystyle=$	$\displaystyle\pi\,\mathrm{m}.$

Answer: $\mbox{At $t=\pi\,\mathrm{s}$},\quad v=2\,\mathrm{m}/\mathrm{s},\quad x=\pi\,% \mathrm{m}$

The graph of $a(t),\,v(t)$ , and $x(t)$ are shown in fig. 10.10 for $0\leq t\leq\pi\,\mathrm{s}$ assuming the given values of $m,\,\lambda$ , and $F_{0}$ . Note the behavior of $v(t)$ and $x(t)$ close to $t=0$ . Since the cart starts from rest, the speed builds up slowly, and the displacement builds up even more slowly because the speed is very low in the beginning.

Filename:sfig8-3-1

Figure 10.10: Plots of acceleration

a(t)

, speed

v(t)

and displacement

x(t)

as functions of time:

a(t)=a_{0}\sin\lambda t

v(t)=\frac{a_{0}}{\lambda}(1-\cos\lambda t)

, and

x(t)=\frac{a_{0}}{\lambda}(t-\frac{1}{\lambda}\sin\lambda t)

where

a_{0}=1\,\mathrm{m}/\mathrm{s}^{2}

and

\lambda=1/\,\mathrm{s}

SAMPLE 10.8 Numerical integration of ODE’s:

1.

Write the second order linear nonhomogeneous differential equation, $\ddot{x}+c\dot{x}+kx=a_{0}\sin\omega t$ , as a set of first order equations that can be used for numerical integration.
2.

Write the second order nonlinear homogeneous differential equation, $\ddot{x}+c\dot{x}^{2}+kx^{3}=0$ , as a set of first order equations that can be used for numerical integration.
3.

Solve the nonlinear equation given in (b) by numerical integration taking $c=0.05,\;k=1,\;x(0)=0$ , and $\dot{x}(0)=0.1$ . Compare this solution with that of the linear equation in (a) by setting $a_{0}=0$ and taking other values to be the same as for (b).

Solution

1.

$\displaystyle\makebox[54.2025pt][l]{If we let}\dot{x}$ $\displaystyle=$ $\displaystyle y,$

$\displaystyle\makebox[54.2025pt][l]{then}\dot{y}$ $\displaystyle=$ $\displaystyle\ddot{x}=-c\dot{x}-kx+a_{0}\sin\omega t$

$\displaystyle=$ $\displaystyle-cy-kx+a_{0}\sin\omega t$

$\hbox to36.135pt{or\hfil}\left\{\begin{array}[]{c}\dot{x}\\ \dot{y}\end{array}\right\}=\left[\begin{array}[]{rr}0&1\\ -k&-c\end{array}\right]\left\{\begin{array}[]{c}{x}\\ {y}\end{array}\right\}+\left\{\begin{array}[]{c}0\\ a_{0}\sin\omega t\end{array}\right\}.$ (10.14)

Equation (10.14) is written in matrix form to show that it is a set of linear first-order ODE’s. In this case linearity means that the dependent variables only appear linearly, not as powers etc.

Filename:sfig8-3-1a

Figure 10.11: Numerical solution of the linear ODE $\ddot{x}+c\dot{x}+kx=0$ with initial conditions $x(0)=0$ and $\dot{x}(0)=0.1$ .

Filename:sfig8-5-wiper

Figure 10.12: Numerical solution of the nonlinear ODE $\ddot{x}+c\dot{x}^{2}+kx^{3}=0$ with initial conditions $x(0)=0$ and $\dot{x}(0)=0.1$ .
2.

$\displaystyle\makebox[36.135pt][l]{If}\dot{x}$ $\displaystyle=$ $\displaystyle y$

$\displaystyle\makebox[36.135pt][l]{then}\dot{y}$ $\displaystyle=$ $\displaystyle\ddot{x}=c\dot{x}^{2}-kx^{3}=-cy^{2}-kx^{3}$

$\displaystyle\hbox to36.135pt{or\hfil}\left\{\begin{array}[]{c}\dot{x}\\ \dot{y}\end{array}\right\}$ $\displaystyle=$ $\displaystyle\left\{\begin{array}[]{c}y\\ -cy^{2}-kx^{3}\end{array}\right\}.$ (10.19)

Equation (10.19) is a set of nonlinear first order ODE’s. It cannot be arranged as Eqn. 10.14 because of the nonlinearity in $x$ and $y$ . It is, however, in an appropriate form for numerical integration.
3.
Now we solve the set of first order equations obtained in (b) using a numerical ODE solver with the following pseudocode.
```
    ODEs = {xdot = y,  ydot = -c y^2 - k x^3}
    IC   = {x(0) = 0, y(0) = 0.1}
    Set  k=1, c=0.05
    Solve ODEs with IC for t=0 to t=200
    Plot x(t) and y(t)
```
The plot obtained from numerical integration using a Runge-Kutta based integrator is shown in fig. 10.12. A similar program used for the equation in (a) with $a_{0}=0$ gives the plot shown in fig. 10.11. The two plots show how a simple nonlinearity changes the response drastically.

Problems for 10.1 Force and motion in 1D

Preparatory Problems

10.1.1 Give three examples of real life objects where you might use the idealization, for dynamic calculations, that the object is a particle in unconstrained 1D motion. Answer: Some possibilities: a) a car with given thrust and drag, b) a person falling vertically during bungy jumping, c) a speaker cone oscillating due to magnetic forces on its coils and resistance from air pressure.

10.1.2 A car is going downhill on a constant slope straight road. For finding out the car’s speed at the end of the road you model it as a particle. For specifying initial velocity, which point on the car would you consider? Answer: All points have equal velocity so all have the same velocity as the center of mass. Any point on the car can be used to measure the car’s position.

10.1.3 The acceleration of a particle is given as a function of time, $a(t)$ . Is this information sufficient to find the speed of the particle at the end of, say, $T$ seconds? Answer: No. You need also to know $v(0)$ . Then $v(T)=v(0)+\int_{0}^{T}a(t)\,dt$ . Knowing $a(t)$ over a given time interval determines the change of $v$ over that interval, but not the value $v(T)$ .

10.1.4 If a particle has constant acceleration, its linear momentum (a) remains constant, (b) changes linearly with time, or (c) changes quadratically with time. Which one is true? Answer: (b) changes linearly in time

10.1.5 In a motorcycle race on a straight track, the speed $v^{*}$ of a motorcyclist at the d=200 $\,\mathrm{m}$ mark is recorded. Given that the rider started from rest, find the initial acceleration of the motorcycle. Assume the acceleration (a) is constant or (b) (challenge) decreases linearly with time to zero at the end. Answer: a) $a={v^{*}}^{2}/(2d)$ ; b) Assume $a=a_{0}-ct$ . If you knew $a_{0}$ and $c$ then you could find $T=a_{0}/c$ , $d=a_{0}^{3}/(3c^{2})$ , and $v^{*}=a_{0}^{2}/(2c)$ . But you are given $v^{*}$ and $d$ and can solve to get $a_{0}=4{v^{*}}^{2}/3d$ .

10.1.6 The force acting on a particle with mass $m$ is a given function of time. If you plot the force vs time and find the area under the graph, from that area can you determine (a) the net displacement of the particle?, (b) the average velocity of the particle?, or (c) the change in linear momentum of the particle? Answer: Only (c), the change in linear momentum. You could find the displacement only if the initial velocity is also given.

10.1.7 If the linear momentum of a body remains constant in time, it must have (pick one): (a) a constant force acting on it, (b) no net force acting on it, or (c) a sinusoidal force acting on it. Answer: (b)

10.1.8 The distance between two points in a bicycle race is 10 km. How many minutes does a bicyclist take to cover this distance if he/she maintains a constant speed of 15 $\,\mathrm{mph}$ . Answer: $t=24.8\,{\rm min}$

10.1.9 A 5 kN constant force acts on a 1 $\,\mathrm{kg}$ object that was initially at rest for 5 seconds and then stops. Find the speed of the object at the end of (a) 5 seconds, and (b) 10 seconds.

10.1.10 Given that $\;\dot{x}=k_{1}+k_{2}t,\ \;k_{1}=1\,{\mathrm{ft/s}},\ \;k_{2}=1\,{\mathrm{ft}}% /{\mathrm{s}}^{2},\;$ and $\;x(0)=1\,\mathrm{ft},\;$ what is the displacement at the end of 10 seconds? Answer: $x(10\,\mathrm{s})=61\,\mathrm{ft}$

10.1.11 Find $x(3\,\mathrm{s})$ given that

\dot{x}=x/(1\,\mathrm{s})\quad\mbox{and}\quad x(0\,\mathrm{s})=1\,\mathrm{m}

or, expressed slightly differently,

\dot{x}=cx\quad\mbox{and}\quad x(0\,\mathrm{s})=x_{0},

where $c=1\,\mathrm{s}^{-1}$ and $x_{0}=1\,\mathrm{m}$ . Make a sketch of $x$ versus $t$ . Answer: $x(3\,\mathrm{s})=20.08\,\mathrm{m}$

10.1.12 A ball of mass $m$ is dropped from rest at a height $h$ above the ground. Find the position and velocity as a function of time (as well as $m$ and $g$ , if needed). Neglect air friction. When does the ball hit the ground? What is the velocity of the ball just before it hits? Answer: $v=-gt$ , $y=h-\frac{1}{2}gt^{2}$ , $y=0\mbox{ when }t=\sqrt{2g/h},v=-\sqrt{2gh}$

10.1.13 The speed of a particle varies sinusoidally as $v=A\sin[ct]$ , where $A=0.5\,\mathrm{m}/\mathrm{s}$ and $c=3\,\mathrm{rad}/\mathrm{s}$ . Let the initial position of the particle be $x(0)=0$ . Find the position of the particle at $t=\pi/2\,\,\mathrm{s}$ . Answer: $x(\pi/2)=0.17\,\mathrm{m}$

10.1.14 The speed of a particle is directly proportional to its position and is given as $\dot{x}=x/\,\mathrm{s}$ . If the initial position, $x(0)=1\,\mathrm{m}$ , how far would the particle be from the origin in 5 seconds? Answer: $x(5\,\mathrm{s})=148.41\,\mathrm{m}$

More-Involved Problems

10.1.15 Consider a force $F(t)$ acting on a cart over a 3 second span. In case (a), the force acts in two impulses of one second duration each as shown in fig. 10.13. In case (b), the force acts continuously for two seconds and then is zero for the last second. Given that the mass of the cart is 10 $\,\mathrm{kg}$ , $v(0)=0$ , and $F_{0}=10\,{\mbox{N}}$ , for each force profile,

(a)

Find the speed of the cart at the end of 3 seconds, and
(b)

Find the distance travelled by the cart in 3 seconds.

Comment on your answers for the two cases. Answer: (a) $v(3\,\mathrm{s})=2\,\mathrm{m}/\mathrm{s}$ in each case. (b) $x(3\,\mathrm{s})=3\,\mathrm{m}$ for case (a), $x(3\,\mathrm{s})=4\,\mathrm{m}$ for case (b). During the one second with no force, in case (b) it has already got up to a higher speed.

Filename:sfig8-5-wiper-a

Figure 10.13:

10.1.16 A car of mass $m$ is accelerated by applying a triangular force profile shown in fig. 10.14(a). Find the speed of the car at $t=T$ seconds. If the same speed is to be achieved at $t=T$ seconds with a sinusoidal force profile, $F(t)=F_{s}\sin\frac{\pi t}{T}$ , find the required force magnitude $F_{s}$ . Is the peak higher or lower? Why? Answer: b) $F_{s}=\frac{\pi}{4}F_{T}$ . This is lower than for (a) because for a given peak force the sinusoidal force is bigger at every instant in time. So, to have the same effect (same impulse) the peak must be lower. Answer: $F_{s}=\frac{\pi}{4}F_{T}$

Filename:sfig8-5-2disks

Figure 10.14:

10.1.17 A particle of mass $m=1\,\mathrm{kg}$ is acted upon by a short duration force given by

F(t)=\begin{cases}F_{0}t/{\rm s}&\text{$0\leq t\leq 1\,\mathrm{s}$}\\ F_{0}(2-t/{\rm s})&\text{$1\,\mathrm{s}<t\leq 2\,\mathrm{s}$}\end{cases}

where $F_{0}=5\,{\mbox{N}}$ . If the particle starts from rest, find the speed of the particle as a function of time. Sketch the given force profile as a function of time and draw the corresponding speed $v(t)$ as a function of time. What is the speed of the particle at $t=2\,\mathrm{s}$ ? Answer: $v(2\,\mathrm{s})=5\,\mathrm{m}/\mathrm{s}$

10.1.18 A ball of mass $m$ is dropped vertically from rest at a height $h$ above the ground. Air resistance causes a drag force on the ball directly proportional to the speed $v$ of the ball, $F_{d}=bv$ . Find the velocity and position of the ball as a function of time. Find the velocity as a function of position. Gravity is non-negligible, of course.

10.1.19 A sinusoidal force acts on a $1\,\mathrm{kg}$ mass as shown in the figure and graph below. The mass is initially still; i. e.,

x(0)=v(0)=0

(a)

What is the velocity of the mass after $2\pi$ seconds?
(b)

What is the position of the mass after $2\pi$ seconds?
(c)

Plot position $x$ versus time $t$ for the motion.

Answer: (a) $v=0$ , (b) $x=10\pi\,\mathrm{m}$

Filename:sfig8-4-1

Figure 10.15:

10.1.20 A motorcycle accelerates from $0\,\mathrm{mph}$ to $60\,\mathrm{mph}$ in 5 seconds. Find the average acceleration in $\,\mathrm{m}/\mathrm{s}^{2}$ . How does this acceleration compare with $g$ , the acceleration of an object falling near the earth’s surface?

10.1.21 A car moves on a straight road with an initial velocity $v_{0}=30\,\mathrm{m}/\mathrm{s}$ . Let its position at $t=0$ be $x=0$ . For the first $5\,\mathrm{s}$ it has no acceleration, and thereafter it brakes with a retarding force that gives it a constant acceleration $a_{x}=-10m/s^{2}$ . Calculate the velocity and the $x$ -coordinate of the car when $t$ = $8\,\mathrm{s}$ and when ${t}$ = $12\,\mathrm{s}$ , and find the distance travelled by the car from start until it comes to a final stop.

10.1.22 A grain of sugar falling through honey has a negative acceleration proportional to the difference between its velocity and its ‘terminal’ velocity, which is a known constant $v_{t}$ . Write this sentence as a differential equation, defining any constants you need. Solve the equation assuming some given initial velocity $v_{0}$ .

10.1.23 The mass-dashpot system shown below is released from rest at $x=0$ . Determine an equation of motion for the particle of mass $m$ that involves only $\dot{x}$ and $x$ (a first-order ordinary differential equation). The dashpot opposes the motion of mass $m$ with a force $F=c\dot{x}$ where $c$ is the damping coefficient of the dashpot. (You can think of the dashpot as a model for linear friction.)

Filename:sfig8-4-1a

Figure 10.16:

10.1.24 Due to gravity, a particle falls in air with a drag force proportional to the speed squared.

1.

Write $\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}$ in terms of variables you clearly define,
2.

find a constant speed motion that satisfies your differential equation,
3.
pick numerical values for your constants and for the initial height. Assume the initial speed is zero
1. (a)
  
  set up the equation for numerical solution,
2. (b)
  
  solve the equation on the computer,
3. (c)
  
  make a plot with your computer solution and show how that plot supports your answer to (2).

10.1.25 In quadratic drag problems, the deceleration is proportional to the square of velocity, i.e., $a=\frac{dv}{dt}=-kv^{2}$ . Assume that a particle with initial velocity $v(0)=v_{0}$ experiences quadratic drag.

(a)

How long does it take for the particle to reduce its speed to half of its initial speed (i.e., find $t$ such that $v(t)=\frac{1}{2}v_{0}$ )?
(b)

Find the position of the particle as a function of velocity. How far does the particle move from its initial position when its velocity drops to half its initial value?

10.1.26 A bullet penetrating flesh slows approximately as it would if penetrating water. The drag on the bullet is about $F_{D}=c\rho_{w}v^{2}A/2$ where $\rho_{w}$ is the density of water, $v$ is the instantaneous speed of the bullet, $A$ is the cross sectional area of the bullet, and $c$ is a drag coefficient which is about $c\approx 1$ . Assume that the bullet has mass $m=\rho_{l}AL$ where $\rho_{l}$ is the density of lead, $A$ is the cross sectional area of the bullet and $L$ is the length of the bullet (approximated as cylindrical). Assume $m=2\textrm{ grams}$ , entering velocity $v_{0}=400\,\mathrm{m}/\mathrm{s}$ , $\rho_{l}/\rho_{w}=11.3$ , and bullet diameter $d=5.7\,{\rm mm}$ .

(a)

Plot the bullet position vs time.
(b)

Assume the bullet has effectively stopped when its speed has dropped to $5\,\mathrm{m}/\mathrm{s}$ , what is its total penetration distance?
(c)

According to the equations implied above, what is the penetration distance in the limit $t\rightarrow\infty$ ?
(d)

How would you change the model to make it more reasonable in its predictions for long time?

10.1.27 A force pulls a particle of mass $m$ towards the origin according to the law (assume same equation works for $x>0,x<0$ )

F=Ax+Bx^{2}+C\dot{x}

Assume $\dot{x}(0)=0$ .
Using numerical solution, find values of $A, B, C, m,$ and $x_{0}$ so that

1.

the mass never crosses the origin,
2.

the mass crosses the origin once,
3.

the mass crosses the origin many times.

[Hint: Vary one parameter at a time and choose a different set of parameter values for each case.]

10.1.28 Do the falling cone experiment: Make two cones, drop them simultaneously and see which falls faster. Explain the result in a way that would convince you if it was written by another student who you did not already trust (and you had not already seen the experiment done).

More details. Copy the drawing below onto two pieces of paper. Make one cone from each page. One has twice the radius as the other. They both have the same conical angle. Which falls faster? Use $F=ma$ to predict the result. Hint 1: the cutout lets you think about how to calculate the area of the paper used to make the cone. Hint 2: The mass of the cone is proportional to its area. The result holds for linear or quadratic drag, or a combination of both. It holds for the steady state or for the transient response.

Answer: The cones fall the same. All of the different forces (gravity, linear drag, quadratic drag) are proportional to either the mass or the area. And, for the cones, mass is proportional to area.

Filename:sfig8-6-2

Figure 10.17:

10.2 Energy methods in 1D

Energy is an important concept in science and engineering. Energy is also a kind of currency in human trade. Energy is also a concept that is somewhat bigger than can be defined inside classical mechanics, when we look at, say, the chemical energy cost of various mechanical tasks.

For a student learning mechanics, energy is first a method, or trick, for solving some simple problems of the type assigned in elementary courses like this one. As problems become more difficult (have more degrees of freedom or include, say, more-than-just-constant friction) energy becomes less useful as a problem solving technique. However, in more advanced mechanics, energy gets a central role again: energy is the central concept in some advanced ways to write equations of motion and for some methods of understanding stability.

Power, work, kinetic energy and potential energy

Before we get to the facts and theorems, we start with some definitions. Here are four words. We will use these definitions, or generalizations of them, throughout dynamics.

Power. The power of a force $F$ is its product with the velocity $v$ of the point on which it is acting,

\mbox{Power}=P=Fv.

This is the 1D version of the more general $P={\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{v}}}$ which we will use once we go on to 2D and 3D dynamics. In full generality, power is a scalar (not a vector). The common units for power are watts ( $1\,\mbox{W}=\,{\mbox{N}}\,\mathrm{m}/\,\mathrm{s}=\,\mbox{J}/\,\mathrm{s}$ ), kilowatts ( $1\mbox{KW}=10^{3}$ W), $\,{\rm lbf}\,\mathrm{ft}/\,\mathrm{s}$ (no special name) and horsepower ( $1\,\mbox{hp}\equiv 550\,{\rm lbf}\,\mathrm{ft}/\,\mathrm{s}\approx 745.7% \approx 746$

^†^†margin: In fourteen hundred and ninety-two,
Columbus sailed the ocean blue,
And that’s twice the watts
in a horsepower too.

\begin{array}[]{rr}746&\\ \underline{\times 2}&\\ 1492&\end{array}

watts).

Example:

A $5\,{\mbox{N}}$ force acting on a particle moving $3\,\mathrm{m}/\mathrm{s}$ has a power of
$P=Fv=(5\,{\mbox{N}})(3\,\mathrm{m}/\mathrm{s})=15\,{\mbox{N}}\,\mathrm{m}/\,% \mathrm{s}=15\,\mbox{W}\approx 0.02\,\mbox{hp}$ .

Work. The work $W$ of a force is most easily defined incrementally ( $\Delta W$ ) for small motions $\Delta x$ of a particle; motions so small that variations in force can be neglected and the force viewed as constant,

\mbox{increment of work}=\Delta W=F\Delta x.

This is a special 1D reduction of the more general $\Delta W={\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot\Delta{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ . Even in 2D and 3D work is a scalar. Often we want to know the work for larger (non-infinitesimal) displacements. We do this by adding up the increments. Using sloppy calculus (implicitly taking the limit of a Riemann sum):

W=\sum\Delta W=\int dW=\int F\,dx,\par\par\mbox{which we can write more % definitely as }

\mbox{Work}=W=\int_{x_{0}}^{x}F(x^{\prime})\,dx^{\prime}.

which is the 1D version of the more general $W=\int{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot d{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ . Common units of work are Joules ( $1\,\mbox{J}=1\,{\mbox{N}}\,\mathrm{m}=1\,\mathrm{kg}\,\mathrm{m}^{2}/\,\mathrm% {s}^{2}$ ), foot-pounds ( $=1\,\mathrm{ft}\,{\rm lbf}$ ) and kilo-watt hours ( $=3.6\cdot 10^{6}\,\mbox{J}$ ).

Example:

The force $F=F_{0}\sin(cx)$ pushes a mass from $x_{0}=0\,\mathrm{m}$ to $x_{1}=\pi\,\mathrm{m}$ where $F_{0}=7\,{\mbox{N}}$ and $c=1/\,\mathrm{m}$ . Then

W=\int_{x_{0}}^{x_{1}}F\,dx=\int_{0}^{\pi\,\mathrm{m}}F_{0}\sin(cx)\,dx=-(F_{0% }/c)\cos(cx)|_{0}^{\pi\,\mathrm{m}}=14\,{\rm N}\mathord{\cdot}{\rm m}

Kinetic energy. The kinetic energy quantifies the motion a little differently than momentum does. In kinetic energy high speed gets extra credit ( $v^{2}$ instead of just $v$ ). Further, for kinetic energy we don’t worry about which way a particle moves. The kinetic energy ${E_{\mathrm{K}}}$ of a particle in 1D is

\mbox{kinetic energy}={E_{\mathrm{K}}}=\frac{1}{2}mv^{2}

In two and three dimensions the formula above applies for one particle (taking $v=|{\accentset{\rightharpoonup}{\boldsymbol{v}}}|$ ). For a collection of particles ${E_{\mathrm{K}}}$ is defined as a sum of ${E_{\mathrm{K}}}$ for each particle separately. In full generality kinetic energy is a scalar. The units of work (force $\times$ distance) and of kinetic energy (mass $\times$ speed²) are the same (mass $\times$ distance ${}^{2}/$ time²) and so are the common measures, namely Joules, foot-pounds and kilo-watt hours.

Example:

A $3\,\mathrm{kg}$ mass moving at a speed of $4\,\mathrm{m}/\mathrm{s}$ has a kinetic energy of
${E_{\mathrm{K}}}=mv^{2}/2=(3\,\mathrm{kg})(4\,\mathrm{m}/\mathrm{s})^{2}/2=24% \,\mathrm{kg}\,\mathrm{m}^{2}/\,\mathrm{s}^{2}=24\,\mbox{J}$ .

Potential energy. This is the most abstract of the definitions. The potential energy ${E_{\mathrm{P}}}$ associated with a force $F$ is defined as that function of $x$ with these properties

\par{E_{\mathrm{P}}}(x)=-\int_{x_{0}}^{x}F(x^{\prime})\,dx^{\prime}\qquad\mbox% {and}\qquad F(x)=-\frac{d}{dx}{E_{\mathrm{P}}}(x)

which people write more indefinitely as ${E_{\mathrm{P}}}=-\int F\,dx$ and $F=-{E_{\mathrm{P}}}^{\prime}$ . In two and three dimensions the concept of potential energy is more subtle still, being defined by a path integral which may or may not be sensible. But it is still a scalar.

Example:

The force $F=c/x^{2}$ is associated with the potential energy
${E_{\mathrm{P}}}=-\int F\,dx=c/x+C_{0}$ .

The datum for potential energy. Potential energy always has an undetermined, and generally irrelevant, integration constant. The integration constant is irrelevant because usually we care about changes in energy. So in the example above we could set $C_{0}=0$ and write ${E_{\mathrm{P}}}=-\int F\,dx=c/x$ . In general we define the datum for potential energy as that position where we set the potential energy to zero.

•

For near-earth gravity the datum is usually set at the height of the ground (so that ${E_{\mathrm{P}}}=mgh$ ), a launch point, or of a conspicuous physical point (say the hinge of a pendulum).
•

For inverse-square gravity the datum is usually set at $r=\infty$ so that formulas are most simple.
•

For springs that datum is usually set at the position where the spring is ‘relaxed’ (un-stretched and at its rest-length), again simplifying the terms in energy equations.

Potential energy is a shortcut for calculating work. From the definition of potential energy we can calculate work of a force in moving a particle from one place to another as:

\mbox{work}=\int_{x_{1}}^{x_{2}}F(x^{\prime})\,dx^{\prime}=-\left({E_{\mathrm{% P}}}_{2}-{E_{\mathrm{P}}}_{1}\right).

Of course you need to know, or find, ${E_{\mathrm{P}}}(x)$ first in order to use this shortcut.

Example:

The work of $F=c/x^{2}$ in moving a mass from $x_{1}$ to $x_{2}$ is

\mbox{work}=\int_{x_{1}}^{x_{2}}F(x^{\prime})\,dx^{\prime}=-\left({E_{\mathrm{% P}}}_{2}-{E_{\mathrm{P}}}_{1}\right)=c/x_{1}-c/x_{2}

Where we used that ${E_{\mathrm{P}}}=c/x$ has the needed property that $\displaystyle F=-\frac{d}{dx}{E_{\mathrm{P}}}$ .

Why all this new language? All of the words above are defined in terms of position, velocity and force. So anything we say about power, work and kinetic and potential energies we could say already using $x$ , $v$ and $F$ . More particularly, we already have two ways of quantifying the motion of a particle, $v$ and $L=mv$ . Why do we need a third, ${E_{\mathrm{K}}}=mv^{2}/2$ ? The answer is this, to simplify the solution of some problems. Various facts and theorems are simpler if commonly appearing groups of terms are given names. And all of the definitions above are common groups. Then, luckily, some of them turn out to be more general than just 1D particle mechanics.

The new vocabulary makes thinking easier. Various so-called ‘one degree of freedom’ problems can be solved by noting that energy is conserved. And features of solutions of more-complex problems can be extracted or checked by making sure that energy balance comes out right.

Power and work

The simplest relation between the quantities we have defined above is that between Power and work:

W=\int F\,dx=\int Fv\frac{dx}{v}=\int Fv\,dt

or more definitely

W=\int_{x_{0}}^{x_{1}}F\,dx=\int_{t_{0}}^{t_{1}}Fv\,dt=\int_{t_{0}}^{t_{1}}P\,dt

Example: Integrate power to get work.

If the power of a force acting on a particle is $P=P_{0}(ct^{2})$ where $P_{0}=10\,\mbox{W}$ and $c=3/\,\mathrm{s}^{2}$ then over 3 seconds the work done by the force is:

	$\displaystyle W$	$\displaystyle=\int_{t_{0}}^{t_{1}}P\,dt=\int_{t_{0}}^{t_{1}}P_{0}(ct^{2})\,dt=% \left.P_{0}ct^{3}/3\right\|_{t_{0}}^{t_{1}}=\left.(10\,\mbox{W})(3/\,\mathrm{s}% ^{2})t^{3}/3\right\|_{0}^{3\,\mathrm{s}}$
		$\displaystyle=270\,\mbox{W}\,\mathrm{s}=270\,\mbox{J}$

Power and rate-of-change of kinetic energy

On the inside cover the third basic law of mechanics is energy balance. Energy balance takes a number of different forms, depending on context.

The power balance equation from the front cover and simplified for a particle is

P=\dot{E_{\mathrm{K}}},

where, recall, $P=Fv$ is the power of the applied force $F$ . The derivation of this result from $F=ma$ for a particle is simple enough, and is good to know. First note the following result from using the chain of differentiation:

\par\frac{d}{dt}\left(v^{2}\right)=2v\frac{dv}{dt}=2v\dot{v}=2va.

When we need to call on this simple kinematics (calculus) result it usually comes to us the other way around. So what you should remember is this formula, one of the basic tricks of the trade:

va=\frac{d}{dt}\left(\frac{v^{2}}{2}\right).

Multiplying both sides by $m$ and substituting in $F=ma$ we get our 1D power balance equation:

Fv=\frac{d}{dt}\left(\frac{mv^{2}}{2}\right).

The power of a given force depends on the speed of the object to which it is applied. When a finite non-zero force is applied to a stationary object the power of the force is zero and so is the rate of change of kinetic energy. If the object accelerates, its speed is increasing, but when the speed is zero, $\dot{E_{\mathrm{K}}}=0$ .

Example:

A constant force $F$ is applied to an initially stationary mass $m$ starting at $t=0$ . Then $v=Ft/m$ , ${E_{\mathrm{K}}}=mv^{2}/2=F^{2}t^{2}/(2m)$ and $P=Fv=F^{2}t/m$ . Note that $\dot{E_{\mathrm{K}}}=P$ and both are zero at $t=0$ .

Work is change of kinetic energy

Integrating the power balance equation in time we get

\par\int P\,dt=\int\dot{E_{\mathrm{K}}}\,dt=\Delta{E_{\mathrm{K}}}\par

(10.20)

More definitely, and also using the work integral, we have that the work of the net force on a particle is the change of its kinetic energy:

\int_{t_{1}}^{t_{2}}\underbrace{Fv}_{P}\,dt=\int_{x_{1}}^{x_{2}}F\,dx={E_{% \mathrm{K}}}_{2}-{E_{\mathrm{K}}}_{1}

Once we remember that

work is change in kinetic energy,

we can use it without deriving it every time from $F=ma$ or from more general energy balance equations.

Example:

A force applied to a particle $m$ varies sinusoidally with position according to $F=F_{0}\cos(cx)$ . At $x=0$ the particle has speed $v=v_{0}$ . Then

	$\displaystyle W=\Delta{E_{\mathrm{K}}}$	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\int_{0}^{x}F(x^{\prime}% )\,dx^{\prime}=\Delta\left(mv^{2}/2\right)$
		$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F_{0}\sin(cx)/c=mv^{2}/2% -mv_{0}^{2}/2$

\mbox{so}\qquad v=\sqrt{v_{0}^{2}+2F_{0}\sin(cx)/(mc)}

The above example illustrates three points you should remember:

•

The work-energy equations always leave the sign of the velocity unknown. You can see this because the derivation involves $v^{2}$ . You can also see it in formulas you get for velocity. They involve a square root, and thus, implicitly a $\pm$ . Whether one, the other or both roots are relevant depends on reasoning that lies outside the energy equation itself.
•

The work-energy equations can generate formulas that, in certain situations, are nonsense: If the initial speed $v_{0}$ is not high enough the particle will not get very far. In particular if $v_{0}^{2}<2F_{0}/(mc)$ the inside of the square root will be negative for some $x$ and the “answer” will be imaginary. These are values of $x$ that the particle will never reach.
•

Here we have apparently solved for something about the motion of a particle. And we have, partially. But to find the $x(t)$ we would have to integrate again. And that next integral is hard. That is, energy balance lets us solve for some aspects of the motion, namely speed vs position, without ever needing to know in detail how position varies with time.

Conservation of energy

Many people leave high-school physics loving conservation of energy. It makes certain special homework problems easy. In the real world the principle is also useful for building intuition, and sometimes also for problem solving. In 1D particle mechanics energy conservation is a theorem

^†^†margin: Be forewarned, generally we cannot think of energy conservation as necessarily applicable nor, if applicable, as derivable from the equations of mechanics.

Recall that if a particle is acted on by a force that varies with position, $F=F(x)$ , then we can define a potential energy ${E_{\mathrm{P}}}=-\int F\,dx$ and that

the work done by the force when the particle moves from $x_{1}$ to $x_{2}$ is

-({E_{\mathrm{P}}}_{2}-{E_{\mathrm{P}}}_{1})=-\Delta{E_{\mathrm{P}}}.

That is, the decrease in ${E_{\mathrm{P}}}$ is the amount of work that the force does. Or, in other words, ${E_{\mathrm{P}}}$ represents a potential to do work. Because work causes an increase in kinetic energy, ${E_{\mathrm{P}}}$ is called the potential energy of the force field.

Now we can compare this result with the work-energy equation 10.20 to find that

-\Delta{E_{\mathrm{P}}}=\Delta{E_{\mathrm{K}}}\quad\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}\quad 0=\Delta\underbrace{({E_{\mathrm{P}}}+{E_{\mathrm{K}}}% )}_{E_{\mathrm{T}}}.

The total energy $E_{\mathrm{T}}$ doesn’t change ( $\Delta E_{\mathrm{T}}=0$ ) and thus is a constant. In other words,

as a particle moves in the presence of a force field with a potential energy, the total energy $E_{\mathrm{T}}={E_{\mathrm{K}}}+{E_{\mathrm{P}}}$ is constant.

This fact goes by the name of conservation of energy.

Filename:sfig8-6-2a

Figure 10.18: Free-body diagram of a falling ball, assuming gravity is the only significant external force acting on the ball.

Example: Falling ball

Consider the ball in the free-body diagram 10.18. If we define gravitational potential energy as minus the work gravity does on a ball while it is lifted from the ground, then

{E_{\mathrm{P}}}=-\int_{0}^{y}(-mg)\,dy^{\prime}=mgy=mgh.

For vertical motion

{E_{\mathrm{K}}}={1\over 2}m\dot{y}^{2}.

So conservation of energy says that in free fall:

\mbox{Constant}={E_{\mathrm{P}}}+{E_{\mathrm{K}}}=mgy+m\dot{y}^{2}/2

which you could also derive directly from $m\ddot{y}=-mg$ .

Using conservation of energy to find equations of motion. On the one hand conservation of energy sometimes gives us a (partial) solution to a mechanics problem. On the other, we can use conservation of energy to find the “equations of motion”. The basic strategy is to take the derivative of the conservation of energy equation.

Example: Falling ball eqns. from energy.

\begin{array}[]{rclcc}E_{\mathrm{T}}=\mbox{constant}\quad\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}\quad 0&=&{d\over dt}E_{\mathrm{T}}\\ &=&{d\over dt}({E_{\mathrm{P}}}+{E_{\mathrm{K}}})\\ &=&{d\over dt}(mgy+m\dot{y}^{2}/2)\\ &=&(mg\dot{y}+m\dot{y}\ddot{y})\\ \quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad m\ddot{y}&=&-mg.\end{array}

We had to assume (and this is just a technical point) that $\dot{y}\neq 0$ in one of the cancellations. We have used energy balance to derive linear-momentum balance.

One can also find equations of motion starting with power balance.

P=\dot{E_{\mathrm{K}}}

as derived here in detail for the case of gravity acting on a particle.

\begin{array}[]{rclp{3 in}}P&=&\frac{d}{dt}\left({E_{\mathrm{K}}}\right)&(% Power balance)\\ {\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot{\accentset{\rightharpoonup}% {\boldsymbol{v}}}&=&\frac{d}{dt}\left({E_{\mathrm{K}}}\right)&(Power of % external force)\\ (-mg\hat{\boldsymbol{\jmath}})\cdot(\dot{y}\hat{\boldsymbol{\jmath}})&=&\frac{% d}{dt}\left[\frac{1}{2}mv^{2}\right]&(expanding terms)\\ -mg\dot{y}&=&\frac{1}{2}m\frac{d}{dt}\left(\dot{y}^{2}\right)&(evaluate dot % product, substitute for $v$)\\[3.0pt] -mg\dot{y}&=&\frac{1}{2}m\left(2\dot{y}\cdot\ddot{y}\right)&(the chain rule)\\ \ddot{y}&=&-g&(cancel terms, switch sides),\end{array}

(10.21)

The potential energy of a spring is $k(\Delta\ell)^{2}/2$ . Besides near-earth gravity, which we already covered ( ${E_{\mathrm{P}}}=mgh$ ), the main elementary use of potential energy is for the stretch of a linear spring.

Integrating $dW=Fdx$ for a linear spring with force on an object $F=-kx$ , where $x$ is the spring stretch, from the rest length, we get

{E_{\mathrm{P}}}=-\int_{0}^{x}F(x^{\prime})\,dx^{\prime}=-\int_{0}^{x}-kx^{% \prime}\,dx^{\prime}={1\over 2}kx^{2}.

(10.22)

In the above example we measured $x$ from the rest position of one end of the spring. But often the natural $x$ coordinate will not be so nicely set up. It is safer to remember the spring’s potential energy in terms of its stretch:

{E_{\mathrm{P}}}=\frac{k\,\Delta\ell^{2}}{2},

(10.23)

where we measure $\Delta\ell=\ell-\ell_{0}$ where $\ell_{0}$ is the spring’s rest length ( $\ell_{0}=$ length when the tension is zero).

Thus for a spring and mass oscillator, the subject of the next section, conservation of energy tells us that $mv^{2}/2+kx^{2}/2=constant$ .

Is energy balance a principle or a calculation trick?

For one dimensional particle motion, momentum balance, power balance, and energy balance can each be derived from either of the others. If we take $F=ma$ as primary, energy calculations are just a convenience of notation or, in the case of the work-energy relation, a useful calculation technique (trick).

Historically, conservation of energy was first noted in particle mechanics problems. But because the position-dependent forces of springs and gravity seemed so fundamental, that they had a description as the derivative of a potential gave the energy relations the smell of something more fundamental. And so it has turned out that energy is an important topic for chemistry, thermodynamics, electrodynamics and sub-atomic physics. It’s not just an analogy, it’s the same energy. Thus energy is the primary currency of exchange between, say, the superficially disparate chemical and mechanical systems.

The exchange of energy between these forms, in the context of particle mechanical models, can give the sense that we are doing the same 1D momentum based mechanics calculations when actually we are using more general energy balance equations, equations that cannot be derived from $F=ma$ .

Terrestrial locomotion: Trains, cars, bicycles and animals

A free-body diagram of an accelerating car, treated as a 1D particle system, is shown in fig. 10.5 on page 10.5. The point represents the car, the force is the propulsion force from the wheel-ground interaction, and for now, we have neglected air friction. Without worrying about details we could say then that the power of the propulsion force is equal to the rate of change of kinetic energy.

Example: Accelerating car

An aggressive $1\,{\rm ton}$ car can accelerate with $a=0.5g$ while going $60\,\mathrm{mph}$ . Neglecting friction and air resistance, the power of the propulsion force is

$\displaystyle P$	$\displaystyle=$	$\displaystyle Fv=mav=(1\,{\rm ton})(0.5g)(60\,\mathrm{mi}/\,{\mathrm{hr}})$
	$\displaystyle=$	$\displaystyle(1\,{\rm ton})(0.5g)(60\,\mathrm{mi}/\,{\mathrm{hr}})\underbrace{% \left(\frac{2000\,{\mathrm{lbm}}}{\,{\rm ton}}\right)}_{1}\underbrace{\left(% \frac{5280\,\mathrm{ft}}{\,{\rm mi}}\right)}_{1}\underbrace{\left(\frac{1\,{% \mathrm{hr}}}{3600\,\mathrm{s}}\right)}_{1}\underbrace{\left(\frac{1\,{\rm lbf% }}{g\cdot\,{\mathrm{lbm}}}\right)}_{1}\underbrace{\left(\frac{1\,\mbox{hp}}{55% 0\,\mathrm{ft}\cdot\,{\rm lbf}/\,\mathrm{s}}\right)}_{1}$
	$\displaystyle=$	$\displaystyle(0.5\cdot 60\cdot 2000\cdot 5280)/(3600\cdot 550)\,\mbox{hp}% \approx 160\,\mbox{hp}\par$

Note the judicious multiple multiplications by 1 so that all units cancel but for horse-power; $\,{\rm ton}$ cancels $\,{\rm ton}$ , $\,{\mathrm{hr}}$ cancels $\,{\mathrm{hr}}$ , $g$ cancels $g$ and so on. The car engine needs to supply this $160\,\mbox{hp}$ plus any internal transmission dissipation. And more still to cover the tire drag and air drag etc.

Such calculations are deceptively simple. Some apparent paradoxes:

•

The propulsive force on the car comes from the interaction of the ground with the car. Are we saying that the (dead-as-a-doormat) ground supplies a power of, say, $160\,\mbox{hp}$ to an accelerating car?
•

The point of application of the force on the car is at the bottom of the tire. That point has no velocity. So the actual power of the ground force on the car (tire) is zero. How is that reconciled with, say, the $160\,\mbox{hp}$ that we get from particle mechanics.

These are legitimate concerns which are discussed further in box 10.2 on page 10.2. The bottom line is that the calculation turns out, perhaps by the demands of dimensional consistency, to be useful and correct.

Drag power. The drag force of air on moving things has an effect on the energy balance. Air drag is important for cars, bicycles and animals that are moving quickly (say, running people). The air drag is proportional to

F_{d}=\frac{1}{2}\rho C_{d}Av^{2}

What are the proportionalities in the drag formula?

•

The cross-sectional area (the area visible from directly in front) $A$ . The bigger the area the more air has to be pushed out of the way. For a car $A\approx 2\,\mathrm{m}^{2}$
•

The density of air $\rho$ . The more mass has to be pushed out of the way, the bigger the force. For rough calculations one can remember that the density of air is about one thousandth that of water $\rho_{air}\approx 1\,\mathrm{kg}/\,\mathrm{m}^{3}$ . But the density varies in human environments from about $1.1\,\mathrm{kg}/\,\mathrm{m}^{3}$ in high-altitude (low pressure), high-temperature (gas expands when hot), humid (water vapor is lighter than air) environments up to about $1.4\,\mathrm{kg}/\,\mathrm{m}^{3}$ in low-lying cold dry places.
•

The relative speed squared $v^{2}$ . The faster you are moving the more air per unit time you must displace, and each bit of air gets displaced with a bigger speed. Typical highway speeds are about $v=30\,\mathrm{m}/\,\mathrm{s}$ ( $\approx 67\,\mathrm{mph}$ ) and a typical human walking speed is about $v=1\,\mathrm{m}/\mathrm{s}$ (about 10% over $\approx 2\,\mathrm{mph}$ ).
•

A shape coefficient, sometimes called a drag coefficient $C$ or $C_{d}$ . Different shapes of the same size, can displace the air more or less as the vehicle passes through. Streamlined shapes have small $C_{d}$
•

One half ( $1/2$ ). Convention has a factor of 1/2. This simplifies the power interpretation.

The drag power is

P=F_{d}v=\frac{1}{2}\rho CAv^{3}

which is a key result: increasing the speed 1% increases the power demand by 3% and doubling the speed multiplies the power demand by a factor of 8. This huge dependence of power on speed motivates smug energy-conservers to drive annoyingly slowly on highways.

Another way of writing the drag equation is

P=C_{d}\times\left(\mbox{ The relative kinetic energy swept by the vehicle per unit time}\right)

How’s that? The volume “swept” by the vehicle per unit time is its area times speed $v A$ . Here area $A$ means the area seen by the wind (the area of a shadow of the object made by a light shining in the wind direction on to a plane whose normal is in the wind direction). The air mass swept per unit time is thus $\rho vA$ . The kinetic energy of the air, measured as moving relative to the vehicle, is $v^{2}/2$ per unit mass. Putting this together we get the swept kinetic energy of the air per unit time is $\rho Av^{3}/2$ .

How big is the drag coefficient $C_{d}$ ? When in doubt take dimensionless constants as 1 and you are usually not too far off. At one extreme is a flat plate whose normal is parallel to the wind direction. Such a plate has $C_{d}\approx 1.25$ . On the other hand, a good airfoil has $C_{d}\approx 0.05$ . People, animals, and bicyclists all have $C_{d}$ close to 1.

Drag on cars. For the worst cars $C_{d}$ is actually almost 1. For typical cars on the street $C_{d}\approx 0.35$ . For the best high-efficiency cars on the market in 2007, $C_{d}\approx 0.25$ . Real marketed cars may one day get drag as low as $C_{d}\approx 0.2$ . And concept cars that are shaped like trout can have drag as low as $C_{d}\approx 0.1$ .

The drag power of a $2\,\mathrm{m}^{2}$ car going $30\,\mathrm{m}/\mathrm{s}$ ( $\approx 67\,\mathrm{mph}=108\,{\mbox{km}}/\,{\mathrm{hr}}$ ) is about

	$\displaystyle P_{drag}$	$\displaystyle=$	$\displaystyle\frac{1}{2}C\rho Av^{3}\approx\frac{1}{2}\cdot 0.35\cdot(1\,% \mathrm{kg}/\,\mathrm{m}^{3})\cdot(2\,\mathrm{m}^{2})\cdot(30\,\mathrm{m}/% \mathrm{s})^{3}$
		$\displaystyle=$	$\displaystyle 9450\,\mathrm{kg}\,\mathrm{m}^{2}/\,\mathrm{s}^{3}=9.45\,\mbox{% KW}\approx 13\,\mbox{hp}$

That is, comparing with the example above, a car that needs $160\,\mbox{hp}$ extra to make a zippy pass needs only $13\,\mbox{hp}$ to move steadily along at a typical highway speed. For the units conversion we used $1\,{\mbox{N}}=1\,\mathrm{kg}\,\mathrm{m}/\,\mathrm{s}^{2}$ , $1\,\mbox{J}=1\,{\mbox{N}}\,\mathrm{m}$ , $1\,\mbox{W}=1\,\mbox{J}/\,\mathrm{s}$ , and $1\,\mbox{KW}\approx 1.34\,\mbox{hp}$ .

Caveat on the drag “law”. While there is some physics in the reasoning behind the drag law $F_{d}=\rho CAv^{2}/2$ the emphasis should be on the word “some”, the whole chaotic nature of turbulent flow is not captured. The quadratic drag law is an empirical fit. For a given shape the $C_{d}$ actually depends on the surface texture. And for a given shape and texture the $C_{d}$ depends on $v$ , the $v^{2}$ doesn’t capture all of the velocity dependence. Nonetheless, the drag law is a reasonable approximation for most engineering purposes where drag is important.

Summary

There are two basic types of energy problems

•

Problems where force or acceleration is given as a function of position ( $a=a(x)$ or $F=F(x)$ ) and energy methods are basically a trick for finding $v(x)$ .
•

Problems where work, energy or power is of interest for its own sake because of, say, interest in engine power, dissipated energy, etc.

Of course the two problem types can also overlap.

Box 10.1 Energetics of locomotion: using particle equations for non-particle systems

On page 10.2 we showed a naive locomotion power example in which we used

P=Fv

where $v$ was the car velocity, $F$ the thrust on the car, and $P$ was ‘the power’ of the locomotion force. We pointed out two issues.

•

How does it make sense for the passive ground, the source of the propulsive force, to supply power?
•

The point of application of the ground force on the car is at the bottom of a wheel, a point that is not moving ( $v=0$ ). So how can $F v$ be other than zero?

The basic issue is that a car is not a particle, it has many moving parts and also some chemistry, so particle equations need to be interpreted with some care.

Particle equations are exact for non-particle systems

The most general form for linear momentum balance, as applied to a complicated system moving and deforming in complicated ways, reduces to equation ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ . That is, so long as we interpret ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ to be the total force on the system, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ to be the acceleration of the center of mass, and $m$ to be the total mass of the system.

The power and energy equations in this chapter have been based on ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ (or their 1D scalar version $F=ma$ ) so apply to any system. But the terms $P$ and ${E_{\mathrm{K}}}$ have meanings that go beyond particle mechanics. So while it is correct that (we derived it from $F=ma$ ),

Fv=\frac{d}{dt}\left(\frac{1}{2}mv^{2}\right)

for non-particle systems it is not correct that $F v$ is the actual power of the force applied nor that $mv^{2}/2$ is the kinetic energy of the system.

To understand the situation depends on understanding multi-body systems where we will see that the power of a force is ${\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot{\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{P}$ where ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ is the velocity of the material point to which the force is applied; and the kinetic energy is larger than $\frac{1}{2}mv^{2}$ because of motion relative to the average motion. Remember to reconsider these issues when you know more.

More general energy balance equations

Without worrying about what we can derive from what, there is no doubt that for any closed system we can write the energy balance equation from the front inside cover of the book, the first law of thermodynamics, as:

\dot{Q}+P=\dot{E_{\mathrm{K}}}+\dot{E_{\mathrm{P}}}+\dot{E}_{\mbox{int}}.

About the ever-shifting sign conventions, here we use $\dot{Q}$ as the heat flow in to the system, $P$ is the power of external forces on the system, $\dot{E_{\mathrm{K}}}$ and $\dot{E_{\mathrm{P}}}$ are the rate of increase of the kinetic and potential energies of the system, and $\dot{E}_{\mbox{int}}$ is the rate of increase of internal energy.

We can consider an accelerating car using this energy equation. For simplicity assume that no external forces do work on the car (the ground certainly does no work, and let’s neglect air friction for now). We can also look at a car on level ground so there are no changes in gravitational potential energy. Finally, even though a car has many moving parts, the bulk of the material goes at the speed of a typical point on the body of the car. Thus the particle formula for kinetic energy is reasonably accurate. Putting this altogether we have

	$\displaystyle{\dot{Q}}+\underbrace{P}_{0}=\frac{d}{dt}\left(\frac{1}{2}mv^{2}% \right)+\underbrace{{E_{\mathrm{P}}}}_{0}+\dot{E}_{\mbox{int}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\frac{d}{dt}\left(\frac{% 1}{2}mv^{2}\right)=-\dot{E}_{\mbox{int}}+\dot{Q}$

The rate of loss of chemical potential energy $-\dot{E}_{\mbox{int}}$ less the heat flow out $-\dot{Q}$ is what we call the power of the engine. Say chemical energy is being lost (used up) at a rate of $-\dot{E}_{\mbox{int}}\dot{=}40\,\mbox{KW}$ . Say the heat flow out the exhaust is $-\dot{Q}=30\,\mbox{KW}$ . Then, with that $40\,\mbox{KW}$ of fuel use and that 25% efficient engine, we would have

\frac{d}{dt}\left(\frac{1}{2}mv^{2}\right)=-\dot{E}_{\mbox{int}}+\dot{Q}=40\,% \mbox{KW}-30\,\mbox{KW}=10\,\mbox{KW}\approx 13\,\mbox{hp}.

But even this is not quite right because it does not take account of the flow of gases in and out of the car. Things can be messy if you look carefully.

What’s the bottom line? In the end, with some sloppiness of thought but not much inaccuracy, we are not far off thinking that the change of kinetic energy of the car has to come from some place. And that place is the work of the engine as supplied by the decrease in chemical potential energy of the fuel. When we write $P=\dot{E_{\mathrm{K}}}$ for a car, the $P$ in that equation is the force applied to the car times the velocity of the car. But that $P$ is not the power supplied by the outside agents on the car (e.g., the passive ground). Rather it is the power of forces inside the car. Never mind that we’re modeling a car as a particle with no internal structure, at least for momentum-balance purposes.

This whole situation can only be properly clarified when we look at the power of internal and external forces in multi-body systems.

SAMPLE 10.9 Which is the best bicycle helmet? Assume a bicyclist moves with speed $25\,\mathrm{mph}$ when her head hits a brick wall. Assume her head is rigid and that it has constant deceleration as it travels through the 2 inches of the bicycle helmet. What is the deceleration? What force is required? (Neglect force from the neck on the head.)

Filename:efig1-2-28

Figure 10.19:

Solution
Solution 1 – Kinematics method 1: We are given the initial speed of $V_{0}$ , a final speed of $0$ , and a constant acceleration $a$ (which is negative) over a given distance of travel $d$ . If we call $t_{c}$ the time when the helmet is fully crushed,

$\displaystyle v(t)$	$\displaystyle=$	$\displaystyle v_{0}+\int_{0}^{t_{c}}a(t^{\prime})dt^{\prime}=v_{0}+at_{c}$
$\displaystyle 0=v(t_{c})$	$\displaystyle=$	$\displaystyle v_{0}+at_{c}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}t_{c}=-v_{0% }/a$	(10.24)
$\displaystyle x(t)$	$\displaystyle=$	$\displaystyle x_{0}+\int_{0}^{t_{c}}v(t^{\prime})dt^{\prime}=0+\int_{0}^{t_{c}% }(v_{0}+at)dt$
$\displaystyle d=x(t_{c})$	$\displaystyle=$	$\displaystyle 0+v_{0}t_{c}+at_{c}^{2}/2$
$\displaystyle d$	$\displaystyle=$	$\displaystyle v_{0}\left(\frac{-v_{0}}{a}\right)+a\left(\frac{v_{0}}{a}\right)% ^{2}/2\hbox to36.135pt{\hfil$\Rightarrow$\hfil}d=\frac{-v_{0}^{2}}{2a}\qquad% \left(\mbox{using (\ref{oneye})}\right)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}a$	$\displaystyle=$	$\displaystyle\frac{-v_{0}^{2}}{2d}$
	$\displaystyle=$	$\displaystyle\frac{-(25\,\mathrm{mph})^{2}}{2\cdot(2\in)}$
	$\displaystyle=$	$\displaystyle\frac{-25^{2}}{4}\cdot\frac{\,\mathrm{mi}^{2}}{\,{\mathrm{hr}}^{2% }\cdot\in}\cdot\underbrace{\left(\frac{5280\,\mathrm{ft}}{\,\mathrm{mi}}\right% )^{2}}_{1}\cdot\underbrace{\left(\frac{1\,{\mathrm{hr}}}{3600\,\mathrm{s}}% \right)^{2}}_{1}\cdot\underbrace{\left(\frac{12\in}{\,\mathrm{ft}}\right)}_{1}% \cdot\underbrace{\left(\frac{1g}{32.2\,{\mathrm{ft}}/{\mathrm{s}}^{2}}\right)}% _{1}$
	$\displaystyle=$	$\displaystyle\frac{-25}{4}\cdot\frac{5280^{2}}{3600^{2}}\cdot 12\cdot\frac{1}{% 32.2}g$
$\displaystyle a$	$\displaystyle=$	$\displaystyle-125g$

To stop from $25\,\mathrm{mph}$ in 2 inches requires an acceleration that is 125 times that of gravity.

Solution 2 – Kinematics method 2:

$\displaystyle\frac{dv}{dt}$	$\displaystyle=$	$\displaystyle a\hbox to36.135pt{\hfil$\Rightarrow$\hfil}dv=adt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}vdv$	$\displaystyle=$	$\displaystyle avdt\hbox to36.135pt{\hfil$\Rightarrow$\hfil}vdv=a\frac{dx}{dt}dt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}vdv$	$\displaystyle=$	$\displaystyle adx$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\int vdv$	$\displaystyle=$	$\displaystyle\int adx$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\Delta\frac{v^{2}}{2}$	$\displaystyle=$	$\displaystyle ax\qquad\qquad\mbox{(since $a=$ constant)}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0-\frac{v_{0}^{2}}{2}$	$\displaystyle=$	$\displaystyle ad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}a=\frac{-v_{0}^{2}}{2% d}\qquad\qquad\mbox{(as before)}$

Solution 3 – Quote formulas:

	$\displaystyle``v$	$\displaystyle=$	$\displaystyle\sqrt{2ad}"$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}a$	$\displaystyle=$	$\displaystyle\frac{v^{2}}{2d}\qquad\mbox{which is right if you know how to % interpret it!}$

Solution 4 – Work-Energy:
Constant acceleration $\Rightarrow$ constant force

Filename:pfigure-blue-118-2

Figure 10.20:

F

is the force of the helmet on the moving head.

Work in	$\displaystyle=$	$\displaystyle\Delta{E_{\mathrm{K}}}$
$\displaystyle-Fd$	$\displaystyle=$	$\displaystyle 0-\frac{mv_{0}^{2}}{2}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F% =\frac{mv_{0}^{2}}{2d}$
$\displaystyle\mbox{But }\ {\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle\Rightarrow$	$\displaystyle-F\hat{\boldsymbol{\imath}}=ma\hat{\boldsymbol{\imath}}% \Rightarrow a=\frac{-F}{m}$
$\displaystyle\mbox{So }\ a$	$\displaystyle=$	$\displaystyle\frac{-v_{0}^{2}}{2d}\qquad\qquad\mbox{(again)}$

Assuming a head mass of $8\,{\mathrm{lbm}}$ , the force on the head during impact is

|F|=\frac{mv_{0}^{2}}{2d}\quad=\quad ma\quad=\quad 8\,{\mathrm{lbm}}\cdot 125g.

Answer: $|F|=1000\,{\rm lbf}$

During a collision in which an $8\,{\mathrm{lbm}}$ head decelerates from $25\,\mathrm{mph}$ to $0$ in 2 inches, the force applied to the head is $1000\,{\rm lbf}$ .

Filename:pfigure-s95f3a

Figure 10.21:

- Note 1:
  
  The way to minimize the peak acceleration when stopping from a given speed over a given distance is to have constant acceleration. The ‘best’ possible helmet, the one we assumed, causes constant deceleration. There is no helmet of any possible material with $2\in$ thickness that could make the deceleration for this collision less than $125g$ or the peak force less than $1000\,{\rm lbf}$ .
- Note 2:
  
  Collisions with head decelerations of $250g$ or greater are often fatal. Even $125g$ usually causes brain injury. So, the best possible helmet does not insure against injury for fast riders hitting solid objects.
- Note 3:
  
  Epidemiological evidence suggests that, on average, chances of serious brain injury are decreased by about a factor of 5 by wearing a helmet.

SAMPLE 10.10

Dissipated energy in viscous drag: A ball of mass $m=1\,\mathrm{kg}$ is dropped from rest from a height $h=100\,\mathrm{m}$ under gravity. The air resistance on the ball is modeled as viscous drag $F_{s}=cv$ where $v$ is the speed of the ball and $c=0.25\,\mathrm{kg}/\,\mathrm{s}$ is the drag coefficient. Find the energy dissipated in overcoming the air resistance during the entire flight of the ball.

Solution

There are various ways in which we could calculate the energy dissipated in viscous drag. The most straightforward way is to compute the work done by the drag force on the body, $\int F_{s}dx$ during the entire flight. This calculation will be very easy if we knew the drag force as a function of position, that is, if we have $F_{s}(x)$ . Unfortunately, we have $F_{s}\equiv F_{s}(v)=cv$ and we do not know $v$ as a function of position. However, we can find the speed $v$ as a function of time by solving the equation of motion $F=ma$ and determine the speed just before the ball hits the ground. Now, we can find the energy of the ball in two positions — just when it starts falling and just before it hits the ground. The difference between the two energies is what is lost or dissipated in overcoming the air resistance.

Let ‘A’ denote position-1 from where the ball is dropped, i.e., $y_{A}=h$ , and ‘B’ denote position-2, a hair above the ground, i.e., $y_{B}=0$ . Taking the ground as the datum for potential energy, we have,

$\displaystyle E_{A}$	$\displaystyle=$	$\displaystyle({E_{\mathrm{K}}}+{E_{\mathrm{P}}})_{A}=\frac{1}{2}m\overbrace{v_% {A}^{2}}^{0}+mg\overbrace{y_{A}}^{h}$
	$\displaystyle=$	$\displaystyle mgh$
$\displaystyle E_{B}$	$\displaystyle=$	$\displaystyle({E_{\mathrm{K}}}+{E_{\mathrm{P}}})_{B}=\frac{1}{2}m\overbrace{v_% {B}^{2}}^{?}+mg\overbrace{y_{B}}^{0}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}mv_{B}^{2}.$

Therefore, the energy dissipated in air resistance is

E_{\rm drag}=\Delta E=E_{A}-E_{B}=mgh-\frac{1}{2}mv_{B}^{2}

(10.25)

Filename:Danef94s1q2

Figure 10.22: Free-body diagram of the falling ball. Note that the drag force

F_{s}=cv

is shown acting downwards. This is because we have assumed

v

to be positive upwards and the drag force always acts in the opposite direction of the velocity.

Now, we just need to find $v_{B}$ . From the free-body diagram shown in fig. 10.22, we have,

$\displaystyle m\overbrace{\ddot{y}}^{\dot{v}}$	$\displaystyle=$	$\displaystyle-mg-cv$
$\displaystyle\hbox to36.135pt{or\hfil}\frac{dv}{dt}$	$\displaystyle=$	$\displaystyle-g-\frac{c}{m}v$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\int_{0}^{v(t)}\frac{dv}% {\tilde{c}v+g}$	$\displaystyle=$	$\displaystyle-\int_{0}^{t}d\tau$
where $\tilde{c}=\frac{c}{m}$ . Thus,
$\displaystyle\left.\frac{1}{\tilde{c}}\ln(\tilde{c}v+g)\right\|_{0}^{v(t)}$	$\displaystyle=$	$\displaystyle-t$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ln\frac{\tilde{c}v(t)+g% }{g}$	$\displaystyle=$	$\displaystyle-\tilde{c}t$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v(t)$	$\displaystyle=$	$\displaystyle\frac{g}{\tilde{c}}(e^{-\tilde{c}t}-1).$	(10.26)

So, we have solved for $v(t)$ . Unfortunately, we cannot find $v_{B}$ from this expression because we do not know what $t$ is when the ball reaches the ground. Thus we need to first find $t_{B}$ and then substitute it in eqn. (10.26) to find $v_{B}$ . From eqn. (10.26), we have,

$\displaystyle\frac{dy}{dt}$	$\displaystyle=$	$\displaystyle\frac{g}{\tilde{c}}(e^{-\tilde{c}t}-1)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\int_{h}^{0}dy$	$\displaystyle=$	$\displaystyle\frac{g}{\tilde{c}}\int_{0}^{t_{B}}(e^{-\tilde{c}t}-1)dt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-h$	$\displaystyle=$	$\displaystyle\frac{g}{\tilde{c}}\left.\left(\frac{e^{-\tilde{c}t}}{-\tilde{c}}% -t\right)\right\|^{t_{f}}_{0}$
	$\displaystyle=$	$\displaystyle\frac{g}{\tilde{c}}\left(-\frac{e^{-\tilde{c}t_{f}}}{\tilde{c}}-t% _{f}+\frac{1}{\tilde{c}}\right)$
$\displaystyle\hbox to36.135pt{or\hfil}\frac{\tilde{c}h}{g}$	$\displaystyle=$	$\displaystyle\frac{1}{\tilde{c}}(e^{-\tilde{c}t_{f}}-1)+t_{f}.$	(10.27)

This turns out to be a transcendental equation^†^†margin: A transcendental equation in $t$ is one where $t$ appears both as an argument of a trigonometric or exponential function and elsewhere. Such equations can almost never be solved by hand in closed form. with no simple solution for $t_{f}$ . We can, however, solve it numerically (either using a computer program, or by trial and error). For the given values of $m$ , $c$ , and $h$ , we solve eqn. (10.27) by trial and error (to locate zero crossing), and find that $t_{f}=5.5495\,\mathrm{s}$ (see fig. 10.23). Substituting $t=t_{f}$ in eqn. (10.26), we get

Filename:pfigure-blue-123-1

Figure 10.23: A graph of

y(t)=h-\frac{mg}{c}\left(\frac{m}{c}e^{-\frac{c}{m}t}+t-\frac{m}{c}\right)

for determining

t_{f}

when

y=0

. This is the same as solving eqn. (10.27) for

t_{f}

$\displaystyle v_{B}$	$\displaystyle=$	$\displaystyle\frac{mg}{c}(e^{-\frac{c}{m}t_{f}}-1)$
	$\displaystyle=$	$\displaystyle\frac{1\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}}{0.25\,% \mathrm{kg}/\,\mathrm{s}}(e^{-\frac{0.25\,\mathrm{kg}/\,\mathrm{s}}{1\,\mathrm% {kg}}\cdot 5.5495\,\mathrm{s}}-1)$
	$\displaystyle=$	$\displaystyle-29.44\,\mathrm{m}/\mathrm{s}.$

Note that $v$ comes out to be negative, which is expected because we assumed $v$ to be positive upwards. The velocity is clearly directed downwards once the ball starts falling. Now, substituting the values of $m$ , $g$ , $h$ , and $v_{B}$ in eqn. (10.25), we get

$\displaystyle E_{\rm drag}$	$\displaystyle=$	$\displaystyle mgh-\frac{1}{2}mv_{B}^{2}$
	$\displaystyle=$	$\displaystyle 1\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot 100\,% \mathrm{m}-\frac{1}{2}\cdot 1\,\mathrm{kg}\cdot(29.44\,\mathrm{m}/\mathrm{s})^% {2}$
	$\displaystyle=$	$\displaystyle 547.64\,{\rm N}\mathord{\cdot}{\rm m}$

Thus more than half of the initial energy is dissipated in air friction. If there were no viscous drag on the ball, its speed just before hitting the ground would be

v_{B}=\sqrt{2gh}=44.29\,\mathrm{m}/\mathrm{s}.

Answer: $E_{\rm drag}=547.64\,{\rm N}\mathord{\cdot}{\rm m}$

SAMPLE 10.11 How much time does it take for a car of mass 800 kg to go from 0 mph to 60 mph, if we assume that the engine delivers a constant power P of 40 horsepower during this period. (1 horsepower = 745.7 W)

Solution

$\displaystyle P$	$\displaystyle=$	$\displaystyle\dot{W}\equiv\frac{dW}{dt}$
$\displaystyle dW$	$\displaystyle=$	$\displaystyle Pdt$
$\displaystyle W_{12}$	$\displaystyle=$	$\displaystyle\int_{t_{0}}^{t_{1}}Pdt=P(t_{1}-t_{0})=P\Delta t$
$\displaystyle\Delta t$	$\displaystyle=$	$\displaystyle\frac{W_{12}}{P}.$

Now, from IIIa in the inside front cover,

$\displaystyle W_{12}$	$\displaystyle=$	$\displaystyle({E_{\mathrm{K}}})_{2}-({E_{\mathrm{K}}})_{1}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})$
	$\displaystyle=$	$\displaystyle\frac{800\,\mathrm{kg}[(60\,\mathrm{mph})^{2}-0]}{2}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}\cdot 800\,\mathrm{kg}\left(60\frac{\,\mathrm{mi}}{\,{% \mathrm{hr}}}\cdot\frac{1.61\times 10^{3}\,\mathrm{m}}{1\,\mathrm{mi}}\cdot% \frac{1\,{\mathrm{hr}}}{3600\,\mathrm{s}}\right)^{2}$
	$\displaystyle=$	$\displaystyle 288.01\times 10^{3}\ \,\mathrm{kg}\cdot\,\mathrm{m}\cdot\,% \mathrm{m}/\mathrm{s}^{2}$
	$\displaystyle=$	$\displaystyle 288\mbox{ KJoule}.$

Therefore,

\Delta t=\frac{288\times 10^{3}\,{\rm J}}{40\times 745.7\mbox{ W}}=9.66\,% \mathrm{s}.

Thus it takes about 10 $\,\mathrm{s}$ to accelerate from a standstill to 60 $\,\mathrm{mph}$ .

Answer: $\Delta t=9.66\,\mathrm{s}$

- Note 1:
  
  This model gives a roughly realistic answer but it is not a realistic model, at least at the start, at time $t_{0}$ . In the model here, the acceleration is infinite at the start (the power jumps from zero to a finite value at the start, when the velocity is zero), something the finite-friction tires would not allow.
- Note 2:
  
  We have been a little sloppy in quoting the energy equation. Since there are no external forces doing work on the car, somewhat more properly we should perhaps have written
  
  $0={\dot{E}_{\mathrm{K}}}+\dot{E}_{\mathrm{int}}+{\dot{E}_{\mathrm{P}}}$
  
  and set $-(\dot{E}_{\mathrm{int}}+{\dot{E}_{\mathrm{P}}})=\mbox{`the engine power' }$ where the engine power is from the decrease in gasoline potential energy $(-{\dot{E}_{\mathrm{P}}}$ is positive $)$ less the increase in ‘heat’ $(\dot{E}_{\mathrm{int}})$ from engine inefficiencies.

SAMPLE 10.12

Filename:pfigure-blue-119-2

Figure 10.24:

Energy of a mass-spring system. A mass $m=2\,\mathrm{kg}$ is attached to a spring with spring constant $k=2\,{\rm kN}/\,\mathrm{m}$ . The relaxed (unstretched) length of the spring is $\ell=40\,\mathrm{cm}$ . The mass is pulled up and released from rest at position A shown in Fig. 10.24. The mass falls by a distance $h=10\,\mathrm{cm}$ before reaching position B, which is the relaxed position of the spring. Find the speed at point B.

Solution The total energy of the mass-spring system at any instant or position consists of the energy stored in the spring and the sum of potential and kinetic energies of the mass. For potential energy of the mass, we need to select a datum where the potential energy is zero. We can select any horizontal plane to be the datum. Let the ground support level of the spring be the datum. Then, at position A,

	Energy in the spring	$\displaystyle=$	$\displaystyle\frac{1}{2}k\mbox{ (stretch)}^{2}=\frac{1}{2}kh^{2}\quad\mbox{(% see eqn.~{}(\ref{eq:springenergy}), page~{}\ref{eq:springenergy})}$
	Energy of the mass	$\displaystyle=$	$\displaystyle{E_{\mathrm{K}}}+{E_{\mathrm{P}}}=\frac{1}{2}m\underbrace{v_{A}^{% 2}}_{0}+mg(\ell+h)=mg(\ell+h).$

Therefore, the total energy at position A

E_{A}=\frac{1}{2}kh^{2}+mg(\ell+h).

Let the speed of the mass at position B be $v_{B}$ . When the mass is at B, the spring is relaxed, i.e., there is no stretch in the spring. Therefore, at position B,

	Energy in the spring	$\displaystyle=$	$\displaystyle\frac{1}{2}k\mbox{ (stretch)}^{2}=0$
	Energy of the mass	$\displaystyle=$	$\displaystyle{E_{\mathrm{K}}}+{E_{\mathrm{P}}}=\frac{1}{2}mv_{B}^{2}+mg\ell,$

and the total energy

E_{B}=\frac{1}{2}mv_{B}^{2}+mg\ell.

Because the net change in the total energy of the system from position A to position B is

$\displaystyle 0$	$\displaystyle=$	$\displaystyle\Delta E$
	$\displaystyle=$	$\displaystyle E_{A}-E_{B}=\frac{1}{2}kh^{2}+mg(\ell+h)-\frac{1}{2}mv_{B}^{2}-mg\ell$
	$\displaystyle=$	$\displaystyle\frac{1}{2}(kh^{2}-mv_{B}^{2})+mgh$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad v_{B}^{2}$	$\displaystyle=$	$\displaystyle kh^{2}/m+2gh$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\|v_{B}\|$	$\displaystyle=$	$\displaystyle\left(kh^{2}/m+2gh\right)^{1/2}$
	$\displaystyle=$	$\displaystyle\left((2000\,{\mbox{N}}/\,\mathrm{m}\cdot(0.1\,\mathrm{m})^{2}/2% \,\mathrm{kg})+2\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot 0.1\,\mathrm{m}% \right)^{1/2}$
	$\displaystyle=$	$\displaystyle 3.46\,\mathrm{m}/\mathrm{s}.$

Answer: $|v_{B}|=3.46\,\mathrm{m}/\mathrm{s}$

Problems for 10.2 Energy methods in 1D

Preparatory Problems

10.2.1 A mass $m$ is at position $x$ moving at velocity $v$ and being acted upon by force $F$ .

For each of the quantities below:

i

give the symbol used for the quantity
ii

describe the quantity in words
iii

give a formula to evaluate the quantity in terms of some or all of $m, x, v$ and $F$ and any other variables you may need.
iv

Give the standard units for the quantity in the SI system.
v

Give the standard units for the quantity in the English system.

(a)

Power
(b)

Kinetic energy
(c)

Work
(d)

Potential energy

10.2.2 Write an equation relating the two words in each of these pairs. If any conditions or descriptions of the situation are needed, give them. If you know more than one equation (or form for a given equation), give all that you know. All should be given in the context of this section: 1D motion.

(a)

work and power
(b)

work and kinetic energy
(c)

power and kinetic energy
(d)

work and potential energy
(e)

potential energy and kinetic energy

10.2.3 A force $F=F_{0}\sin(ct)$ acts on a particle with mass $m=3\,\mathrm{kg}$ which has position $x=3\,\mathrm{m}$ , velocity $v=5\,\mathrm{m}/\mathrm{s}$ at $t=2\,\mathrm{s}$ . $F_{0}=4\,{\mbox{N}}$ and $c=2/\,\mathrm{s}$ . At $t=2\,\mathrm{s}$ evaluate (give numbers and units):

(a)

$a$ ,
(b)

${E_{\mathrm{K}}}$ ,
(c)

$P$ ,
(d)

$\dot{E_{\mathrm{K}}}$ ,
(e)

the rate at which the force is doing work.

Answer: (a) $-1.008\,\mathrm{m}/\mathrm{s}$ , (b) 37.5 Joules, (c), (d) and (e) $-15.12$ Watts

10.2.4 A force only depends on position according to $F=C_{0}+C_{1}x$ where $C_{0}$ and $C_{1}$ are constants. What is the work done by this force when the point to which it is applied moves from $x_{1}$ to $x_{2}$ ? Answer in terms of some or all of $C_{0},C_{1},x_{1}$ and $x_{2}$ .

10.2.5 Find the potential ${E_{\mathrm{P}}}$ associated with each of these force fields.

(a)

$F=0$ .
(b)

$F=F_{0}$ (=constant).
(c)

$F=kx$ .
(d)

$F=A\sin(x/x_{0})$ .
(e)

$F=c/x^{2}$ .

10.2.6 Consider a spring-mass system with $m=2\,\mathrm{kg}$ and $k=5\,{\rm N}/{\rm m}$ . The mass is pulled to the right a distance $x=x_{0}=0.5\,\mathrm{m}$ from the unstretched position and released from rest. No external forces act on the mass.

(a)

What are the initial potential and kinetic energy of the system?
(b)

What is the potential and kinetic energy of the system as the mass passes through the static equilibrium (unstretched spring) position?
(c)

What is the speed of the mass when it passes through the static equilibrium position?

Filename:pfigure-s95q14

Figure 10.25:

10.2.7 A mass $m$ is held in place by a spring whose restoring force is $T(x)=kx$ . Derive the equation of motion of the system (that is, find the acceleration $a$ in terms of $x$ ).

10.2.8 The peak propulsion force on a 4-wheel-drive car is about $\mu mg$ where $\mu\approx 1$ for rubber on road (a bit more for fancy racing tires). Assume a car starts from rest at position zero. Answer the following questions with symbols and with numbers (using $\mu=1$ , $m=1000\,\mathrm{kg}$ , and $g=10\,\mathrm{m}/\mathrm{s}^{2}$ ).

(a)

What is the minimum distance required to reach $v_{1}=60\,\mathrm{mph}$ ?
(b)

What is the extra distance required to get from $v_{1}=60\,\mathrm{mph}$ up to $v_{2}=70\,\mathrm{mph}$ ?
(c)

What is the peak power used by the engine in getting up to $v_{1}=60\,\mathrm{mph}$ (assuming no dissipation and no air friction)?

10.2.9 A car (mass $m=1000\,\mathrm{kg}$ ) traveling at speed $v_{0}=30\,\mathrm{m}/\mathrm{s}$ crashes into a brick wall and comes to a stop as the front end of the car compresses a distance $d=1\,\mathrm{m}$ . Answer with symbols and numbers. Assume constant deceleration during the crash. Neglect the mass of the crushing region of the car.

(a)

What is the total energy dissipated in the crash?
(b)

What is the force of the car on the wall?
(c)

What is the force of the wall on the car?
(d)

What is the deceleration of the car passengers (assuming they are strapped in and move with the bulk of the car). Answer in $g$ ’s?
(e)

Assuming an $m_{p}=50\,\mathrm{kg}$ person, what is the force of the seat belts on the person (answer in body weight).
(f)

If a parent was holding a $15\,\mathrm{kg}$ child on his lap, what force would he need to hold on to the child through the crash (answer in N and in number of child body weights).

More-Involved Problems

10.2.10 A kid ( $m=90\,{\mathrm{lbm}}$ ) stands on a $h=10\,\mathrm{ft}$ wall and jumps down, accelerating with $g=32\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ . Upon hitting the ground with straight legs, she bends them so her body slows to a stop over a distance $d=1\,\mathrm{ft}$ . Neglect the mass of her legs. Assume constant deceleration as she brakes the fall.

(a)

What is the total distance her body falls? Answer: 11 ft
(b)

What is the potential energy lost?
(c)

How much work must be absorbed by her legs?
(d)

What is the force of her legs on her body? Answer in symbols, numbers and numbers of body weight (i.e., find $F/mg$ ).

10.2.11

In traditional archery, when pulling an arrow back the force starts from 0 and increases approximately linearly up to the peak ‘draw force’ $F_{draw}$ . The draw force varies from about $F_{draw}=25\,{\rm lbf}$ for a bow made for a small person to about $F_{draw}=75\,{\rm lbf}$ for a bow made for a big strong person. The distance the arrow is pulled back, the draw length $\ell_{draw}$ , varies from about $\ell_{draw}=2\,\mathrm{ft}$ for a small adult to about $30\,{\mathrm{inch}}$ for a big adult. An arrow has mass of about 300 grain (1 grain $\approx 64.8{\rm milli}\,\hbox{gm}$ , so an arrow has mass of about $19.44\approx 20\,\hbox{gm}\approx 3/4\,\mbox{ounce}$ ). Give all answers in symbols and numbers.

(a)

What is the range of speeds you can expect an arrow to fly?
(b)

What is the range of heights an arrow might go if shot straight up (it’s a bad approximation, but for this problem neglect air friction)?

10.2.12 A big person ( $m=100\,\mathrm{kg}$ ) jumps on a trampoline which we model as a linear spring with stiffness $k$ . You know that the trampoline deflects $d_{0}=20\,\mathrm{cm}$ under the stationary weight $m g$ of the person (use $g=10\,\mathrm{m}/\mathrm{s}^{2}$ ). Assume there is no dissipation and the person is jumping repeatedly a height $h=1\,\mathrm{m}$ above the unloaded surface of the trampoline. Give all answers with symbols and numbers.

(a)

What is the stiffness $k$ of the spring (answer in terms of some or all of $m, g$ and $d_{0}$ ).
(b)

What is the maximum deflection of the trampoline during these jumps?
(c)

What is the peak force of the trampoline on the jumper? (answer in symbols, Newtons, and numbers of body weights).

10.2.13

For the car of problem 10.25 what is the average power required to reach speed $v_{1}$ ? There are two plausible ways to calculate this power:

	$\displaystyle\bar{P}_{1}$	$\displaystyle=$	$\displaystyle\int_{0}^{x}P(x^{\prime})\,dx^{\prime}/x\qquad\mbox{and}\qquad$
	$\displaystyle\bar{P}_{2}$	$\displaystyle=$	$\displaystyle\int_{0}^{t}P(t^{\prime})\,dt^{\prime}/t\ .$

Use both. Do the two methods give the same answer? If so, why, and will the answers be the same for all problems? If not, why not, in what cases will the answers agree, and, when they differ, which one is right?

10.2.14 For problem 10.25 which answers would change, and in which way, if the deceleration was not exactly constant during the crash? That is, for which quantities would the answer be bigger, which smaller, which the same, for which would the answer depend on the nature of the non-constant acceleration?

10.2.15 The earth’s gravitational pull on a mass $m$ is $F=-\frac{mgR^{2}}{r^{2}}$ , where $m g$ is the pull at the surface of the earth and $R$ is the radius of the earth. Assume a ballistic rocket is shot straight up with a launch velocity of $v_{0}$ (measured in a ‘fixed’ not-rotating-with-the-earth frame). Assume the rocket goes in a straight radial line as the earth turns underneath it (relative to the surface of the earth this rocket would be launched somewhat to the West to cancel the earth’s rotation). Assume the period of active thrust is negligibly short (hence the word ballistic: “relating to or characteristic of the motion of objects moving under their own momentum and the force of gravity”).

(a)

Solve for $v$ as a function of $r$ (and some or all of $m, g, R$ and $v_{0}$ ).
(b)

Find the maximum height the rocket reaches.
(c)

Find the ‘escape velocity’ $v_{escape}$ , the minimum launch speed needed for the rocket to never return.
(d)

On one graph plot height ( $r$ or $r-R$ ) vs $t$ for a $v_{0}$ just below $v_{escape}$ and for $v_{0}$ just greater than $v_{escape}$ . If you use numerical methods to make this plot use $g=10\,\mathrm{m}/\mathrm{s}^{2}$ , $R=6400\,{\mbox{km}}$ , and $m=1\,\mathrm{kg}$ . Make sure your axes are such that you can see a clear qualitative difference between the two cases.

10.2.16 The power available to a very strong accelerating cyclist over short periods of time (up to, say, about 1 minute) is about $1$ horsepower. Assume a rider starts from rest and uses this constant power. Assume a mass (bike + rider) of $\,150\,{\mathrm{lbm}}$ , a realistic drag force of $\,.006\,{\rm lbf}/(\,\mathrm{ft}/\,\mathrm{s})^{2}v^{2}$ . Neglect other drag forces.

(a)

What is the peak (steady state) speed of the cyclist?
(b)

Using analytic or numerical methods make an accurate plot of speed vs. time.
(c)

What is the acceleration as $t\rightarrow\infty$ in this solution?
(d)

What is the acceleration as $t\rightarrow 0$ in your solution?
(e)

How would you improve the model to fix the problem with the answer above?

10.3 A mass and spring: the harmonic oscillator

Filename:Danef94s3q2

Figure 10.26: A spring mass system. For simplicity in 1D we only show forces along the motion. Although it would be more complete to show gravity and support forces, it would add clutter.

Filename:bikefork1-ang-accel

Figure 10.27: Oscillations.

A mass and spring moving in one dimension have the following governing differential equation and solution

m\ddot{x}+kx=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}x=C\cos\left(\sqrt{(k/m% )}t-\phi\right).

The rest of this section is a discussion of, and elaboration of, this fact.

Any system with mass and elasticity can oscillate. Because lots of materials are elastic in normal conditions and all real things have mass, most things will oscillate if provoked.

The simplest, and also the most important, example is the spring-mass system of fig. 10.26. When the mass is on the right ( $x>0$ ) it accelerates to the left; when it is on the left ( $x<0$ ) it accelerates to the right. So it goes back and forth. As a general rule,

Oscillations (vibrations) happen when a ‘restoring’ force pulls something back towards a rest position from both sides.

Vibrations occur in the strings of cellos and sitars; the air columns in clarinets, trumpets and organ pipes and the wood blocks in a marimba. A system vibrating like this, whether literally a spring and mass or something more subtle, is called a ‘harmonic oscillator’. More specifically,

a harmonic oscillator has persistent (non-decaying) oscillations which are sinusoidal in time (e.g., $x=\sin t$ ).

With varying degrees of approximation, car suspensions, buildings responding to earthquakes, earthquake faults themselves, quartz timing crystals and vibrating machines of all kinds are also modeled as mass-spring harmonic oscillators. The subject ‘vibration theory’ is based on the harmonic oscillator.

The unforced oscillation of a spring and mass is the basic model for all vibrating systems.

Even structures which you think of as rigid (for example, when doing statics) will vibrate if encouraged to do so by the shaking of an unbalanced motor, the rumbling of a truck, a party upstairs, or the ground motion of an earthquake. And the vibrations of one thing can excite oscillations of another. This mutual excitement of fluids and solids can cause music, as in the vibrations in a clarinet reed (fig. 10.28), or trouble. The Tacoma Narrows bridge infamously collapsed because of the mutual excitement of the air and bridge.

All music, even if electrically powered, is mechanical vibrations (at least of the air in your ear), and so are all annoying sounds. Vibrations are the main function of a vibrating massager, and the main defect of a squeaking hinge. Mechanical vibrations in pendula or quartz crystals are used to measure time, but vibrations can cause a machine to go out of control (e.g., bicycle shimmy), or a bridge to collapse. So the generation of good vibrations and the inhibition of bad vibrations is an important application of dynamics.

The harmonic oscillator

The mother of all vibrating machines is the simple harmonic oscillator of fig. 10.26. The mass slides on a frictionless surface. The spring is relaxed at $x=0$ . The spring is thus stretched from $\ell_{0}$ to $\ell_{0}+\Delta\ell$ , a stretch of $\Delta\ell=x$ .

A free-body diagram of the mass, cut ‘free’ from the spring in its extended state, is shown in the lower part of fig. 10.26.

Linear momentum balance in the $x$ direction (the balance of forces) gives:

	$\displaystyle\sum F_{x}$	$\displaystyle=$	$\displaystyle\dot{L}_{x}$
	$\displaystyle-kx$	$\displaystyle=$	$\displaystyle m\ddot{x}.$

Rearranging, we get one of the most famous and useful differential equations of all time

^†^†margin: Caution: If you make a sign error you might get

\ddot{x}-(k/m)x=0

\ddot{x}=(k/m)x

. That not-so-innocent error changes the equation solutions from oscillations to exponential blow up. See appendix 3.1 on page 3.1.

Refer to caption — Figure 10.28: Blowing grass. If you squeeze a blade of grass between your parallel thumbs and then blow in the slot between the left and right thumb joints, at the grass, you can make a squeal sound that is similar (kind of) to that of a clarinet. This can be done with a strip of paper between the lips too. These are self-excited oscillations. The elastic restoring force is related to the tension in the grass which you can control by straightening and bending your thumbs. With practice you can adjust the pitch. Music, kind of. Photo by Katja Ojala.

\ddot{x}+{k\over m}x=0.

(10.28)

The harmonic oscillator ODE. This ubiquitous is a 2nd order linear constant-coefficient homogeneous ordinary differential equation. Basically you can read this as that it is a ‘simple simple simple simple simple’ differential equation. It’s 2nd order rather than, say 9th order. It’s linear as opposed to non-linear. It’s constant coefficient as opposed to having crazy functions of time where the constant $k/m$ now sits. It’s homogeneous (zero on the right hand side) instead of having crazy functions of time on the right hand side. And it involves ordinary rather than partial derivatives.

\par\frac{\text{Ordinary `$d$', rather than partial $\partial$ derivatives}{d}% ^{2}}{dt^{2}}\par x\text{Linear: $x$ always appears alone, never as, say $x^{2% }$ or $\sin x$}{}+\par\text{Constant coefficient, not, say, $\sin t$ or $t^{2}% $ here}{\lambda^{2}}x=\par 0\text{Homogeneous means zero on the right hand % side}{}.\par

(10.29)

We have used $\lambda^{2}$ instead of $k/m$ because Eqn. (10.29) shows up all over the place, both in and out of dynamics. Thus $x$ doesn’t necessarily represent displacement. In electronics, for example, $x$ might represent a voltage and the term corresponding to $k/m$ might be $1/LC$ , where $C$ is a capacitance and $L$ an inductance. But even in dynamics eqn. (10.29) applies to other things besides a single spring and mass. For example, $x$ might represent rotation of a pendulum.

Why $\lambda^{2}$ instead of just $\lambda$ ?

Two reasons:

1.

It shows that $\lambda^{2}$ is positive,
2.

In the solution we need the square root of this coefficient, so it is convenient to start with $\sqrt{\lambda^{2}}=\lambda$ .

For the spring-block system, $\lambda^{2}$ is $k/m$ and in other problems $\lambda^{2}$ is a combination of other physical quantities.

Why $\lambda$ instead of another letter? In a large book we can’t avoid all notational conflicts, there are more quantities of interest than there are letters. The other common choices, $p$ and $\omega$ , are more problematic, however

^†^†margin: Notation:

p

\omega

\lambda

. Many books use

p^{2}

\omega^{2}

in the place we have put

\lambda^{2}

. But using

\omega

(‘omega’) can lead to confusion because we will later use

\omega

for angular velocity. If one is studying vibrations of a rotating shaft then there would be two very different

\omega

’s in the problem. One, the coefficient of a differential equation and, the other, the angular velocity.
To add to the confusion, simple harmonic oscillations and circular motion have a deep connection, so the coincidence of notation is not accidental. Deep connection or not, the

\omega

in the harmonic oscillator equation is not the same thing as the

\omega

describing angular motion of a physical object. We avoid this confusion by using

\lambda

instead of

\omega

here. (Also note that this

\lambda

is unrelated to the magnitude of the unit vector

\hat{\boldsymbol{\lambda}}

Solution of the harmonic oscillator differential equation

We need to find a solution to the differential equation 10.29. A solution is a function $x(t)$ whose second derivative is the negative of the original function multiplied by the constant $\lambda^{2}$ . Although math classes may spend some time on finding the solution to such equations (see 10.3 on page 10.3, for starters it is ok to remember the solution like this: when the mass is on the left it accelerates right when it’s on the right it accelerates left. So it goes back and forth, so it seems like a sine wave. And it is. Until you know this (which should be soon) you can look it up in appendix 3.1 on page 3.1.

The general solution is

\begin{array}[]{lrclclcc}&x(t)&=&A\ \cos(\lambda\ t)&+&B\ \sin(\lambda\ t),\\ \mbox{or}&x(t)&=&C_{1}\cos(\lambda\ t)&+&C_{2}\sin(\lambda\ t).\\ \end{array}

(10.30)

This sum of two sine waves

^†^†margin: A graph of the cosine function is also a sine wave.

is a solution of differential equation 10.29 for any values of the constants $A$ (or $C_{1}$ ) and $B$ (or $C_{2}$ ).

Checking the solution in detail

What does it mean to say “ $u=C_{1}\sin(\lambda\ t)+\ C_{2}\cos(\lambda\ t)$ satisfies the equation: $\ddot{u}=-\lambda^{2}u$ ?”

You satisfy a differential equation by feeding it a function that fully eliminates it. If you plug a candidate solution into a differential equation and get $0=0$ you have satisfied (solved) the equation.

Whether or not you learn to derive this solution, you should remember it and be able to check it.

To check if a function is a solution, plug it into the differential equation and see if the right side is equal to the left, like this.

Example: Detailed check of solution

$\displaystyle{d^{2}\over dt^{2}}u$	$\displaystyle\text{Is this true for the given $u(t)$?}=$	$\displaystyle-\lambda^{2}u$
$\displaystyle{d^{2}\over dt^{2}}\overbrace{[C_{1}\sin(\lambda\ t)+\ C_{2}\cos(% \lambda\ t)]}^{u(t)}$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}$	$\displaystyle-\lambda^{2}\overbrace{[C_{1}\sin(\lambda\ t)+\ C_{2}\cos(\lambda% \ t)]}^{u(t)}$
$\displaystyle{d\over dt}\left({d\over dt}[C_{1}\sin(\lambda\ t)+\ C_{2}\cos(% \lambda\ t)]\right)$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}$	$\displaystyle-\lambda^{2}[C_{1}\sin(\lambda\ t)+\ C_{2}\cos(\lambda\ t)]$
$\displaystyle{d\over dt}[C_{1}\lambda\cos(\lambda\ t)-\ C_{2}\lambda\sin(% \lambda\ t)]$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}$	$\displaystyle-\lambda^{2}[C_{1}\sin(\lambda\ t)+C_{2}\cos(\lambda\ t)]$
$\displaystyle\overbrace{-C_{1}\lambda^{2}\sin(\lambda t)-C_{2}\lambda^{2}\cos(% \lambda\ t)}^{\ddot{u}}$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}\text{The equation $\ddot{u}=-% \lambda^{2}u$ {\em does} hold with the given $u(t)$. Right and left sides match. This shows at a glance in the next line.}$	$\displaystyle-\lambda^{2}\overbrace{[C_{1}\sin(\lambda\ t)+C_{2}\cos(\lambda\ % t)]}^{u(t)}$
$\displaystyle 0$	$\displaystyle\mathrel{\mathop{=}\limits^{\mbox{\tt:)\vphantom{$\displaystyle% \sum$}}}}$	$\displaystyle 0\qquad\mbox{(Satisfied.)}\par$

Filename:bikefork1-alt

Figure 10.29: Position versus time for an undamped, unforced harmonic oscillator.

x

is the position of the mass,

t

is time.

Whatever the constants $C_{1}$ and $C_{2}$ , the proposed solution eqn. (LABEL:harmoscillate) satisfies the differential equation eqn. (10.29).

Uniqueness. Maybe there are other solutions to this differential equation than eqn. (LABEL:harmoscillate)? There are not, as the mathematics-for-its-own-sake inclined student can learn to prove elsewhere.

^†^†margin: A plausibility argument for uniqueness goes like this. If you release a mass from a given position

x_{0}

at a given speed

v_{0}

it will move in a definite way and no other way. This is a special case of what is called “determinism”. But all solutions have some position and speed at

t=0

and we can find a

C_{1}

and

C_{2}

in eqn. (LABEL:harmoscillate) to match each such. Thus we have found the motion for every possible situation, and there can be no others.

The only possible motion of a spring and mass is a sinusoidal oscillation.

Angular frequency, period, and frequency

There are three common ways to measure the ‘speed’ of oscillation: angular frequency, period, and frequency. The simplest of these is angular frequency $\lambda=\sqrt{(k/m)}$ , sometimes called circular frequency.

The period $T$ is the amount of time that it takes to complete one oscillation. One oscillation of both the sine function and the cosine function occurs when the argument of the function advances by $2\pi$ , that is when

\lambda T=2\pi,\qquad\mbox{so}\qquad T={2\pi\over\lambda}={2\pi\over\sqrt{(k/m% )}},

Some people memorize these formulas in high school.

The frequency $f$ , without the modifiers ‘angular’ or ‘circular’, is usually taken to mean the reciprocal of the period

f={1\over T}={\lambda\over 2\pi}={\sqrt{(k/m)}\over 2\pi}.

Typically, frequency $f$ is measured in cycles per second or Hertz and the angular frequency $\lambda$ in radians per second. A computer or watch quartz timing crystal has mechanical vibrations at a frequency of millions of cycles per second, some molecules about a million times faster than that. On the other extreme, the free vibrations of the whole earth have frequencies of thousandths of a cycle per second (i.e. thousands of seconds per cycle). The slowest vibration mode of the earth has a period of about 54 minutes

^†^†margin: Why does the earth oscillate? First because it can. It has both mass and an ‘elastic restoring force’. The elastic restoring force comes from a combination of two things: 1) the elasticity of rock and 2) the self-gravitation of the earth trying to bundle itself into a ball. What gets the earth started oscillating? Mostly big earthquakes.

An oscillation is called ‘fast’ if its frequency or angular frequency is high and it is called ‘slow’ if the period is long.

Amplitude. The amplitude of the sine wave that results from the addition of the sine function and the cosine function is given by the square root of the sum of the squares of the two amplitudes. That is, the amplitude of the resulting sine wave is $\sqrt{A^{2}+B^{2}}$ . Another way of describing this sum is through the trigonometric identity:

A\cos(\lambda\ t)+B\sin(\lambda\ t)=R\cos(\lambda\ t-\phi),

(10.31)

where $R=\sqrt{A^{2}+B^{2}}$ and $\tan\phi=B/A$ (see box 10.3 on page 10.3).

Note that an oscillating particle could have a high velocity even if the vibrations are ‘slow’, so long as the amplitude is high enough. And it could have a small velocity even if the oscillations are ‘fast’, if the amplitude is low enough. In common usage ‘speed’ of oscillations has to do with frequency, not particle velocity.

Box 10.2 $A\cos(\lambda t)+B\sin(\lambda t)=R\cos(\lambda t-\phi)$

derivation and visualization

Here we show that a cosine function and a sine function add to a new sine wave. By sine wave we mean a function whose shape is the same as the sine function, though it may be displaced along the time axis. For example $\cos t$ and $\cos(t-const)$ are both sine waves.

The trig identity approach. The quickest approach is to start with the function $f(t)=R\cos(\lambda t-\phi)$ and use the trig identity for cosines angle addition:

\cos(\theta-\phi)=\cos\theta\cos\phi+\sin\theta\sin\phi.

Thus:

	$\displaystyle R\cos(\lambda t-\phi)$	$\displaystyle=$	$\displaystyle R\cos\lambda t\cos\phi+R\sin\lambda t\sin\phi$
		$\displaystyle=$	$\displaystyle A\cos\lambda t+B\sin\lambda t.$

We can run the reasoning from right to left and set $A=R\cos\phi$ and $B=R\sin\phi$ and then solve for $R$ and $\phi$ in terms of $A$ and $B$ . Thus demonstrating the title of this box. If you have trouble remembering the trig identity, you can derive it using the picture to the right. Trigonometry is the one subject in which circular reasoning is good.

The geometric approach. Consider the line segment $A$ spinning in circles about the origin at rate $\lambda$ ; that is, the angle the segment makes with the positive $x$ axis is $\lambda t$ . The projection of that segment onto the $x$ axis is $A\cos(\lambda t)$ , a sine wave. Now consider the segment labeled $B$ in the figure, glued at a right angle to $A$ . The length of its projection on the $x$ -axis is $B\sin(\lambda t)$ . So, the sum of these two projections is $A\cos(\lambda t)+B\sin(\lambda t)$ . The two segments $A$ and $B$ make up a right triangle with diagonal $R=\sqrt{A^{2}+B^{2}}$ .

The projection or ‘shadow’ of $R$ on the $x$ axis is the same as the sum of the shadows of $A$ and $B$ . The angle it makes with the $x$ axis is $\lambda t-\phi$ where one can see from the triangle drawn that $\phi=\arctan{(B/A)}$ . So, by adding the shadow lengths, we see

A\cos(\lambda t)+B\sin(\lambda t)=\sqrt{A^{2}+B^{2}}\cos(\lambda t-\phi)\ .

The function $f(t)=R\cos(\lambda t-\phi)$ is a sine wave. In particular it is the cosine function with a maximum at $\lambda t-\phi$ .

Filename:tfigure12-circle

Initial conditions determine the constants $A$ and $B$

The constants $A$ and $B$ in equation LABEL:harmoscillate could have any value. Or, equivalently, the amplitude $R$ and phase $\phi$ in equation 10.31 could be anything. These are determined by the way motion is started, the initial conditions. The following two special initial conditions are worth getting a feel for.

Release from rest.

The simplest motion is release from rest, meaning the initial velocity of the mass is zero. We find the motion from the general solution

x(t)=A\cos(\sqrt{(k/m)}\ t)+B\sin(\sqrt{(k/m)}\ t).

At $t=0$ , this general solution has to agree with the initial condition that $x(0)=x_{0}$ and the initial velocity is $v(0)=v_{0}=0$ . In this case

x(0)=x_{0}\qquad\mbox{and}\qquad v(0)=0\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}A=x_{0}\qquad\mbox{and}\qquad B=0.

Filename:bikefork-alt

Figure 10.30: The position of a mass as a function of time if

k=50\,{\rm N}/{\rm m}

m=0.5\,\mathrm{kg}

x(0)=1\,\mathrm{cm}

and

v(0)=0

Example: Release from rest

The mass in fig. 10.26 is $0.5\,\mathrm{kg}$ , the spring constant is $k=50\,{\rm N}/{\rm m}$ , and the initial displacement is $1\,\mathrm{cm}$ , then

x(0)=A\underbrace{\cos(0)}_{1}+B\underbrace{\sin(0)}_{0}=A\quad\hbox to36.135% pt{\hfil$\Rightarrow$\hfil}\quad A=1\,\mathrm{cm}.

The initial velocity must also match, so

v(t)=\dot{x}(t)=-A\sqrt{(k/m)}\sin(\sqrt{(k/m)}\ t)+B\sqrt{(k/m)}\cos(\sqrt{(k% /m)}\ t).

Evaluating at $t=0$ and matching with the initial condition $v_{0}=0$ ^†^†margin: It is tempting, but wrong, to evaluate $x(t)$ at $t=0$ and then differentiate to get $v(0)$ . Why wrong? Because $x(0)$ is just a number, differentiating it would always give zero, even when the initial velocity is not zero.

v(0)=-A\sqrt{(k/m)}\underbrace{\sin(0)}_{0}+B\sqrt{(k/m)}\underbrace{\cos(0)}_% {1}=B\sqrt{(k/m)}\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad B=0.

Substituting in $k=50\,{\rm N}/{\rm m}$ and $m=0.5\,\mathrm{kg}$ , we get

x(t)=1\cdot\cos\left(\sqrt{\underbrace{\left(\frac{50\,{\rm N}/{\rm m}}{0.5\,% \mathrm{kg}}\right)}_{100\,\mathrm{s}^{-2}}}\ \ t\right)\,\mathrm{cm}=1\cos({1% 0t/\,\mathrm{s}})\,\mathrm{cm}

which is plotted in fig. 10.30.

Filename:pfigure-blue-128-1

Figure 10.31: The position of a mass as a function of time if

k=50\,{\rm N}/{\rm m}

m=0.5\,\mathrm{kg}

x(0)=0

and

v(0)=10\ \,\mathrm{cm}/\,\mathrm{s}

Initially at motion. The case where the initial position is zero but there is some initial velocity is similar. How could this happen? Say we start paying attention just after a still mass has been hit by a hammer.

Example:

Again use $m=0.5\,\mathrm{kg}$ and $k=50\,{\rm N}/{\rm m}$ . But now use $x(0)=0$ and $v(0)=10{\,{\mathrm{cm}}/{\mathrm{s}}}$ . Following the same procedure we get

x(t)=B\sin(\sqrt{(k/m)}\ t)

with $B\sqrt{(k/m)}=10{\,{\mathrm{cm}}/{\mathrm{s}}}\quad\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}\quad B=1\,\mathrm{cm}$ .

The motion, $\displaystyle x(t)=(1\,\mathrm{cm})\cdot\sin\left(\frac{0.1t}{s}\right)$ , is shown in fig. 10.31.

Energy

The harmonic oscillator conserves energy, as one can check with either analytical or numerical solutions. Conversely, we could start with the idea of energy conservation to find the governing differential equations.

Conservation of energy

The harmonic oscillator is friction free. So the total mechanical energy, the sum of the kinetic energy ${E_{\mathrm{K}}}={1\over 2}mv^{2}$ and the potential energy (from eqn. (10.23)) ${E_{\mathrm{P}}}={1\over 2}k(\Delta L)^{2}$ , is constant in time.

E_{\mathrm{T}}={E_{\mathrm{K}}}+{E_{\mathrm{P}}}=\mbox{constant}.

Energy oscillations.

As the mass moves, energy is exchanged back and forth between kinetic and potential energies. At the extremes in the displacement, where the spring is most stretched, the potential energy is at a maximum and the kinetic energy is zero. When the mass passes through the center position the spring is relaxed, the potential energy is at a minimum (zero) and the mass is at its peak speed, and the kinetic energy is at its peak.

Motion details. Let’s assume the block in fig. 10.32 is released from rest at $x=x_{A}>0$ . The mass begins to move to the left and the spring does positive work on the mass since the motion and the force are in the same direction. After the block passes through the rest point $x=O$ , it does work on the spring until it comes to rest at its left extreme. The spring then commences to do work on the block again as the block gains kinetic energy in its rightward motion. The block then passes through the rest position and does work on the spring until its kinetic energy is all used up and it is back in its rest position.

Note that the potential and kinetic energy each have two local maxima and minima for each oscillation of the mass, thus their plots are sine-waves with twice the frequency of the basic oscillation.

Filename:tfigure8-syst-bods

Figure 10.32: Harmonic oscillator. At

t=0

the mass is released from rest at a. The spring is relaxed at

x=0

(points b and d). Some things to note: The acceleration curve is proportional to the negative of the displacement curve; The displacement is at a maximum or minimum when the velocity is zero; The velocity is at a maximum or minimum when the displacement is zero; The kinetic and potential energy fluctuate at twice the frequency as the position; The motion is an ellipse in the cross plot of velocity vs. position.

Example: Check.

Using the special case where the motion starts from rest (i.e., $x(t)=A\cos(\sqrt{k/m}\ t)$ ), we can check that the total energy really is constant.

$\displaystyle E_{\mathrm{T}}$	$\displaystyle=$	$\displaystyle{E_{\mathrm{P}}}+{E_{\mathrm{K}}}$
	$\displaystyle=$	$\displaystyle{1\over 2}kx^{2}+{1\over 2}mv^{2}$
	$\displaystyle=$	$\displaystyle{1\over 2}k(\underbrace{A\cos(\sqrt{k/m}t)}_{x})^{2}+{1\over 2}m(% \underbrace{A\sqrt{k/m}\sin(\sqrt{k/m}t)}_{v})^{2}$
	$\displaystyle=$	$\displaystyle{1\over 2}kA^{2}\underbrace{\{\cos^{2}(\sqrt{k/m}t)+\sin^{2}(% \sqrt{k/m}t)\}}_{1}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}kA^{2}=\quad\mbox{initial energy in spring}$

which does not change with time.

Using energy to derive the oscillator equation

Conversely we can start with energy balance and derive the equations of motion. Starting from $E_{T}=$ constant, we get

$\displaystyle 0$	$\displaystyle=$	$\displaystyle{d\over dt}E_{\mathrm{T}}\ \ =\ {d\over dt}({E_{\mathrm{P}}}+{E_{% \mathrm{K}}})$
	$\displaystyle=$	$\displaystyle{d\over dt}({1\over 2}kx^{2}+{1\over 2}mv^{2})$
	$\displaystyle=$	$\displaystyle kx\underbrace{\dot{x}}_{v}+mv\underbrace{\dot{v}}_{a}$
	$\displaystyle=$	$\displaystyle kx\cancel{v}+m\cancel{v}\underbrace{a}_{\ddot{x}}$
$\displaystyle 0$	$\displaystyle=$	$\displaystyle kx+m\ddot{x}$

which is the harmonic oscillator equation. A technical defect of this derivation is that it does not apply at the instants when $v=0$ (that is, $0\cdot x=0\cdot y$ does not imply that $x=y$ ). Thus, technically, from this derivation we only know the differential equation holds for those times when $v\neq 0$ . Nonetheless, it gives the right equation for all times.

Similarly, power balance also leads to the harmonic oscillator equation. Referring to the FBD in fig. 10.26, the equation of power balance for the block during its motion after release is:

$\displaystyle\underbrace{P}_{\text{\hfil Power in}}$	$\displaystyle=$	$\displaystyle\underbrace{\dot{{E_{\mathrm{K}}}}}_{\text{Rate of change of % kinetic energy}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{spring}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}(\frac{1}{2}mv_{A}^{2})$
$\displaystyle-kx_{A}\hat{\boldsymbol{\imath}}\cdot\dot{x}_{A}\hat{\boldsymbol{% \imath}}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}(\frac{1}{2}m\dot{x}_{A}^{2})$
$\displaystyle-kx_{A}\dot{x}_{A}$	$\displaystyle=$	$\displaystyle m\dot{x}_{A}\ddot{x}_{A}\ .$

Dividing both sides by $\dot{x}_{A}$ (assuming it is not zero), we again get our friend,

-kx_{A}=m\ddot{x}_{A}\qquad\mbox{ or }\qquad m\ddot{x}_{A}+kx_{A}=0.

Filename:sfig8-7-2

Figure 10.33: Spring and mass with gravity. The position can be measured by

x

y

z

Dealing with offsets

Consider a mass hanging from a spring, as in fig. 10.33. If we assume statics, we can find the equilibrium stretch of the spring as $mg/k$ . For dynamics we can keep track of the position of a hanging mass at least 3 ways.

$\displaystyle y$		$\displaystyle=\textrm{distance down from the ceiling},$
$\displaystyle z$	$\displaystyle=y-\ell_{0}$	$\displaystyle=\textrm{distance the spring is stretched, or}$
$\displaystyle x$	$\displaystyle=z-mg/k=y-\ell_{0}-mg/k$	$\displaystyle=\textrm{distance below static equilib}$

At the equilibrium position $y=\ell_{0}+mg/k,z=mg/k$ and $x=0$ .

For $x$ , $y$ and $z$ we can define velocity $v$ and acceleration $a$ as (down is positive)

v=\dot{y}=\dot{z}=\dot{x}\hbox to36.135pt{\hfil and\hfil}a=\ddot{y}=\ddot{z}=% \ddot{x}.

Using the free-body diagram shown we can write $\sum F_{i}=ma$ for each of these cases as

$\displaystyle m\ddot{y}$	$\displaystyle=-k\Delta\ell+mg$	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m\ddot{y}$	$\displaystyle=-k(\underbrace{y-\ell_{0}}_{\Delta\ell})$	$\displaystyle+$	$\displaystyle mg$
$\displaystyle m\ddot{z}$	$\displaystyle=-k\Delta\ell+mg$	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m\ddot{z}$	$\displaystyle=-k\underbrace{z}_{\Delta\ell}$	$\displaystyle+$	$\displaystyle mg$
$\displaystyle m\ddot{y}$	$\displaystyle=-k\Delta\ell+mg$	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m\ddot{x}$	$\displaystyle=-k(\underbrace{x+mg/k}_{\Delta\ell})$	$\displaystyle+$	$\displaystyle mg,$	(10.32)

where in each of the three cases we had to solve for $\Delta\ell$ in terms of $x$ , $y$ and $z$ respectively.

$\displaystyle m\ddot{y}+ky$	$\displaystyle=k\ell_{0}+mg$
$\displaystyle m\ddot{z}+kz$	$\displaystyle=mg$
$\displaystyle m\ddot{x}+kx$	$\displaystyle=0.$	(10.33)

All three equations 10.3 must describe the same motion. But they are three different differential equations with three different solutions. From a course in differential equations you know you can write the solutions as

General solution	$\displaystyle=\textrm{general \text@underline{h}omogeneous solution}+\textrm{% any \text@underline{p}articular solution}$
$\displaystyle y(t)$	$\displaystyle=\underbrace{A\cos\sqrt{(k/m)t}+B\sin\sqrt{(k/m)t}}_{y_{h}}+% \underbrace{\ell_{0}+mg/k}_{y_{p}}$
$\displaystyle z(t)$	$\displaystyle=\underbrace{A\cos\sqrt{(k/m)t}+B\sin\sqrt{(k/m)t}}_{z_{h}}+% \underbrace{mg/k}_{z_{p}}$
$\displaystyle x(t)$	$\displaystyle=\underbrace{A\cos\sqrt{(k/m)t}+B\sin\sqrt{(k/m)t}}_{x_{h}}$	(10.34)

That is, $x$ has simple sinusoidal motions but $y$ and $z$ have sine waves plus a constant. This gives us two basic choices for finding the motion:

1.

Try to simplify the ODE by picking the coordinate that makes the math the easiest (in this case $x$ ), or
2.

Pick any convenient coordinate and use the math of homogeneous and particular solutions to find the motions.

See more discussion of constant forcing in section 11.2 on page 11.2.

Numerical solution

Most numerical differential equation solvers depend on writing the equations in first order form. We do this by defining $v=\dot{x}$ . Thus

m\ddot{x}+kx=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m\dot{v}+kx=0.

Combining the definition of $v$ with the differential equation we get the set of two coupled first order equations

	$\displaystyle\dot{x}$	$\displaystyle=v$
	$\displaystyle\dot{v}$	$\displaystyle=-\frac{k}{m}x$		(10.35)

We can think of this as

\dot{z}=f(z)

where $z$ is the list of two numbers $z(1)=x$ and $z(2)=v$ so

\frac{d}{dt}\left[\begin{array}[]{cc}z_{1}\\ z_{2}\end{array}\right]=\left[\begin{array}[]{cc}z_{2}\\ -\frac{k}{m}z_{1}\end{array}\right].

which is a form easy to use with any numerical ODE solver.

Energy check. Note that there is no mention of energy in setting up the numerical solution. Yet we claim that the solution conserves energy. What does this mean?

After we find the numerical solution we have numerical values for $x$ at a sequence of times and for $v$ at a sequence of times. We can use these lists to calculate

E_{\mathrm{T}}=kx^{2}/2+mv^{2}/2.

at all of these times. The claim is that this list of numbers is a list of the same number, again and again. That is, a property of the solution is that it ‘conserves’ energy, meaning that it finds that the total energy is a constant.

This observation is hugely useful in numerical calculations. Why? Because we know a priori that energy should be conserved. If the numerical solution does not give this constancy of energy it tells us we have made a mistake. How constant? Of course there are numerical errors, so we expect the energy to only be constant to the same accuracy as the numerical solution (typical errors are one part per thousand or less).

If it’s wrong then it’s wrong. Isn’t the energy check merely saying: ‘If the solution gives the wrong answer we know that the solution is wrong.’? Actually, it’s a bit more than that because the idea of energy conservation transcends the details of the actual motion. In this case, for example, we can check energy conservation without ever knowing that the position and velocity oscillate exactly as sine waves. In more complicated problems we might never know formulas for the motions. Yet we might know we have energy conservation. So, even not knowing anything about the solution in detail, we can use energy conservation to check, for example, a numerical solution. If a numerical solution doesn’t conserve energy, when the system conserves energy, then we can’t trust anything about the numerical solution.

Box 10.3 Finding solutions to the harmonic oscillator equation

In the same way that you know, say, that $a^{2}+b^{2}=c^{2}$

for a right triangle, you should know that

\ddot{x}+\lambda^{2}x=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\textrm{a sine% wave}.

Knowing these does not depend on knowing how to derive them. But should you be both insecure and math-inclined here are some derivations of the latter.

Note that the methods below still involve guesses of sorts. That is how it is for differential equations, most of them no-one can solve. And the ones that people can solve they solve by guessing.

Method I: separation of variables

Start with

\ddot{x}+\lambda^{2}x=0

(10.36)

and multiply both sides by $\dot{x}$ (that is, guess that this is a good thing to try) to get

\dot{x}\ddot{x}+\lambda^{2}x\dot{x}=0.

Notice that

\frac{d}{dt}\frac{\dot{x}^{2}}{2}=\dot{x}\ddot{x}\hbox to36.135pt{\hfil and% \hfil}\frac{d}{dt}\frac{x^{2}}{2}=x\dot{x}

\frac{d}{dt}\left\{m\dot{x}^{2}+kx^{2}\right\}=0.

The expression in braces is thus a constant, call it $C^{2}$ to show that it is positive,

\dot{x}^{2}/\lambda^{2}+x^{2}=C^{2}.

We could also have jumped to this result by conservation of energy

	$\displaystyle{E_{\mathrm{K}}}+{E_{\mathrm{P}}}$	$\displaystyle=E_{\mathrm{T}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m\frac{\dot{x}^{2}}{2}+k% \frac{x^{2}}{2}$	$\displaystyle=E_{\mathrm{T}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{x}^{2}/\lambda^{2}+% x^{2}$	$\displaystyle=2E_{\mathrm{T}}/k=C^{2}.$

Now we separate variables to write

dx/\sqrt{C^{2}-x^{2}}=\lambda dt.

Then integrate. How? Either substitute $x=C\sin\theta$ , or guess the solution, or look up the integral on your symbolic calculator, a symbolic math program or the internet. The result is

	$\displaystyle\cos^{-1}(x/C)$	$\displaystyle=$	$\displaystyle\lambda t-c_{2}$
	$\displaystyle\mbox{so}\qquad\qquad\qquad x$	$\displaystyle=$	$\displaystyle C\cos(\lambda t-c_{2})$

where $C$ and $c_{2}$ are arbitrary constants. That’s one form of the general solution of the harmonic oscillator equation. You can plug it back into eqn. (10.29) and see that you get satisfaction, that is $0=0$ , for all values of $C,c_{2}$ and $t$ . The other standard form

x=A\cos\lambda t+B\sin\lambda t

follows from the reasoning in box 10.3 on page 10.3. This is a solution for any value of the constants $A$ and $B$ .

Method II: complex variables

For all linear constant-coefficient homogeneous ordinary differential equations, including eqn. (10.36), a great guess, just one term, is that

x(t)=e^{\alpha t}.

Plugging this guess in and cancelling the common non-zero factor $e^{\alpha t}$ from each term gives

\alpha^{2}+\lambda^{2}=0.

So the guess implies that

\alpha=\pm\sqrt{-\lambda^{2}}.

That’s the square root of a negative number. So

\alpha=\pm\lambda i\qquad\textrm{where}\qquad i=\sqrt{-1}.

Our guess has given us two complex solutions:

x_{1}(t)=e^{\imath\lambda t}\hbox to36.135pt{\hfil and\hfil}x_{2}(t)=e^{-% \imath\lambda t}.

You can check that both of these satisfy eqn. (10.36). We can multiply both solutions by arbitrary constants and add them and check that we still have a solution, namely the two term complex solution

x(t)=Ce^{\imath\lambda t}+De^{-\imath\lambda t}.

We then use the Euler formula (The‘most remarkable … astounding … jewel’ - Richard Feynman) that relates exponentials to sine waves

e^{i\lambda t}=\cos\lambda t+i\sin\lambda t.

We get a solution with 4 terms: Further, $C$ and $D$ can be complex so $C=C_{1}+C_{2}i$ and $D=D_{1}+D_{2}i$ so

	$\displaystyle x(t)=$		$\displaystyle(C_{1}+iC_{2})$	$\displaystyle[\cos(\phantom{-}\lambda t)+i\sin(\phantom{-}\lambda t)]$		(10.37)
		$\displaystyle+$	$\displaystyle(D_{1}+iD_{2})$	$\displaystyle[\cos(-\lambda t)+i\sin(-\lambda t)].$		(10.38)

valid for any values of the constants $C_{1},C_{2},D_{1}$ and $D_{2}$ . Multiplying this out we get, from our simple one-term guess that $x=e^{\alpha t}$ , 8 additive terms. By using the simplifying rules that $\sin-\lambda t=-\sin\lambda t$ and $\cos-\lambda t=\cos\lambda t$ these collapse to four terms. Down here in the simple real world we only care about the real part of this solution (or we simply pick $C_{1}=D_{1}$ and $C_{2}=D_{2}$ to cancel the imaginary terms). We are then left with the two terms

x(t)=(\underbrace{C_{1}+D_{1}}_{A})\cos\lambda t+(\underbrace{-C_{2}-D_{2}}_{B% })\sin\lambda t.

which, after defining the new constants $A=C_{1}+D_{1}$ and $B=-C_{2}-D_{2}$ , is our general (real) solution to the harmonic oscillator equation. Using box 10.3 on page 10.3 we can write this as one sine wave,

x(t)=C\cos(\lambda t-{\phi}).

We are stuck with $\lambda$ but $C$ and $\phi=c_{2}$ are arbitrary constants. One simple guess has given us $1\rightarrow 2\rightarrow 4\rightarrow 8\rightarrow 4\rightarrow 2\rightarrow 1$ solution.

SAMPLE 10.13

Filename:sfig8-7-2a

Figure 10.34:

A block of mass $m=20\,\mathrm{kg}$ is attached to two identical springs each with spring constant $k=1\,$ kN/m. The block slides on a horizontal surface without any friction.

1.

Find the equation of motion of the block.
2.

What is the oscillation frequency of the block?
3.

How much time does the block take to go back and forth 10 times?

Solution

1.
Filename:sfig8-7-2again

Figure 10.35:

The free-body diagram of the block is shown in Figure 10.35. The linear momentum balance, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ , for the block gives

$-2kx\hat{\boldsymbol{\imath}}+(N-mg)\hat{\boldsymbol{\jmath}}=m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}$

Dotting both sides with $\hat{\boldsymbol{\imath}}$ we have,

$\displaystyle-2kx=ma_{x}$ $\displaystyle=$ $\displaystyle m\ddot{x}$ (10.39)

$\displaystyle\mbox{or}\quad m\ddot{x}+2kx$ $\displaystyle=$ $\displaystyle 0$ (10.40)

$\displaystyle\mbox{or}\quad\ddot{x}+\frac{2k}{m}x$ $\displaystyle=$ $\displaystyle 0.$ (10.41)

Answer: $\ddot{x}+\frac{2k}{m}x=0$
2.

Comparing Eqn. (10.41) with the standard harmonic oscillator equation, $\ddot{x}+\lambda^{2}x=0$ , where $\lambda$ is the oscillation frequency, we get

$\displaystyle\lambda^{2}$ $\displaystyle=$ $\displaystyle\frac{2k}{m}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\lambda$ $\displaystyle=$ $\displaystyle\sqrt{\frac{2k}{m}}$

$\displaystyle=$ $\displaystyle\sqrt{\frac{2\mathord{\cdot}(1\,\mbox{kN/m})}{20\,\mathrm{kg}}}$

$\displaystyle=$ $\displaystyle 10\,\mathrm{rad}/\mathrm{s}.$

Answer: $\lambda=10\,\mathrm{rad}/\mathrm{s}$
3.

Time period of oscillation $T=\frac{2\pi}{\lambda}=\frac{2\pi}{10\,\mathrm{rad}/\mathrm{s}}=\frac{\pi}{5}% \,\mathrm{s}$ . Since the time period represents the time the mass takes to go back and forth just once, the time it takes to go back and forth 10 times (i.e., to complete 10 cycles of motion) is

$t=10T=10\mathord{\cdot}\frac{\pi}{5}\,\mathrm{s}=2\pi\,\mathrm{s}.$

Answer: $t=2\pi\,\mathrm{s}$

SAMPLE 10.14

Filename:sfig8-7-2disks

Figure 10.36:

Simple harmonic motion of a buoy. A cylinder of cross sectional area $A$ and mass $M$ is in static equilibrium inside a fluid of specific weight $\gamma$ when $L_{o}$ length of the cylinder is submerged in the fluid. From this position, the cylinder is pushed down vertically by a small amount $x$ and let go. Assume that the only forces acting on the cylinder are gravity and the buoyant force and assume that the buoy’s motion is purely vertical. Derive the equation of motion of the cylinder using Linear Momentum Balance. What is the period of oscillation of the cylinder?

Solution The free-body diagram of the cylinder is shown in Fig. 10.37 where $F_{B}$ represents the buoyant force (see the hydrostatics chapter starting on 9). Before the cylinder is pushed down by $x$ , the linear momentum balance of the cylinder gives

F_{B}-Mg=M\underbrace{a}_{0}\;=0\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}% \quad F_{B}=Mg

Now $F_{B}$ = (volume of the displaced fluid) $\cdot$ (its specific weight) $=AL_{o}\gamma$ . Thus,

AL_{o}\gamma=Mg.

(10.42)

Filename:sfig8-4-4

Figure 10.37:

Now, when the cylinder is pushed down by an amount $x$ ,

F^{\prime}_{B}=\mbox{ new buoyant force }=(L_{o}+x)A\gamma.

Therefore, from LMB we get

$\displaystyle F^{\prime}_{B}-Mg$	$\displaystyle=$	$\displaystyle-M\ddot{x}$
$\displaystyle\hbox to36.135pt{or\hfil}(L_{o}+x)A\gamma-Mg$	$\displaystyle=$	$\displaystyle-M\ddot{x}$
$\displaystyle\hbox to36.135pt{or\hfil}M\ddot{x}+A\gamma x$	$\displaystyle=$	$\displaystyle\overbrace{-AL_{o}\gamma+Mg}^{=0\mbox{ from (\ref{eq16.1}).}}$
$\displaystyle\hbox to36.135pt{or\hfil}M\ddot{x}+A\gamma x$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{or\hfil}\ddot{x}+\frac{A\gamma}{M}x$	$\displaystyle=$	$\displaystyle 0.$

Answer: $\ddot{x}+\frac{A\gamma}{M}x=0$

Comparing this equation with the standard simple harmonic equation (e.g., eqn.(g), in the box on ODE’s on page 3.1).

\mbox{The circular frequency }\lambda=\sqrt{\frac{A\gamma}{M}},

\hbox to57.81621pt{Therefore,\hfil}\mbox{the period of oscillation }T=\frac{2% \pi}{\lambda}=2\pi\sqrt{\frac{M}{A\gamma}}

Answer: $T=2\pi\sqrt{\frac{M}{A\gamma}}$

Comments: This calculation inaccurately uses fluid statics to calculate the dynamics of a buoy; the pressure used in this calculation assumes fluid statics when actually the fluid is moving. One common partial correction is to use ‘added mass’ to account for fluid that moves more-or-less with the cylinder. The added mass is usually something like one-half the mass of the displaced fluid, that is one half the mass of the bouy. Another missing effect is the fluid damping. This would be added as a drag force proportional to the velocity or the velocity squared.

SAMPLE 10.15

Filename:sfig8-4-4a

Figure 10.38:

A spring-mass system executes simple harmonic motion: $x(t)=A\cos(\lambda t-\phi)$ . The system starts with initial conditions $x(0)=25\,{\rm mm}$ and $\dot{x}(0)=160\,{\rm mm}/\,\mathrm{s}$ and oscillates at the rate of 2 cycles/sec.

1.

Find the time period of oscillation and the oscillation frequency $\lambda$ .
2.

Find the amplitude of oscillation $A$ and the phase angle $\phi$ .
3.

Find the displacement, velocity, and acceleration of the mass at $t=1.5\,\mathrm{s}$ .
4.

Find the maximum speed and acceleration of the system.
5.

Draw an accurate plot of displacement $v s .$ time of the system and label all relevant quantities. What does $\phi$ signify in this plot?

Solution

1.

We are given $f=2\;$ Hz. Therefore, the time period of oscillation is

$T=\frac{1}{f}=\frac{1}{2\mbox{ Hz}}=0.5\,\mathrm{s},$

and the oscillation frequency $\lambda=2\pi f=4\pi\,\mathrm{rad}/\mathrm{s}.$

Answer: $T=0.5\,\mathrm{s},\quad\lambda=4\pi\,\mathrm{rad}/\mathrm{s}.$
2.

The displacement $x(t)$ of the mass is given by

$x(t)=A\cos(\lambda t-\phi).$

Therefore the velocity (actually the speed) is

$\dot{x}(t)=-A\lambda\sin(\lambda t-\phi)$

At $t=0$ , we have

$\displaystyle x(0)$ $\displaystyle=$ $\displaystyle A\cos(-\phi)=A\cos\phi$ (10.43)

$\displaystyle\dot{x}(0)$ $\displaystyle=$ $\displaystyle-A\lambda\sin(-\phi)=A\lambda\sin\phi$ (10.44)

By squaring Eqn (10.43) and adding it to the square of [Eqn (10.44) divided by $\lambda$ ], we get

$\displaystyle A^{2}\cos^{2}\phi+\frac{A^{2}\lambda^{2}\sin^{2}\phi}{\lambda^{2}}$ $\displaystyle=$ $\displaystyle A^{2}=x^{2}(0)+\frac{\dot{x}^{2}(0)}{\lambda^{2}}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad A$ $\displaystyle=$ $\displaystyle\sqrt{(25\,{\rm mm})^{2}+\frac{(160\,{\rm mm}/\,\mathrm{s})^{2}}{% (4\pi\,\mathrm{rad}/\mathrm{s})^{2}}}$

$\displaystyle=$ $\displaystyle 28.06\,{\rm mm}.$

Substituting the value of $A$ in Eqn (10.43), we get

$\displaystyle\phi$ $\displaystyle=$ $\displaystyle\cos^{-1}\frac{x(0)}{A}$

$\displaystyle=$ $\displaystyle\cos^{-1}\frac{25\,{\rm mm}}{28.06\,{\rm mm}}$

$\displaystyle=$ $\displaystyle 0.471\,{\rm rad}\approx 27^{{}^{\circ}}.$

Answer: $A=28.06\,{\rm mm}.\quad\phi=0.471\,{\rm rad}.$

The displacement, velocity, and acceleration of the mass at any time $t$ can now be calculated as follows

$\displaystyle x(t)$	$\displaystyle=$	$\displaystyle A\cos(\lambda t-\phi)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}x(1.5\,\mathrm{s})$	$\displaystyle=$	$\displaystyle 28.06\,{\rm mm}\mathord{\cdot}\cos(6\pi-0.471)$
	$\displaystyle=$	$\displaystyle 25\,{\rm mm}.$
$\displaystyle\dot{x}(t)$	$\displaystyle=$	$\displaystyle-A\lambda\sin(\lambda t-\phi)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{x}(1.5\,\mathrm{s})$	$\displaystyle=$	$\displaystyle 28.06\,{\rm mm}\mathord{\cdot}(4\pi\,\mathrm{rad}/\mathrm{s})% \mathord{\cdot}\sin(6\pi-0.471)$
	$\displaystyle=$	$\displaystyle 160\,{\rm mm}/\,\mathrm{s}.$
$\displaystyle\ddot{x}(t)$	$\displaystyle=$	$\displaystyle-A\lambda^{2}\cos(\lambda t-\phi)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{x}(1.5\,\mathrm{s})$	$\displaystyle=$	$\displaystyle 28.06\,{\rm mm}\mathord{\cdot}(4\pi\,\mathrm{rad}/\mathrm{s})^{2% }\mathord{\cdot}\cos(6\pi-0.471)$
	$\displaystyle=$	$\displaystyle-3.95\times 10^{3}\,{\rm mm}/\,\mathrm{s}^{2}$
	$\displaystyle=$	$\displaystyle-3.95\,\mathrm{m}/\,\mathrm{s}^{2}.$

^†^†margin: We can find the displacement and velocity at

t=1.5\,\mathrm{s}

without any differentiation. Note that the system completes 2 cycles in 1 second, implying that it will complete 3 cycles in 1.5 seconds. Therefore, at

t=1.5\,\mathrm{s}

, it has the same displacement and velocity as it had at

t=0\,\mathrm{s}

Answer: $x(1.5\,\mathrm{s})=25\,{\rm mm}.\quad\dot{x}(1.5\,\mathrm{s})=160\,{\rm mm}/\,% \mathrm{s}.\quad\ddot{x}(1.5\,\mathrm{s})=-3.93\,\mathrm{m}/\,\mathrm{s}^{2}.$

4.

Maximum speed:

$\displaystyle|\dot{x}_{\rm{max}}|$ $\displaystyle=$ $\displaystyle A\lambda=(28.06\,{\rm mm})\mathord{\cdot}(4\pi\,\mathrm{rad}/% \mathrm{s})=0.35\,\mathrm{m}/\mathrm{s}.$

Maximum acceleration:

$\displaystyle|\ddot{x}_{\rm{max}}|$ $\displaystyle=$ $\displaystyle A\lambda^{2}=(28.06\,{\rm mm})\mathord{\cdot}(4\pi\,\mathrm{rad}% /\mathrm{s})^{2}=4.43\,\mathrm{m}/\mathrm{s}^{2}.$

Answer: $|\dot{x}_{\rm{max}}|=0.35\,\mathrm{m}/\mathrm{s},\quad|\ddot{x}_{\rm{max}}|=4.% 43\,\mathrm{m}/\mathrm{s}^{2}.$
5.

The plot of $x(t)$ versus $t$ is shown in Fig. 10.39. The phase angle $\phi$ represents the shift in $\cos(\lambda t)$ to the right by an amount $\frac{\phi}{\lambda}$ .

Filename:sfig8-4-4b

Figure 10.39:

Problems for 10.3 Simple Harmonic Oscillator

Preparatory Problems

10.3.1 The basic model.

(a)

Draw a spring ( $k$ ) mass ( $m$ ) system in a configuration where the spring is stretched.
(b)

On the drawing indicate the variable $x$ .
(c)

Draw a free-body diagram of the mass.
(d)

Write the equation of linear momentum balance for the mass.
(e)

Rearrange the momentum balance equation to get the harmonic oscillator equation in standard form.
(f)

Write the general solution to the harmonic oscillator equation in two different ways (one as a sum of a sine and cosine function and one as a phase shifted sine or cosine function).
(g)

What is the natural frequency of this system?
(h)

What is the period?
(i)

What is the frequency (or circular frequency)?
(j)

Find the solution for the special case that the mass is released from rest at $x(0)=x_{0}$ .
- •
  
  give the analytic expression.
- •
  
  plot the position vs time for at least one whole cycle of motion.
- •
  
  with the same time scale, plot velocity vs time (what is the peak velocity).
- •
  
  with the same time scale, plot both the potential and kinetic energies vs time.
(k)

Find the solution for the special case that the mass is launched at $v_{0}$ from the rest position (just the analytic form, no need to repeat all the parts just above).

10.3.2 Does the function $x=C_{1}e^{\lambda t}+C_{2}e^{-\lambda t}$ satisfy the harmonic oscillator equation $\ddot{x}+\lambda^{2}x=0$ for any, possibly special, values of $C_{1}$ and $C_{2}$ ? Show that it does or does not.

10.3.3 Given that $\ddot{x}=-cx$ , with $c=1/\,\mathrm{s}^{2}$ , $x(0)=1\,\mathrm{m}$ , and $\dot{x}(0)=0$ find:

(a)

$x(\pi\,\mathrm{s})=$ ?
(b)

$\dot{x}(\pi\,\mathrm{s})=$ ?

10.3.4 Given that $\ddot{x}+\lambda^{2}x=C_{0}$ , $x(0)=x_{0}$ , and $\dot{x}(0)=0$ , find the value of $x$ at $t=\pi/\lambda\ \,\mathrm{s}$ .

More-Involved Problems

10.3.5 A spring and mass system is shown in the figure.

(a)

First, as a review, let $k_{1}$ , $k_{2}$ , and $k_{3}$ equal zero and $k_{4}$ be non-zero. What is the natural frequency of this system?
(b)

Now, let all the springs have non-zero stiffness. What is the stiffness of a single spring equivalent to the combination of $k_{1},k_{2},k_{3},k_{4}$ ? What is the frequency of oscillation of mass $M$ ?

Filename:sfig8-4-4c

Figure 10.40:

10.3.6 Mass $m$ hangs from a spring with constant $k$ and which has the length $l_{0}$ when it is relaxed (i.e., when no mass is attached). It only moves vertically.

(a)

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

Write the equation of linear momentum balance. Answer: $mg-k(x-\ell_{0})=m\ddot{x}$
(c)

Reduce this equation to a standard differential equation in $x$ , the position $x$ of the mass. Answer: $\ddot{x}+\frac{k}{m}x=g+\frac{k\ell_{0}}{m}$
(d)

Verify that one solution is that $x(t)$ is constant at $x=l_{0}+mg/k$ .
(e)

What is the meaning of that solution? (That is, describe in words what is going on.) Answer: This solution is the static equilibrium position; i.e., when the mass is hanging at rest, its weight is exactly balanced by the upwards force of the spring at this constant position $x$ .
(f)

Define a new variable $\hat{x}=x-(l_{0}+mg/k)$ . Substitute $x=\hat{x}+(l_{0}+mg/k)$ into your differential equation and note that the equation is simpler in terms of the variable $\hat{x}$ . Answer: ${\ddot{\hat{x}}}+\frac{k}{m}\hat{x}=0$
(g)

Assume that the mass is released from an an initial position of $x=D$ . What is the motion of the mass?

Answer: $x(t)=\left[D-\left(\ell_{0}+\frac{mg}{k}\right)\right]\cos\sqrt{\frac{k}{m}}\ % t+(\ell_{0}+\frac{mg}{k})$
(h)

What is the period of oscillation of this oscillating mass? Answer: period= $2\pi\sqrt{\frac{m}{k}}$ .
(i)

Why might this solution not make physical sense for a long, soft spring if the initial stretch is large. In other words, what is wrong with this solution if $D>\ell_{0}+2mg/k$ ? Answer: If the initial position $D$ is more than $\ell_{0}+2mg/k$ , then the spring is in compression for part of the motion. A floppy spring would buckle when in compression.

Filename:pfigure-s94h13p2

Figure 10.41:

10.3.7 One egg-drop contestant used a structure which held the egg (mass $m$ ) at the center with rubber bands. Consider the 2D model shown. The springs are linear with spring constants $k$ . After falling a height $h$ the frame hits the ground on a flat edge. Assume small motions (deflection $\ll$ side-length) and that the springs do not buckle.

(a)

what is the vibration frequency after impact?
(b)

What is the maximum vertical deflection of the egg (relative to its equilibrium position)?

Filename:pfigure-f93f5

Figure 10.42:

10.3.8 A person jumps on a trampoline. The trampoline is modeled as having an effective vertical undamped linear spring with stiffness $k=200\,{\rm lbf}/\,\mathrm{ft}$ . The person is modeled as a rigid mass $m=150\,{\mathrm{lbm}}$ . $g=32.2\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ .

(a)

What is the period of motion if the person’s motion is so small that her feet never leave the trampoline? Answer: period $=\frac{2\pi}{\sqrt{\frac{k}{m}}}=0.96\,\mathrm{s}$
(b)

What is the maximum amplitude of motion (amplitude of the sine wave) for which her feet never leave the trampoline? Answer: maximum amplitude= $0.75\,\mathrm{ft}$
(c)

(harder) If she repeatedly jumps so that her feet clear the trampoline by a height $h=5\,\mathrm{ft}$ , what is the period of this motion (note, the contact time is not exactly half of a vibration period)? [Hint, a neat graph of height vs time will help.]

Answer: period $=2\sqrt{\frac{2h}{g}}+\par\sqrt{\frac{m}{k}}\left[\pi+2\tan^{-1}\sqrt{\frac{mg% }{2kh}}\right]\approx 1.64\,\mathrm{s}$ .

Filename:pfigure-s94h13p3

Figure 10.43:

A person jumps on a trampoline.

10.4 Coupled motions in 1D

Thinking of a car, a plane, a person on a bicycle or a satellite as a single particle is often edifying, and sufficient for many engineering purposes. However, the one-particle model is also often inadequate. That the parts of a machine or structure move relative to each other is obviously sometimes important; many important engineering systems have parts that move independently.

Here we begin the study of independent, but coupled, motions of parts. The independent motions are coupled in that the motion of each part may affect the motion of the others.

Filename:p-s96-p3-3

Figure 10.44: The vertical motions of a car and wheel might be considered as coupled motions in 1D.

Example: Car suspension.

A model of a car suspension treats the wheel as one particle and the car as another. The wheel is coupled to the ground by a tire and to the car by the suspension. In a first analysis the only motion to consider would be vertical for both the wheel and the car. Think of the ground as moving up and down and ‘forcing’ the motion of the car and wheel system.

Still using one-dimensional mechanics, we consider systems that can be modeled as two or more particles. Such one-dimensional coupled motion analysis is common in engineering practice in situations where there are connected parts that all move in about the same direction, but the parts do not move the same amount or necessarily at the same time. Many of the ideas generalize to systems where parts, each with one degree of freedom, are coupled together. Many generalizations apply even if each degree of freedom is quite different from the others. These generalizations to more general coupled motions come later in the book.

The primary goal in this section is to develop two skills:

•

To write correct equations of motion for a line of particles connected to each other with springs and dashpots, and
•

To simulate the motions of such systems on a computer.
•

(the third of the two things, really implicit in the first two) To use the simulation results to find errors in the equations.

The concept of ‘normal modes’ is postponed to Section 11.3

The simplest way of dealing with the coupled motion of two or more particles is

•

to write ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ for each particle and then
•

to use the forces on the free-body diagrams to evaluate the forces.

Because the most common models for the interaction forces are springs and dashpots (see chapter 3), one needs to account for the relative positions and velocities of the particles.

Relative motion in one dimension

If the position of A is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm A}$ , and B’s position is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm B}$ , then B’s position relative to A is

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B}-{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{A}.

Relative velocity and acceleration are similarly defined by subtraction, or by differentiating the above expression, as

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}={\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B}-{\accentset{\rightharpoonup}{\boldsymbol% {v}}}_{A}\mbox{ and }{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B/A}={% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}-{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{A}.

Filename:bikefork1-ang-mom

Figure 10.45: The relative position of points A and B in one dimension.

In one dimension, the relative position diagram of fig. 1.5 on page 1.5 becomes fig. 10.45.

${\accentset{\rightharpoonup}{\boldsymbol{r}}}=x\hat{\boldsymbol{\imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=v\hat{\boldsymbol{\imath}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=a\hat{\boldsymbol{\imath}}$ . So, we can write,

$\displaystyle x_{B/A}$	$\displaystyle\equiv$	$\displaystyle x_{B}-x_{A},$
$\displaystyle v_{B/A}$	$\displaystyle\equiv$	$\displaystyle v_{B}-v_{A}={d\over dt}x_{B/A},\mbox{ and}$
$\displaystyle a_{B/A}$	$\displaystyle\equiv$	$\displaystyle a_{B}-a_{A}={d\over dt}v_{B/A}={d^{2}\over dt^{2}}x_{B/A}.$

An alternative notation, discussed in Chapter 2, is $x_{AB}$ where the directed line AB is equivalent to the position of B relative to A:

x_{AB}=x_{B/A}

Example: Two masses connected by a spring.

Consider the two masses on a frictionless support (fig. 10.46). Assume the spring is un-stretched when $x_{1}=x_{2}=0$ . After drawing free-body diagrams of the two masses we can write ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ for each mass:

Filename:bikefork-ang-mom

Figure 10.46:

mass 1:	${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}_{1}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T\hat{% \boldsymbol{\imath}}=m_{1}\ddot{x}_{1}\hat{\boldsymbol{\imath}}$
mass 2:	${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{2}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}_{2}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-T\hat{% \boldsymbol{\imath}}=m_{2}\ddot{x}_{2}\hat{\boldsymbol{\imath}}$

(10.45)

The stretch of the spring is

\Delta\ell=x_{2}-x_{1}

\mbox{so}\qquad T=k\Delta\ell=k(x_{2}-x_{1}).

(10.46)

Combining (10.45) and (10.46) we get

$\displaystyle\ddot{x}_{1}$	=	$\left(\frac{1}{m_{1}}\right)k(x_{2}-x_{1})$
$\displaystyle\ddot{x}_{2}$	=	$\left(\frac{1}{m_{2}}\right)(-k(x_{2}-x_{1}))$

(10.47)

Note:

Take care with signs when setting up this type of problem. You should check, for example, that if $x_{2}>x_{1}$ , mass 1 accelerates to the right ( $\ddot{x}_{1}>0$ ) and mass 2 accelerates to the left( $\ddot{x}_{2}<0$ ). It is easy to make sign errors. You’ve been warned!

The differential equations that result from writing ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ for the separate particles are coupled second-order equations. The equations are ‘coupled’ in that the equation for $m_{1}$ , say, includes the position $x_{2}$ or velocity $v_{2}$ of mass 2. Such systems of second order coupled equations are often solved on a computer by writing them as a system of first-order equations. You have two first-order equations for each of the second order equations because of the addition of equations like, for example, $\dot{x}_{17}=v_{17}$ .

Example: Writing second-order ODEs as first-order ODEs.

Refer again to fig. 10.46 If we define $v_{1}=\dot{x}_{1}$ and $v_{2}=\dot{x}_{2}$ we can rewrite equation 10.47 as

$\displaystyle\dot{x}_{1}$	$\displaystyle=$	$\displaystyle v_{1}$
$\displaystyle\dot{v}_{1}$	$\displaystyle=$	$\displaystyle\left(\frac{1}{m_{1}}\right)k(x_{2}-x_{1})$
$\displaystyle\dot{x}_{2}$	$\displaystyle=$	$\displaystyle v_{2}$
$\displaystyle\dot{v}_{2}$	$\displaystyle=$	$\displaystyle\left(\frac{1}{m_{2}}\right)\left(-k\right)(x_{2}-x_{1})$

or, defining $z_{1}=x_{1},z_{2}=v_{1},z_{3}=x_{2},z_{4}=v_{2}$ , we get

Most numerical solutions depend on specifying numerical values for the various constants and initial conditions.

Example: computer solution

If we take, in consistent units, $m_{1}=1,k=1,m_{2}=1,x_{1}(0)=0,x_{2}(0)=0,v_{1}(0)=1$ , and $v_{2}(0)=0$ , we can set up a well defined computer problem (please see the preface for a discussion of the computer notation). This problem corresponds to finding the motion just after the left mass was hit on the left side with a hammer:

Filename:summer95f-5-a

Figure 10.47: Plot of the position of the left mass vs. time.

   ODEs = {z1dot =  z2
           z2dot = -z1 + z3
           z3dot =  z4
           z4dot =  z1 - z3}
   ICs  = {z1(0) =0, z2(0)=1, z3(0)=0, z4(0)=0}
   solve ODEs with ICs from t=0 to t=10
   plot z1 vs t.

This yields the plot shown in fig. 10.47.

The same methods work for problems involving connections with dashpots.

Example: Multi-DOF system with a dashpot.

Filename:pfigure4-2-rp10

Figure 10.48: A system with 3 masses, 4 springs and a dashpot. Also shown is free-body diagram of mass B.

Consider $m_{B}$ in fig. 10.48. Using the free-body diagram shown linear momentum balance gives

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$
$\displaystyle\left\{(-T_{k_{4}}-T_{k_{2}}+T_{c_{1}}+T_{k_{3}})\hat{\boldsymbol% {\imath}}\right.$	$\displaystyle=$	$\displaystyle\left.ma_{B}\hat{\boldsymbol{\imath}}\phantom{T_{k_{c}}}\hskip-10% .84006pt\right\}$
$\displaystyle\left\{\right\}\cdot\hat{\boldsymbol{\imath}}\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}-\underbrace{T_{k_{4}}}_{k_{4}x_{B}}-\underbrace{T_{k_% {2}}}_{k_{2}(x_{B}-x_{A})}+\underbrace{T_{c_{1}}}_{c_{1}(\dot{x}_{D}-\dot{x}_{% B})}+\underbrace{T_{k_{3}}}_{k_{3}(x_{D}-x_{B})}$	$\displaystyle=$	$\displaystyle m\underbrace{a_{B}}_{\ddot{x}_{B}}$
$\displaystyle k_{2}x_{A}-(k_{2}+k_{4}+k_{3})x_{B}+k_{3}x_{D}-c_{1}\dot{x}_{B}+% c_{1}\dot{x}_{D}$	$\displaystyle=$	$\displaystyle m\ddot{x}_{B}$

Similar equations could be written for masses A and C. Some things to note

•

We assumed zeros for the displacements so that the system is in static equilibrium if $x_{A}=x_{B}=x_{D}=0$ .
•

We have taken the sign convention that tension is positive for all springs and dashpots.
•

All of the spring coefficients of $x_{B}$ have a minus sign in front. That is because all springs, whether to the right or the left of mass B, provide a restoring force if mass B is displaced.
•

All of the spring coefficients of $x_{A}$ and $x_{D}$ make a positive contribution because motion to the right of mass A or mass D causes a force to the right on mass B.

As for the example above, for any system of masses, linear springs and linear dashpots the set of momentum balance equations can be written in the form

[M]\ddot{\boldsymbol{x}}+[C]\dot{\boldsymbol{x}}+[K]\boldsymbol{x}=\boldsymbol% {0}

(10.49)

where $\boldsymbol{x}$ is a list of positions of the masses. The mass matrix $[M]$ is diagonal because each equation corresponds to $F=ma$ for one mass. The damping and stiffness matrices $[C]$ and $[K]$ are symmetric because, as Jim Marley said, ‘every action has a reaction’; if motion of mass 7 causes a stretch on the spring between it and mass 19 then motion of mass 19 causes a stretch on the same spring, similarly affecting mass 7. So row 7 column 19 has the same entry as row 19 column 7. As noted in the example below, the diagonal elements of $[M]$ , $[C]$ and $[K]$ are positive (or zero).

Example: Matrix form

When the three momentum balance equations for fig. 10.48 are written, one for each mass, they can be assembled in matrix form as

\underbrace{\left[\begin{array}[]{ccc}m_{A}&0&0\\ 0&m_{B}&0\\ 0&0&m_{C}\end{array}\right]}_{[M]}\underbrace{\left[\begin{array}[]{c}\ddot{x}% _{A}\\ \ddot{x}_{B}\\ \ddot{x}_{D}\end{array}\right]}_{\ddot{\boldsymbol{x}}}+\underbrace{\left[% \begin{array}[]{ccc}0&0&0\\ 0&c_{1}&-c_{1}\\ 0&-c_{1}&c_{1}\end{array}\right]}_{[C]}\underbrace{\left[\begin{array}[]{c}% \dot{x}_{A}\\ \dot{x}_{B}\\ \dot{x}_{D}\end{array}\right]}_{\dot{\boldsymbol{x}}}+\underbrace{\left[\begin% {array}[]{ccc}(k_{1}+k_{2})&-k_{2}&0\\ -k_{2}&(k_{2}+k_{4}+k_{3})&-k_{3}\\ 0&-k_{3}&k_{3}\end{array}\right]}_{[K]}\underbrace{\left[\begin{array}[]{c}x_{% A}\\ x_{B}\\ x_{D}\end{array}\right]}_{\boldsymbol{x}}=\underbrace{\left[\begin{array}[]{c}% 0\\ 0\\ 0\\ \end{array}\right]}_{\boldsymbol{0}}.

The equation for $m_{B}$ worked out at the start of this example corresponds to the second row of these matrices.

This form is convenient for numerical solution if it is written as

	$\displaystyle\dot{\boldsymbol{x}}$	$\displaystyle=$	$\displaystyle\boldsymbol{v}$
	$\displaystyle\dot{\boldsymbol{v}}$	$\displaystyle=$	$\displaystyle-[M]^{-1}\left[[C]\boldsymbol{v}+[K]\boldsymbol{x}\right]\par$

For the three mass example this would represent 6 first-order differential equations.

Center of mass

For both theoretical and practical reasons it is often useful to pay attention to the motion of the average position of mass in the system. This average position is called the center-of-mass. For a collection of particles in one dimension the center-of-mass is

x_{\rm CM}={\sum x_{i}m_{i}\over m_{\mathrm{tot}}},\par

(10.50)

where $m_{\mathrm{tot}}=\sum m_{i}$ is the total mass of the system. The velocity and acceleration of the center-of-mass are found by differentiation to be

v_{\rm CM}={\sum v_{i}m_{i}\over m_{\mathrm{tot}}}\hbox to36.135pt{\hfil and% \hfil}a_{\rm CM}={\sum a_{i}m_{i}\over m_{\mathrm{tot}}}.

(10.51)

If we imagine a system of interconnected masses and add the ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ equations from all the separate masses we can get on the left hand side only the forces from the outside; the interaction forces cancel because they come in equal and opposite (action and reaction) pairs. So we get:

\sum F_{\rm external}=\sum a_{i}m_{i}=m_{\mathrm{tot}}a_{\rm CM}.

(10.52)

So the center-of-mass of a system (a system that may be deforming wildly) obeys the same simple governing equation as a single particle. Although our demonstration here was for particles in one dimension. The result holds for any bodies of any type in 1,2, or 3 dimensions.

SAMPLE 10.16 For the given quantities and initial conditions, find $x_{1}(t)$ and $x_{2}(t)$ . Assume the spring is unstretched at its rest length of $L_{0}$ when $x_{1}=x_{2}$ .

Filename:pfigure-blue-125-2

Figure 10.49:

$m_{1}=1\,\mathrm{kg},$	$m_{2}=2\,\mathrm{kg},$	$k=3\,{\rm N}/{\rm m},$	$c=5\,{\mbox{N}}/(\,\mathrm{m}/\mathrm{s})$
$x_{1}(0)=1\,\mathrm{m},$	$\dot{x}_{1}(0)=0,$	$x_{2}(0)=2\,\mathrm{m},$	$\dot{x}_{2}(0)=0.$

Solution The free-body diagrams of all components of the given system are shown below.

Filename:pfigure-blue-68-1

Figure 10.50:

The spring and dashpot laws give

T_{1}=c\dot{x}_{1}\qquad T_{2}=k(x_{2}-x_{1}).

(10.53)

The linear momentum balance for the two masses gives

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}$
mass 1:	$\displaystyle-T_{1}\hat{\boldsymbol{\imath}}+T_{2}\hat{\boldsymbol{\imath}}=m_% {1}\ddot{x}_{1}\hat{\boldsymbol{\imath}}$	(10.54)
mass 2:	$\displaystyle-T_{2}\hat{\boldsymbol{\imath}}=m_{2}\ddot{x}_{2}\hat{\boldsymbol% {\imath}}.$

Applying the constitutive laws (10.53) to the momentum balance equations (10.54) gives

	$\displaystyle\ddot{x}_{1}$	$\displaystyle=$	$\displaystyle[k(x_{2}-x_{1})-c\dot{x}_{1}]/m_{1}$
	$\displaystyle\ddot{x}_{2}$	$\displaystyle=$	$\displaystyle[-k(x_{2}-x_{1})]/m_{2}.$

Defining $z_{1}=x_{1},z_{2}=\dot{x}_{1},z_{3}=x_{2},z_{4}=\dot{x}_{2}$ gives

$\displaystyle\dot{z}_{1}$	$\displaystyle=$	$\displaystyle z_{2}$
$\displaystyle\dot{z}_{2}$	$\displaystyle=$	$\displaystyle[k(z_{3}-z_{1})-cz_{2}]/m_{1}$
$\displaystyle\dot{z}_{3}$	$\displaystyle=$	$\displaystyle z_{4}$
$\displaystyle\dot{z}_{4}$	$\displaystyle=$	$\displaystyle[-k(z_{3}-z_{1})]/m_{2}.$

The initial conditions are

z_{1}(0)=1\,\mathrm{m},

z_{2}(0)=0,

z_{3}(0)=2\,\mathrm{m},

z_{4}(0)=0.

We are now set for numerical solution. Solving these equations numerically, we plot $x_{1}(t)$ and $x_{2}(t)$ as shown in fig. 10.51. From the solution, it is clear that both the masses settle down to the equilibrium position $x_{1}=x_{2}=1\,\mathrm{m}$ after the oscillations die down. In this position, the spring exerts no force as it is unstretched. Also note that the two masses move in the opposite direction immediately after being set into motion as they must because of the opposite accelerations.

Filename:pfigure-blue-58-1

Figure 10.51: Plot of

x_{1}(t)

and

x_{2}(t)

obtained from the numerical solution of the equations of motion. Note that with

x_{1}(0)=1\,\mathrm{m}

and

x_{2}(0)=2\,\mathrm{m}

, both masses settle down to

x_{1}=x_{2}=1\,\mathrm{m}

after the oscillations die.

SAMPLE 10.17

Filename:pfigure-blue-157-1

Figure 10.52:

Flight of a toy hopper. A hopper model ^†^†margin: You can think of it as a model of a hopping robot or a hopping animal where the foot mass is modeled with $m_{2}$ and the body mass with $m_{1}$ . The spring models the elasticity of the system. If you restrict this model to one dimensional vertical motion under gravity, it will hop, once released from some initial height. The number of hops depends on the initial conditions. Theoretically, it can hop forever with suitable initial conditions. is made of two masses $m_{1}=0.4\,\mathrm{kg}$ and $m_{2}=1\,\mathrm{kg}$ , and a spring with stiffness $k=100\,{\rm N}/{\rm m}$ as shown in fig. 10.52. The unstretched length of the spring is $\ell_{0}=1\,\mathrm{m}$ . The model is released from rest from the configuration shown in the figure with $y_{1}=25.5\,\mathrm{m}$ and $y_{2}=24\,\mathrm{m}$ .

1.

Find and plot $y_{1}(t)$ and $y_{2}(t)$ for $t=0$ to $2\,\mathrm{s}$ .
2.

Plot the motion of $m_{1}$ and $m_{2}$ with respect to the center-of-mass of the hopper during the same time interval.
3.

Plot the motion of the center-of-mass of the hopper from the solution obtained for $y_{1}(t)$ and $y_{2}(t)$ and compare it with analytical values obtained by integrating the center-of-mass motion directly.

Solution The free-body diagrams of the two masses are shown in fig. 10.53. From the linear momentum balance in the $y$ direction, we can write the equations of motion at once.

Filename:summer95f-5

Figure 10.53: Free-body diagram of the two masses

m_{1}

and

m_{2}

$\displaystyle m_{1}\ddot{y}_{1}$	$\displaystyle=$	$\displaystyle-k(y_{1}-y_{2}-\ell_{0})-m_{1}g$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{y}_{1}$	$\displaystyle=$	$\displaystyle-\frac{k}{m_{1}}(y_{1}-y_{2})+\frac{k\ell_{0}}{m_{1}}-g$	(10.55)
$\displaystyle m_{2}\ddot{y}_{2}$	$\displaystyle=$	$\displaystyle k(y_{1}-y_{2}-\ell_{0})-m_{2}g$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{y}_{2}$	$\displaystyle=$	$\displaystyle\frac{k}{m_{2}}(y_{1}-y_{2})-\frac{k\ell_{0}}{m_{2}}-g.$	(10.56)

1.

The equations of motion obtained above are coupled linear differential equations of second order. We can solve for $y_{1}(t)$ and $y_{2}(t)$ by numerical integration of these equations. As we have shown in previous examples, we first need to set up these equations as a set of first order equations.
Letting $\dot{y}_{1}=v_{1}$ and $\dot{y}_{2}=v_{2}$ , we get

$\displaystyle\dot{y}_{1}$ $\displaystyle=$ $\displaystyle v_{1}$

$\displaystyle\dot{v}_{1}$ $\displaystyle=$ $\displaystyle-\frac{k}{m_{1}}(y_{1}-y_{2})+\frac{k\ell_{0}}{m_{1}}-g$

$\displaystyle\dot{y}_{2}$ $\displaystyle=$ $\displaystyle v_{2}$

$\displaystyle\dot{v}_{2}$ $\displaystyle=$ $\displaystyle\frac{k}{m_{2}}(y_{1}-y_{2})-\frac{k\ell_{0}}{m_{2}}-g.$

Now we solve this set of equations numerically using some ODE solver and the following pseudocode.
```
    ODEs = {y1dot = v1,
            v1dot = -k/m1*(y1-y2-l0) - g,
            y2dot = v2,
            v1dot = k/m1*(y1-y2-l0) - g}
    IC   = {y1(0)=25.5, v1(0)=0, y2(0)=24, v2(0)=0}
    Set  k=100, m1=0.4, m2=1, l0=1
    Solve ODEs with IC for t=0 to t=2
    Plot y1(t) and y2(t)
```
The solution obtained thus is shown in fig. 10.54.
Filename:pfigure-blue-127-2

Figure 10.54: Numerically obtained solutions $y_{1}(t)$ and $y_{2}(t)$
2.

We can find the motion of $m_{1}$ and $m_{2}$ with respect to the center-of-mass by subtracting the motion of the center-of-mass, $y_{\rm cm}$ from $y_{1}$ and $y_{2}$ . Since,

Filename:pfigure-s94h13p4

Figure 10.55: Numerically obtained solutions $y_{1/{\rm cm}}(t)$ and $y_{1/{\rm cm}}(t)$ .

$y_{\rm cm}=\frac{m_{1}y_{1}+m_{2}y_{2}}{m_{1}+m_{2}}$ (10.57)

we get,

$\displaystyle y_{1/{\rm cm}}$ $\displaystyle=$ $\displaystyle y_{1}-y_{\rm cm}=\frac{m_{2}}{m_{1}+m_{2}}(y_{1}-y_{2})$

$\displaystyle y_{2/{\rm cm}}$ $\displaystyle=$ $\displaystyle y_{2}-y_{\rm cm}=-\frac{m_{1}}{m_{1}+m_{2}}(y_{1}-y_{2}).$

The relative motions thus obtained are shown in fig. 10.55. We note that the motions of $m_{1}$ and $m_{2}$ , as seen by an observer sitting at the center-of-mass, are simple harmonic oscillations.
3.

We can find the center-of-mass motion $y_{\rm cm}(t)$ from $y_{1}$ and $y_{2}$ by using eqn. (10.57). The solution obtained thus is shown as a solid line in

fig. 10.57.

We can also solve for the center-of-mass motion analytically by first writing the equation of motion of the center-of-mass and then integrating it analytically.

Filename:pfigure-blue-90-2

Figure 10.56: Free-body diagram of the hopper as a single system. The spring force does not show up here since it becomes an internal force to the system

The free-body diagram of the hopper as a single system is shown in fig. 10.56. The linear momentum balance for the system in the vertical direction gives

$\displaystyle(m_{1}+m_{2})\ddot{y}_{\rm cm}$ $\displaystyle=$ $\displaystyle-m_{1}g-m_{2}g$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{y}_{\rm cm}$ $\displaystyle=$ $\displaystyle-g.$

We recognize this equation as the equation of motion of a freely falling body under gravity. We can integrate this equation twice to get

$y_{\rm cm}(t)=y_{\rm cm}(0)+\dot{y}_{\rm cm}(0)t-\frac{1}{2}gt^{2}.$

Noting that $y_{\rm cm}(0)=24.43\,\mathrm{m}$ (from eqn. (10.57)), and $\dot{y}_{\rm cm}(0)=0$ (the system is released from rest), we get

$y_{\rm cm}(t)=24.43\,\mathrm{m}-\frac{1}{2}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{% 2}\cdot t^{2}.$

The values obtained for the center-of-mass position from the above expression are shown in fig. 10.57 by small circles.

Filename:s92f1p7

Figure 10.57: Numerically obtained solution for the position of the center-of-mass, $y_{\rm cm}(t)$ .

SAMPLE 10.18

Filename:twodisks-ang-mom

Figure 10.58: Mr. P spots an interesting object.

Conservation of linear momentum. Mr. P with mass $m_{p}=200\,{\mathrm{lbm}}$ is standing on a cart with frictionless and massless wheels. The cart weighs half as much as Mr. P. Standing at one end of the cart, Mr. P spots an interesting object at the other end of the cart. Mr. P decides to walk to the other end of the cart to pick up the object. How far does he find himself from the object after he reaches the end of the cart?

Solution From your own experience in small boats perhaps, you know that when Mr. P walks to the left the cart moves to the right. Here, we want to find how far the cart moves.

Consider the cart and Mr. P together to be the system of interest. The free-body diagram of the system is shown in Fig. 10.59(a).

Filename:pfigure-blue-90-1

Figure 10.59: (a) Free-body diagram of Mr. P-and-the-cart system. (b) The cart has moved to the right by distance

x

when Mr. P reaches the other end.

From the diagram it is clear that there are no external forces in the $x$ -direction. Therefore,

\dot{L}_{x}=\sum F_{x}=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}L_{x}=\mbox{constant}

that is, the linear momentum of the system in the $x$ -direction is ‘conserved’. But the initial linear momentum of the system is zero. Therefore,

L_{x}=m_{tot}(v_{cm})_{x}=0\;\mbox{all the time}\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}(v_{cm})_{x}=0\;\mbox{all the time.}

Because the horizontal velocity of the center-of-mass is always zero, the center-of-mass does not change its horizontal position. Now let $x_{cm}$ and ${x^{\prime}}_{cm}$ be the $x$ -coordinates of the center-of-mass of the system at the beginning and at the end, respectively. Then,

{x^{\prime}}_{cm}=x_{cm}.

Now, from the given dimensions and the stipulated position at the end in Fig. 10.59(b),

x_{cm}=\frac{m_{c}x_{G}+m_{p}x_{p}}{m_{c}+m_{p}}\hbox to36.135pt{\hfil and% \hfil}{x^{\prime}}_{cm}=\frac{m_{c}(x_{G}+x)+m_{p}x}{m_{c}+m_{p}}.

Equating the two distances we get,

$\displaystyle{m_{c}x_{G}}+m_{p}x_{p}$	$\displaystyle=$	$\displaystyle m_{c}(x_{G}+x)+m_{p}x$
	$\displaystyle=$	$\displaystyle{m_{c}x_{G}}+x(m_{c}+m_{p})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}x$	$\displaystyle=$	$\displaystyle\frac{m_{p}x_{p}}{m_{c}+m_{p}}$
	$\displaystyle=$	$\displaystyle\frac{200\,{\mathrm{lbm}}\cdot 10\,\mathrm{ft}}{300\,{\mathrm{lbm% }}}=6\frac{2}{3}\,\mathrm{ft}.$

Answer: $6.67\,\mathrm{ft}$

[Note: if Mr. P and the cart have the same mass, the cart moves to the right the same distance Mr. P moves to the left.]

Problems for 10.4 Coupled motion in 1D

The primary emphasis of this section is setting up correct differential equations (without sign errors) and solving these equations on the computer.

Preparatory Problems

10.4.1 Write the following set of coupled second order ODE’s as a system of first order ODE’s.

	$\displaystyle\ddot{x}_{1}$	$\displaystyle=$	$\displaystyle k_{2}(x_{2}-x_{1})-k_{1}x_{1}$
	$\displaystyle\ddot{x}_{2}$	$\displaystyle=$	$\displaystyle k_{3}x_{2}-k_{2}(x_{2}-x_{1})$

10.4.2 The solution of a set of second order differential equations is:

	$\displaystyle\xi(t)=A\sin\omega t+B\cos\omega t+\xi^{\ast}$
	$\displaystyle\dot{\xi}(t)=A\omega\cos\omega t-B\omega\sin\omega t,$

where $A$ and $B$ are constants to be determined from initial conditions and $\xi^{\ast}$ is a known constant. Assume $A$ and $B$ are the only unknowns.

(a)

Write the equations in matrix form which you would need to solve in order to find $A$ and $B$ in terms of $\xi(0)$ and $\dot{\xi}(0)$ .
(b)

Solve the equations in symbols.
(c)

Solve for the numerical constants $A$ and $B$ using the matrix form, if $\xi(0)=0,\;\dot{\xi}(0)=0.5,\;\omega=0.5\,\mathrm{rad}/\mathrm{s}$ and $\xi^{\ast}=0.2$ .

10.4.3 A set of first order linear differential equations is given:

	$\displaystyle\dot{x}_{1}=x_{2}$
	$\displaystyle\dot{x}_{2}+kx_{1}+cx_{2}=0.$

Write these equations in the form $\dot{\accentset{\rightharpoonup}{\boldsymbol{x}}}=[\boldsymbol{A}]{\accentset{% \rightharpoonup}{\boldsymbol{x}}}$ , where ${\accentset{\rightharpoonup}{\boldsymbol{x}}}=\left\{\begin{array}[]{c}x_{1}\\ x_{2}\end{array}\right\}$ .

10.4.4 Write the following pair of coupled ODE’s as a set of first order ODE’s.

	$\displaystyle\ddot{x}_{1}+x_{1}$	$\displaystyle=$	$\displaystyle\dot{x}_{2}\sin t$
	$\displaystyle\ddot{x}_{2}+x_{2}$	$\displaystyle=$	$\displaystyle\dot{x}_{1}\cos t$

10.4.5 The following set of differential equations can be written in first order form, and in particular, in matrix form $\dot{\accentset{\rightharpoonup}{\boldsymbol{x}}}=[\boldsymbol{A}]{\accentset{% \rightharpoonup}{\boldsymbol{x}}}+{\accentset{\rightharpoonup}{\boldsymbol{c}}}$ . In general equations of motion are not so simple, but linear cases like this are prevalent in the analytic study of dynamical systems.

	$\displaystyle\dot{x}_{1}=x_{3}$
	$\displaystyle\dot{x}_{2}=x_{4}$
	$\displaystyle\dot{x}_{3}+5\Omega^{2}x_{1}-4\Omega^{2}x_{2}=2\Omega^{2}v_{1}^{\ast}$
	$\displaystyle\dot{x}_{4}-4\Omega^{2}x_{1}+5\Omega^{2}x_{2}=-\Omega^{2}v_{1}^{\ast}$

10.4.6 Write each of the following equations as a system of first order ODE’s.

(a)

$\ddot{\theta}+\lambda^{2}\theta=\cos t,$
(b)

$\ddot{x}+2p\dot{x}+kx=0,$
(c)

$\ddot{x}+2c\dot{x}+k\sin x=0.$

10.4.7 A train moves at a constant absolute velocity $v\hat{\boldsymbol{\imath}}$ . A passenger, idealized as a point mass, walks at an absolute velocity $u\hat{\boldsymbol{\imath}}$ , where $u>v$ . What is the velocity of the passenger relative to the train?

10.4.8 Two equal masses, each denoted by the letter $m$ , are on an air track. One mass is connected by a spring to the end of the track. The other mass is connected by a spring to the first mass. The two spring constants are equal and represented by the letter $k$ . In the rest configuration (springs are relaxed) the masses are a distance $\ell$ apart. Motion of the two masses $x_{1}$ and $x_{2}$ is measured relative to this configuration.

(a)

Write the potential energy of the system for arbitrary displacements $x_{1}$ and $x_{2}$ at some time $t$ .
(b)

Write the kinetic energy of the system at the same time $t$ in terms of $\dot{x}_{1}$ , $\dot{x}_{2}$ , $m$ , and $k$ .
(c)

Write the total energy of the system.
(d)

Draw a free-body diagram for each mass.
(e)

Write the equation of linear momentum balance for each mass.

Filename:pfigure-blue-110-1

Figure 10.60:

10.4.9

For the three-mass system shown, draw a free-body diagram of each mass. Write the spring forces in terms of the displacements $x_{1}$ , $x_{2}$ , and $x_{3}$ .

Filename:pfigure-blue-107-1

Figure 10.61:

10.4.10 The springs shown are relaxed when $x_{A}=x_{B}=x_{D}=0$ . In terms of some or all of $m_{A}$ , $m_{B}$ , $m_{D}$ , $x_{A}$ , $x_{B}$ , $x_{D}$ , $\dot{x}_{A},\dot{x}_{B},\dot{x}_{C},$ and $k_{1}$ , $k_{2}$ , $k_{3}$ , $k_{4}$ and $c_{1}$ , find the acceleration of block B.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=\ddot{x}_{B}\hat{\boldsymbol% {\imath}}=\frac{1}{m_{B}}[-k_{4}x_{B}-k_{2}(x_{B}-x_{A})+c_{1}(\dot{x}_{D}-% \dot{x}_{B})+k_{3}(x_{D}-x_{B})]\hat{\boldsymbol{\imath}}$ .

Filename:pfigure-s94h14p4

Figure 10.62:

10.4.11 A system of three masses, four springs, and one damper are connected as shown. Assume that all the springs are relaxed when $x_{A}=x_{B}=x_{D}=0$ . Given $k_{1}$ , $k_{2}$ , $k_{3}$ , $k_{4}$ , $c_{1}$ , $m_{A}$ , $m_{B}$ , $m_{D}$ , $x_{A}$ , $x_{B}$ , $x_{D}$ , $\dot{x}_{A}$ , $\dot{x}_{B}$ , and $\dot{x}_{D}$ , find the acceleration of mass B, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=\ddot{x}_{B}\hat{\boldsymbol% {\imath}}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=\ddot{x}_{B}\hat{\boldsymbol% {\imath}}=\frac{1}{m_{B}}[-k_{4}x_{B}-c_{1}(\dot{x}_{B}-\dot{x}_{A})+(k_{2}+k_% {3})(x_{D}-x_{B})]$ .

Filename:pfigure-blue-112-1

Figure 10.63:

More-Involved Problems

10.4.12 A massless spring with constant k is held compressed a distance $\delta$ from its relaxed length by a thread connecting blocks A and B which are still on a frictionless table. The blocks have mass $m_{A}$ and $m_{B}$ , respectively. The thread is suddenly but gently cut, the blocks fly apart and the spring falls to the ground. Find the speed of block A as it slides away. Answer: $v_{A}=\sqrt{\frac{m_{B}k\delta^{2}}{m_{A}^{2}+m_{B}m_{A}}}$ .

Filename:tfigure8-alt-app2c

Figure 10.64:

10.4.13 In the system below the masses are in equilibrium with the springs when $x_{1}=x_{2}=0$ .

(a)

First do problem 10.60.
(b)

Pick parameter values and initial conditions of your choice and simulate a motion of this system. Make a plot of the motion of, say, one of the masses vs time,
(c)

Explain how your plot does or does not make sense in terms of your understanding of this system. Is the initial motion in the right direction? Are the solutions periodic? Bounded? etc.

Filename:sfig4-6-4a

Figure 10.65:

10.4.14 Two masses are connected to fixed supports and each other with the three springs and dashpot shown. The force $F$ acts on mass 2. The displacements $x_{1}$ and $x_{2}$ are defined so that $x_{1}=x_{2}=0$ when the springs are unstretched. The ground is frictionless. The governing equations for the system shown can be written in first order form if we define $v_{1}\equiv\dot{x}_{1}$ and $v_{2}\equiv\dot{x}_{2}$ .

(a)

Write the governing equations in a neat first order form. Your equations should be in terms of any or all of the constants $m_{1}$ , $m_{2}$ , $k_{1}$ , $k_{2}$ , $k_{3}$ , $C$ , the constant force $F$ , and $t$ . Getting the signs right is important.
(b)

Write computer commands to find and plot $v_{1}(t)$ for 10 units of time. Make up appropriate initial conditions.
(c)

For constants and initial conditions of your choosing, plot $x_{1}$ vs $t$ for enough time so that decaying erratic oscillations can be observed.

Filename:sfig4-6-3

Figure 10.66:

10.4.15 The three beads of masses $m$ , $2m$ , and $m$ connected by massless linear springs of constant $k$ slide freely on a straight rod. Let $x_{i}$ denote the displacement of the $i^{th}$ bead from its equilibrium position at rest.

(a)

Write expressions for the total kinetic and potential energies.
(b)

Write an expression for the total linear momentum.
(c)

Draw free-body diagrams for the beads and use Newton’s second law to derive the equations for motion for the system.
(d)

Verify that total energy and linear momentum are both conserved.
(e)

Show that the center of mass must either remain at rest or move at constant velocity.
(f)

What can you say about vibratory (sinusoidal) motions of the system?

Filename:sfig4-6-3a

Figure 10.67:

10.4.16 Two blocks with masses $M$ and $m$ are connected by a spring with constant $k$ and free length $\ell_{0}$ that can sustain compression. Mass $M$ is resting on the ground at the start. There is gravity. The upwards vertical displacement of mass $m$ is $x$ , which is zero when the spring is at its rest length and $M$ is on the ground.

(a)

For what value of $x$ is the system in static equilibrium?
(b)

Find a differential equation governing the motion of $m$ assuming $M$ remains on the ground.
(c)

Draw a free-body diagram of $M$ .
(d)

For what value of $x$ is $M$ on the verge of lifting off the ground.
(e)

Defining $y$ as the height of the lower mass, write two coupled differential equations for the motion of $m$ and $M$ if both masses are in the air.
(f)

Find the value of $x<0$ so that if the system is started from rest with that $x$ and $y=0$ that the ground reaction force on $M$ just goes to zero.
(g)

Starting here, this problem is more of a project than a typical homework problem. Assume $x(t=0$ ) is less than the value computed above. Write a computer program that integrates the equations of motion until $M$ lifts off and then switches to integrating the equations for the two masses in the air.
(h)

modify your program so that if $M$ hits the ground again, it sticks until the ground reaction force goes to zero again.
(i)

By playing around, this way or that, see if you can find a special value for $x(t=0)$ so that the bouncing continues indefinitely. (This is a perhaps surprising result, that a system with plastic collisions can continue to bounce indefinitely.)

Filename:pfigure4-4-rp12

Figure 10.68:

10.5 Collisions in 1D

Filename:pfigure-blue-38-2

Figure 10.69: Here cars are shown colliding. A free-body diagram of the right car shows the collision force and should not show other forces which are negligibly small. Here they are shown as negligibly small forces to give the idea that they may be much smaller than the collision force. The wheel reaction forces are neglected because of the spring compliance of the suspension and tires.

Filename:pfigure-blue-49-2

Figure 10.70: Just before a collision is called “-”, and just after is called “+”. The only forces that show on a collisional free-body diagram are those that are large and part of the impact. Either a force or an impulse may be shown. This figure exaggerates the difference between the before (-) and after (+) states. In analysis we assume that there is no change in the body’s position or orientation from just before to just after the collision. The only net changes caused by the collision are the body’s velocity and rotation rate.

Sometimes things interact in a sudden manner, like two cars in a head-on crash or a dropped cell-phone hitting the floor. Some sudden interactions are intentional, for example in sports the banging of racquets, bats, clubs, sticks, hands and legs with balls, pucks and bodies. And in machines there are sometimes intentionally sudden interactions like the clicking of a ratchet and the flip of an electric light switch. More esoteric ‘sudden’ interactions include those between subatomic particles in an accelerator and near passes of satellites with planets.

When two solids bump into each other a nearly discontinuous change in their velocities and/or angular velocities is needed to keep the bodies from interpenetrating. This sudden change in velocity demands large interaction. In the case of subatomic particles near nuclei and satellites near planets there might be no contact, but nonetheless there are large forces when the interaction distances get small. Estimating the effects of these large yet short-lived forces is the central problem in collision mechanics.

Two objects are said to collide when some interaction force or moment between them becomes so large that other forces acting on the bodies become negligible. For example, in a car collision the force of interaction at the bumpers may be many times the weight of the car or the reaction forces acting on the wheels. And so short acting that, although velocities change, positions change negligibly during the collision.

Collisional free-body diagrams The analysis of collisions is a little different than the analysis of smooth motions, but still depends on free-body diagrams (See fig. 10.69). Knowing which forces to include and which to ignore in a collisional free-body-diagram is a subtle issue.

Some rules of thumb:

•

ignore forces from gravity, springs, and at places where contact is broken in the collision, and
•

include forces at places where new contact is made, or where contact is maintained.

The elementary analysis of rigid body collisions is based on these ideas:

I.

Collision forces are big, so non-collisional forces are neglected in collisional free-body diagrams.
II.

Collision forces are of short duration, so the position and orientation of the colliding bodies do not change during the collision.

What happens during a collision

During a collision between what would generally be called “rigid” bodies things get wild. There are huge contact forces and stresses in the regions near the nominally

^†^†margin: Nominally means “in name”. That is, what one calls “contacting points” are not points at all, but regions of complex interaction.

contacting points, there could be plastic deformation, fracture, and frictional slip. Elastic waves may travel all over the body, reflect and scatter this way and that. Altogether the contact interaction during the collision is the result of very complex deformations (see fig. 10.70).

Deformations (the lack of rigidity) give rise to the forces between colliding bodies. So what could the phrase “rigid-object collisions” mean? It is an oxymoron. Trying to understand the collision forces in detail, and how they are related to deformations, is way beyond this book. Actually, there is no unified theory of collisions so you can’t read about it in any book. Loosely one might imagine that during part of the collision material is being squeezed, this is called the compression phase and later on it expands back in a restitution phase. But the realities of collisions are not necessarily so simple; the forces and deformations can vary in complex ways.

Soon after the collision, however, the vibrations often die out, each object may have negligible permanent change in shape, and the object returns to motions that are well described by rigid-object kinematics. To find out the net effect of the collision forces we use this one key idea:

III.

The laws of mechanics apply during collisions even though rigid-object kinematics does not.

While the motions during a collision may be wildly complex, the general linear and angular momentum balance laws are still applicable. Rather than applying these laws to understand the details during a collision, we use them to summarize the overall result of the collision.

That is, in rigid-object collision analysis we do not pay attention to how the forces vary in time, or to the detailed trajectories, velocities or accelerations of any material points. Rather, we focus on the net change in the velocities of the colliding bodies that the collision forces cause. Thus, instead of using the differential-equation form of the linear momentum balance, angular-momentum balance and energy equations (Ia, IIa, and IIIa from the inside front cover) we use the time integrated forms (Ib, IIb, and IIIb). All that we note about a collisional force is its net impulse

{\accentset{\rightharpoonup}{\boldsymbol{P}}}_{\rm coll}=\int_{\rm collision\ % time}{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\rm coll}\,dt

in terms of which we have, for one object experiencing this impulse at point C

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{P}}}_{\rm coll}$	$\displaystyle=$	$\displaystyle\Delta{\accentset{\rightharpoonup}{\boldsymbol{L}}},$	(10.58)
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/0}}% \times{\accentset{\rightharpoonup}{\boldsymbol{P}}}_{\rm coll}$	$\displaystyle=$	$\displaystyle\Delta{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}},\quad\hbox{and}$	(10.59)
Collisional dissipation	$\displaystyle=$	$\displaystyle\Delta{E_{\mathrm{K}}}.$	(10.60)

Most often the first two of these, the impulse-momentum equations are used to find the motion after collision. The energy equation is just a check to make sure that the collisional dissipation is positive (otherwise the collision would be an energy source).

Extra assumptions are needed

The momentum balance equations, with the assumptions already discussed, are never enough in themselves to determine the outcome of a collision. The extra assumptions come in various forms. To minimize the algebra we discuss the issues first with one-dimensional collisions.

One dimensional collisions

Here we only consider collisions in the context of one-dimensional mechanics: all motion is constrained to one direction of motion by forces which we ignore. Only momentum and forces in, say, the $\hat{\boldsymbol{\imath}}$ direction are included.

Filename:pfigure4-4-rp13

Figure 10.71: Before the collisions the masses have velocities to the right of

v_{1}^{-}

and

v_{2}^{-}

. After the collision the velocities are

v_{1}^{+}

and

v_{2}^{+}

. During the collision the impulse

P

acts to the right on mass 2 and to the left on mass 1.

Example: 1-D collisions

Consider two masses which collide along their common line of motion. All velocities and momenta are positive if to the right and $P$ is the impulse on mass 2 from mass 1. The relevant impulse-momentum relations are

\begin{array}[]{lrclc}\hbox{For mass 1}&-P&=&m_{1}(v_{1}^{+}-v_{1}^{-}),\\ \hbox{For mass 2}&P&=&m_{2}(v_{2}^{+}-v_{2}^{-}),\hbox{and}\\ \hbox{For the system}&0&=&(m_{1}v_{1}^{+}+m_{2}v_{2}^{+})-(m_{1}v_{1}^{-}+m_{2% }v_{2}^{-}).\end{array}

The third equation comes from a free-body diagram of the system (i.e., conservation of momentum) or by adding the first two equations. In any case, given the masses and initial velocities we have only two independent equations and we have three unknowns: $v_{1}^{+},v_{2}^{+}$ and $P$ . Momentum balance is not enough to determine the outcome of a collision.

To “close” (make solvable) the set of equations one needs to make extra assumptions.

Sticking collisions

The simplest assumption is that the masses stick together after the collision so

v_{1}^{+}=v_{2}^{+}.

Such a collision is sometimes called a perfectly plastic, a perfectly inelastic, or a dead collision. Algebraic manipulations of the momentum equations and the “sticking” constitutive law give

	$\displaystyle v_{1}^{+}=v_{2}^{+}$	$\displaystyle=$	$\displaystyle(m_{1}v_{1}^{-}+m_{2}v_{2}^{-})/m_{\mathrm{tot}}\qquad(\hbox{% where}\ m_{\mathrm{tot}}=m_{1}+m_{2})\quad\hbox{and}$
	$\displaystyle P$	$\displaystyle=$	$\displaystyle(v_{1}^{-}-v_{2}^{-})m_{\rm coll}\qquad(\hbox{where}\ m_{\rm coll% }={m_{1}m_{2}\over m_{1}+m_{2}}).$

The collisional mass or contact mass $m_{\rm coll}$

m_{\rm coll}\quad=\quad{m_{1}m_{2}\over m_{1}+m_{2}}\quad=\quad{1\over{1\over m% _{1}}+{1\over m_{2}}}

is not the mass of anything. It is just a quantity that shows up repeatedly in collision calculations and theory. It is the reciprocal of the sum of the reciprocals of the two masses. If one mass is much bigger than the other, the contact mass is $m_{\rm coll}\approx$ the smaller of the two masses. It is the proportionality constant relating the interaction force and the relative acceleration of the particles during the collision

m_{\rm coll}(a_{2}-a_{1})=F\qquad\mbox{(with $F$ being the force of body 1 on % body 2)}

and is thus related to the effective mass of box 14.1 on page 14.1.

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Figure 10.72: The rate of approach of two about-to-collide points is

v_{1}-v_{2}

. The rate of separation after collision is

v_{2}^{\prime}-v_{1}^{\prime}

. The simplest collision law says

v_{2}^{\prime}-v_{1}^{\prime}=e(v_{1}-v_{2})

, where

e

is the coefficient of restitution.

More general 1-D collisions

The momentum equations can be re-arranged to better get at the essence of the situation which is that

•

In the collision the system’s center-of-mass velocity is unchanged, and
•

The effect of the collision is to change the difference between the two mass velocities.

So we define the center-of-mass velocity $v_{\rm cm}$ and the velocity difference $v_{\rm rel}$ as

v_{\rm cm}\equiv(m_{1}v_{1}+m_{2}v_{2})/m_{\mathrm{tot}}\hbox to36.135pt{\hfil and% \hfil}v_{\rm rel}\equiv v_{2}-v_{1}.

Note that before a collision the masses are approaching each other so $v_{1}^{-}>v_{2}^{-}$ and $v_{\rm rel}^{-}<0$ . A little more algebra shows that for any $P$ ,

$\displaystyle v_{2}^{+}$	$\displaystyle=$	$\displaystyle v_{\rm cm}+{m_{1}\over m_{1}+m_{2}}v_{\rm rel}^{+},$
$\displaystyle v_{1}^{+}$	$\displaystyle=$	$\displaystyle v_{\rm cm}-{m_{2}\over m_{1}+m_{2}}v_{\rm rel}^{+},\quad\hbox{and}$
$\displaystyle P$	$\displaystyle=$	$\displaystyle(v_{\rm rel}^{+}-v_{\rm rel}^{-})m_{\rm coll}$

That is, $P$ acts on $v_{\rm rel}$ as if $v_{\rm rel}$ were the velocity of an object with mass $m_{\rm coll}$ . If $P=0$ the equations above are a long winded way of saying that nothing happened, $v_{1}^{+}=v_{1}^{-}$ and $v_{2}^{+}=v_{2}^{-}$ , and the masses pass right through each other.

If $P=-v_{\rm rel}^{-}m_{\rm coll}$ there is a sticking collision.

Elastic collisions

Application of the above formulas will show that if

P=-2v_{\rm rel}^{-}m_{\rm coll}

then the kinetic energy of the system after the collision is the same as the kinetic energy before. That is

	$\displaystyle{E_{\mathrm{K}}}^{+}$	$\displaystyle=$	$\displaystyle{E_{\mathrm{K}}}^{-}$
	$\displaystyle{m_{1}{v_{1}^{+}}^{2}+m_{2}{v_{2}^{+}}^{2}\over 2}$	$\displaystyle=$	$\displaystyle{m_{1}{v_{1}^{-}}^{2}+m_{2}{v_{2}^{-}}^{2}\over 2}.$

Also, $v_{\rm rel}^{+}=-v_{\rm rel}^{-}$ , the relative velocity maintains its magnitude and reverses its sign.

The coefficient of restitution

We have that as $P$ ranges from $-v_{\rm rel}^{-}m_{\rm coll}$ to $-2v_{\rm rel}^{-}m_{\rm coll}$ , the collision ranges from sticking to an energy conserving reversal of relative velocities. The coefficient of restitution $e$ is introduced as a way of interpolating between these cases.

The most commonly used collision law can be summarized with this simple equation,

\overbrace{(v_{b}^{\prime}-v_{a}^{\prime})}^{\text{The speed with which % colliding points are separating after the collision.}}=\underbrace{e}_{\text{The coefficient of % restitution, assumed to be a constant for given materials.}}\overbrace{(v_{a}-v_{b})}^{\text{The speed at which colliding % points are approaching before collision.}},

(10.61)

Or, more simply expressed, the collision law can be defined by either of the following two equations

	$\displaystyle v_{\rm rel}^{+}$	$\displaystyle=$	$\displaystyle-ev_{\rm rel}^{-}\quad\hbox{or}$
	$\displaystyle P$	$\displaystyle=$	$\displaystyle-(1+e)v_{\rm rel}^{-}m_{\rm coll}.$

If $e=0$ we have a sticking collision. If $e=1$ we have an energy conserving elastic collision. If $e$ is between 0 and 1 the collision is somewhere between as dead and as alive as it can be^†^†margin: A common mistake is to take $e$ as a material property. It is not. $e$ generally depends on the shapes and sizes of the contacting objects also (see box 10.5 on page 10.5). .

In words the collision equation is: the rate of separation is proportional to the rate of approach. The coefficient $e$ is called Newton’s (see box 10.5) or Poisson’s coefficient of restitution.

Somewhat of a miracle is that a given pair of objects seems to have a coefficient of restitution that is roughly independent of the velocities. This is the result of a conspiracy by all kinds of deformation mechanisms that we don’t really understand. But that $e$ is a constant for a given pair of bodies is only an approximation that has roughly the same status (accuracy) as, say, the friction coefficient. Much lower status than the momentum balance equations.

^†^†margin: What did Newton say about collisions? Newton swung spheres at the ends of string, banged them into each other, and measured their bounce. He took account of air friction. At the point of this quote he has already discussed momentum conservation. Here is his statement of, and justification for, what we now call “Newton’s law of collisions” (eqn. (10.61)):

“In bodies imperfectly elastic the velocity of the return is to be diminished together with the elastic force; because that force (except when the parts of bodies are bruised by their impact, or suffer some such extension as happens under the strokes of a hammer) is (as far as I can perceive) certain and determined, and makes bodies to return one from the other with a relative velocity, which is in a given ratio to that relative velocity with which they met. This I tried in balls of wool, made up tightly, and strongly compressed. For, first, by letting go the pendula’s bodies, and measuring their reflection, I determined the quantity of their elastic force; and then, according to this force, estimated the reflections that ought to happen in other cases of impact. And with this computation other experiments made afterwards did accordingly agree; the balls always receding one from the other with a relative velocity, which was to the relative velocity to which they met, as about 5 to 9. Balls of steel returned with almost the same velocity; those of cork with a velocity something less; but in balls of glass the proportion was as about 15 to 16. ” (Newton’s Principia Motte’s translation revised, by Florian Cajori, Univ. of CA press, page 25, 1947)

Box 10.4 The axial collision of elastic rods: the unusual disappearance of vibrations

This box is an aside for the curious. No key skills are covered.

One can try to understand the stresses and deformations during a collision. This generally leads to the solution of partial differential equations. But those equations depend on material behavior that is usually not well-modeled. So even carefully generated numerical solutions may be far from reality.

To get a sense of the complexity we consider an ideal simple system, one that was somewhat controversial amongst the great 19th century scientists Cauchy, Poisson and Saint-Venant (so said E.J. Routh in 1905).

Two identical linear elastic rods. One uniform linear elastic rod with length $\ell$ is stationary. An identical rod approaches it with speed $v$ from the left.

Filename:tfigure-collidingrods

The system has no damping so no matter how the rods shake and vibrate, their elastic potential energy plus kinetic energy is constant.

Using the wave equation (see last paragraph) we can find the motions illustrated above. The pictures exaggerate the compression in the bar (For most materials the compression wouldn’t be visible).

First, the undeformed left rods make contact, and a compressional sound wave starts spreading to the left and right. Between the wave fronts, in both rods, the material is compressed material moving at speed $v/2$ to the right. To the right of the right wave front the material is still. To the left of the left-moving wave front the material continues to move at $v$ . When the wave fronts meet, the ends of their respective bars, the bars are compressed and all material is going to the right at $v/2$ . Then both wave-fronts reflect off the ends of the bars and head back towards the contact point. To the left of the now right-moving wave front (on the left bar) the material is stationary and uncompressed. To the right of the left-moving wave front (on the right bar) the material is uncompressed but moving to the right at speed $v$ . Finally, the waves meet in the center and the bars separate. The right bar is now uniformly moving to the right at speed $v$ and the left bar is stationary.

The result of this collision is that all of the momentum of the left bar is transferred to the right bar. The separation velocity is equal in magnitude to the approach velocity. The coefficient of restitution $e$ is 1, and the kinetic energy of the system is the same after the collision as it was before.

Note that the collision itself was quick. The wave-fronts move at the speed of sound, typically about 1000 $\,\mathrm{m}/\mathrm{s}$ for metals. So for 1 meter metal rods the collision takes a few thousandths of a second. But during that few thousandths of a second, the initial energy was partitioned into elastic strain energy and kinetic energy in different time-changing regions of the bar.

Despite all the complicated details, we predict a totally ‘elastic’ collision. This may seem natural for collisions of elastic objects.

An elastic rod hits a rigid wall If you drop a 3 foot wooden dowel straight down on a thick concrete or stone floor it bounces quite well. Why? A wave analysis like that described above shows that a wave traveling from the first contact at the floor travels up to the top and reflects back to the bottom, leaving the rod moving uniformly up after the collisions just as fast as it was moving down before. Of course a wooden dowel is not perfectly described by the simple wave theory. And the ground is not perfectly rigid. So a real dowel’s collision is not perfectly elastic.

But again we find that if we assume an elastic material we predict an elastic collision. Maybe no surprise. But the previous two examples are completely misleading! These are maybe the only examples where a detailed elastic theory predicts an elastic collision. More commonly it’s more like the next example.

Rods of different length If the rods have length $\ell_{1}$ and $\ell_{2}>\ell_{1}$ then the collision works out differently.

Filename:tfigure-collidingrods2

When the reflection from the left end of the left rod comes back to the contact point, the rods separate. The left rod is stationary but the right rod has waves moving up and back. The average speed of the right rod is $(\ell_{1}/\ell_{2})v$ so the effective coefficient of restitution is $e=\ell_{1}/\ell_{2}<1$ . Later, after the vibrations have died out, the energy of the system will be less than initially. Or, even if the waves don’t die out, the kinetic energy that can be accounted for in rigid-body mechanics is lost to remnant vibrations. Thus a totally elastic system leads to inelastic collisions. It is wrong to think that the restitution constant $e$ depends on material; it also depends on the shapes and sizes of the objects. The amount of vibrational energy left after separation depends on shape and size.

For experts only: the wave equation In one-dimensional linear elasticity the displacement $u$ to the right, of a point at location $x$ on one or the other rod follows this partial differential equation:

{\partial^{2}u\over\partial t^{2}}={E\over\rho}{\partial^{2}u\over\partial x^{% 2}}.

That is, the collision mechanics in detail is the finding of $u(x,t)$ that solves the wave equation above with the given initial conditions (one bar is moving the other isn’t) and the boundary conditions (the ends of the bars have no stresses but where they are in contact they can have equal compressive stresses). The solution is most easily found by constructing right and left going waves that add to meet the initial conditions and boundary conditions (Routh).

SAMPLE 10.19

Filename:pfigure-blue-52-2

Figure 10.73:

Collision without energy loss: A block of mass $m_{1}=2\,\mathrm{kg}$ moves with speed $v_{1}=0.5\,\mathrm{m}/\mathrm{s}$ along the $x$ -axis on a frictionless level ground behind another block of mass $m_{2}=10\,\mathrm{kg}$ moving at a speed $v_{2}=0.2\,\mathrm{m}/\mathrm{s}$ in the same direction. The first block collides with the second block. Given that there is no loss of energy in this collision, find the speeds of the two blocks immediately after the collision.

Solution We are given the speeds of two blocks (of known masses) just before the collision. It is also given that there is no loss of energy in the collision. We have to find the speed of the two masses immediately after collision.

We know that the linear momentum of the system consisting of the two blocks is conserved during the collision. Thus, if $v_{1}^{-}$ and $v_{2}^{-}$ are the speeds of the two masses just before the collision and $v_{1}^{+}$ and $v_{2}^{+}$ are their respective speeds immediately after the collision, then we have

m_{1}v_{1}^{+}+m_{2}v_{2}^{+}=m_{1}v_{1}^{-}+m_{2}v_{2}^{-}

(10.62)

Since there is no loss of energy in the collision, the energy of the system is conserved. Thus, $E^{-}=E^{+}$ , or

\frac{1}{2}m_{1}(v_{1}^{+})^{2}+\frac{1}{2}m_{2}(v_{2}^{+})^{2}=\frac{1}{2}m_{% 1}(v_{1}^{-})^{2}+\frac{1}{2}m_{2}(v_{2}^{-})^{2}.

(10.63)

Thus, we have two equations (eqn. (10.62) and eqn. (10.63)) in two unknowns, $v_{1}^{+}$ and $v_{2}^{+}$ , and hence we can solve for them. It is now only a question in algebra. From eqn. (10.63), we have

$\displaystyle m_{1}\left[(v_{1}^{+})^{2}-(v_{1}^{-})^{2}\right]$	$\displaystyle=$	$\displaystyle m_{2}\left[(v_{2}^{-})^{2}-(v_{2}^{+})^{2}\right]$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m_{1}(v_{1}^{+}+v_{1}^{-% })(v_{1}^{+}-v_{1}^{-})$	$\displaystyle=$	$\displaystyle m_{2}(v_{2}^{-}+v_{2}^{+})(v_{2}^{-}-v_{2}^{+})$	(10.64)

$\displaystyle v_{1}^{+}+v_{1}^{-}$	$\displaystyle=$	$\displaystyle v_{2}^{+}+v_{2}^{-}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v_{1}^{+}-v_{2}^{+}$	$\displaystyle=$	$\displaystyle v_{2}^{-}-v_{1}^{-}.$	(10.65)

Multiplying the above equation by $m_{1}$ and subtracting from eqn. (10.62), we get

	$\displaystyle(m_{1}+m_{2})v_{2}^{+}$	$\displaystyle=$	$\displaystyle 2m_{1}v_{1}^{-}+v_{2}^{-}(m_{2}-m_{1})$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v_{2}^{+}$	$\displaystyle=$	$\displaystyle\frac{2m_{1}}{m_{1}+m_{2}}v_{1}^{-}+\frac{m_{2}-m_{1}}{m_{1}+m_{2% }}v_{2}^{-}.$

Now substituting the given values, $m_{1}=2\,\mathrm{kg},\,m_{2}=10\,\mathrm{kg},\,v_{1}^{-}=0.5\,\mathrm{m}/% \mathrm{s}$ and $v_{2}^{-}=0.2\,\mathrm{m}/\mathrm{s}$ above, we get $v_{2}^{+}=0.3\,\mathrm{m}/\mathrm{s}$ . Further, substituting the values of $v_{2}^{+}$ in eqn. (10.65), we get $v_{1}^{+}=0$ , i.e., the first mass comes to a halt!

Answer: $v_{1}^{+}=0$ and $v_{2}^{+}=0.3\,\mathrm{m}/\mathrm{s}$

Comments: : Note that rather than using energy conservation equation directly as we did above, we could have used the given energy information t o set $e=1$ (perfectly elastic collision) in eqn. (10.61) to get $v_{2}^{+}-v_{1}^{+}=-v_{2}^{-}+v_{1}^{-}$ (rather than deriving it as we did above). We can then solve this equation along with eqn. (10.62) to solve for $v_{1}^{+}$ and $v_{2}^{+}$ .

SAMPLE 10.20

Filename:pfigure4-4-rp16

Figure 10.74:

Estimating peak force in a collision: A metal ball of mass $m=0.5\,\mathrm{kg}$ strikes a stationary surface S₁ with velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=10\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}}$ and rebounds with velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=-9\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}}$ . In a different experiment the same ball strikes another stationary surface S₂ with the same initial velocity and has the same rebound velocity. The contact times during the two experiments were different: $0.1\,\mathrm{s}$ and $0.001\,\mathrm{s}$ respectively. Assuming that the collisional force between the ball and the two surfaces can be modeled as $F(t)=\frac{F_{0}}{2}(1+\cos\frac{2\pi t}{T})$ (see fig. 10.74) where $-T/2\leq t\leq T/2$ and $T$ is the contact time, find the peak force $F_{0}$ in each case.

Solution Let the collisional impulse acting on the ball be ${\accentset{\rightharpoonup}{\boldsymbol{P}}}$ (see fig. 10.75) given by

{\accentset{\rightharpoonup}{\boldsymbol{P}}}=\int_{-T/2}^{T/2}{\accentset{% \rightharpoonup}{\boldsymbol{F}}}(t)dt.

Filename:pfigure4-4-rp14

Figure 10.75: Free-body diagram of the ball during collision with collisional impulse

{\accentset{\rightharpoonup}{\boldsymbol{P}}}

acting on it.

From impulse-momentum relationship, we have

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

Since in the case of each surface, $\Delta{\accentset{\rightharpoonup}{\boldsymbol{v}}}$ is the same ( ${\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}-{\accentset{\rightharpoonup}% {\boldsymbol{v}}}^{-}=-19\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}$ ), the change in linear momentum $\Delta L=m\Delta{\accentset{\rightharpoonup}{\boldsymbol{v}}}$ is also the same. Hence, the impulse acting on the ball in each case has to be the same. Now, let ${\accentset{\rightharpoonup}{\boldsymbol{P}}}_{1}$ and ${\accentset{\rightharpoonup}{\boldsymbol{P}}}_{2}$ be the impulses acting on the ball during the collision with surface S₁ and S₂ respectively. Then,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{P}}}_{1}$	$\displaystyle=$	$\displaystyle-\int_{-T_{1}/2}^{T_{1}/2}{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{1}(t)dt\,\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-\int_{-T_{1}/2}^{T_{1}/2}\frac{(F_{0})_{1}}{2}\left(1+\cos\frac{% 2\pi t}{T_{1}}\right)dt\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-\frac{(F_{0})_{1}T_{1}}{2}\,\hat{\boldsymbol{\imath}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{P}}}_{2}$	$\displaystyle=$	$\displaystyle-\frac{(F_{0})_{2}T_{2}}{2}\,\hat{\boldsymbol{\imath}}.$

Now, setting ${\accentset{\rightharpoonup}{\boldsymbol{P}}}_{1}=\Delta L$ , we get

$\displaystyle-\frac{(F_{0})_{1}T_{1}}{2}\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle-m\Delta v\,\hat{\boldsymbol{\imath}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}(F_{0})_{1}$	$\displaystyle=$	$\displaystyle\frac{2m\Delta v}{T_{1}}$
	$\displaystyle=$	$\displaystyle\frac{2\cdot 0.5\,\mathrm{kg}\cdot 19\,\mathrm{m}/\mathrm{s}}{0.1% \,\mathrm{s}}$
	$\displaystyle=$	$\displaystyle 190\,{\mbox{N}}.$

$\displaystyle(F_{0})_{2}$	$\displaystyle=$	$\displaystyle\frac{2m\Delta v}{T_{2}}$
	$\displaystyle=$	$\displaystyle\frac{2\cdot 0.5\,\mathrm{kg}\cdot 19\,\mathrm{m}/\mathrm{s}}{0.0% 01\,\mathrm{s}}$
	$\displaystyle=$	$\displaystyle 19000\,{\mbox{N}}.$

Clearly, the peak force is inversely proportional to the collision time. In fact, it is easy to see that for the given model of the impulsive force, the peak force $F_{0}=\frac{2m\Delta v}{T}$ . Thus if the change in momentum is constant, then the peak force varies as $1/T$ .

Answer: $(F_{0})_{1}=190\,{\mbox{N}}\quad{\rm and}\quad(F_{0})_{2}=19000\,{\mbox{N}}$

SAMPLE 10.21

Filename:pfigure4-4-rp15

Figure 10.76:

A two-ball multiple collision experiment: A tennis ball of approximate mass $m_{1}=60\,\hbox{gm}$ and a basketball of approximate mass $m_{2}=600\,\hbox{gm}$ are used in a fun collision experiment. The two balls are held in air, one on top of the other with a tiny gap between them, at a height $h$ from the ground as shown in the figure. The two balls are released simultaneously from rest. The coefficient of restitution between the tennis ball and the basketball is $e_{1}=0.6$ and that between the basketball and the floor is $e_{2}=0.9$ . Assume that the collision between the two balls takes place immediately after the basketball rebounds from the floor. Find the height of the tennis ball flight in terms of $h$ as a result of the collision.

Solution

We need to track two separate collisions here — one between the basketball and the floor, and second, between the tennis ball and the basketball. We can find the relevant vertical velocities before and after the collisions to determine the velocity of the tennis ball’s flight which we can use to find the height of the flight. We will assume upward velocities to be positive.

Collision-1: Just before the basketball hits the floor, let its vertical velocity be $v_{2}^{-}$ and let the tennis ball’s speed at the same instant be $v_{1}^{-}$ . Since both balls undergo free fall from height $h$ before attaining these speeds, we have $v_{1}^{-}=v_{2}^{-}=-\sqrt{2gh}$ .

Filename:tfigure4-spherical-rotaxis

Figure 10.77: Collision of the basketball with the ground.

Now let $v_{2}^{+}$ be the speed of the basketball immediately after the collision with the ground (see fig. 10.77). Then, $v_{2}^{+}=-e_{2}v_{2}^{-}=e_{2}\sqrt{2gh}$ .

Collision-2: We assume that the second collision, the collision between the tennis ball and the basketball, takes place immediately after the first collision. Hence, the velocity of the tennis ball just before the collision with the basketball can be assumed to be $v_{1}^{-}=-\sqrt{2gh}$ . The second collision is shown in fig. 10.78. The after impact velocities of the two balls are $v_{1}^{+}$ and $v_{2}^{++}$ . Now, from collision law, we have

Filename:tfigure5-7

Figure 10.78: Collision of the tennis ball with the basketball.

\displaystyle v_{1}^{+}-v_{2}^{++}=-e_{1}(v_{1}^{-}-v_{2}^{+})=-e_{1}(-\sqrt{2% gh}-e_{2}\sqrt{2gh})=\sqrt{2gh}e_{1}(1+e_{2}).

(10.66)

The conservation of linear momentum for the two-ball system gives

$\displaystyle m_{1}v_{1}^{+}+m_{2}v_{2}^{++}$	$\displaystyle=$	$\displaystyle m_{1}v_{1}^{-}+m_{2}v_{2}^{+})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v_{1}^{+}+\frac{m_{2}}{m% _{1}}v_{2}^{++}$	$\displaystyle=$	$\displaystyle v_{1}^{-}+\frac{m_{2}}{m_{1}}v_{2}^{+}$

$\displaystyle v_{1}^{+}+Mv_{2}^{++}$	$\displaystyle=$	$\displaystyle\sqrt{2gh}(Me_{2}-1).$	(10.67)

Now solving eqn. (10.66) and eqn. (10.67) simultaneously, we get

v_{1}^{+}=\frac{\sqrt{2gh}}{1+M}\left[M(e_{1}+e_{2}+e_{1}e_{2})-1\right].

This is the velocity with which the tennis ball takes off on its vertical flight. Let the height of this flight be $h_{f}$ . Then, from constant acceleration motion formula, we get $(v_{1}^{+})^{2}=2gh_{f}$ , or $h_{f}=(v_{1}^{+})^{2}/2g$ . Thus, from the derived expression for $v_{1}^{+}$ above, we get

h_{f}=\frac{h}{(1+M)^{2}}\left[M(e_{1}+e_{2}+e_{1}e_{2})-1\right]^{2}.

Substituting $M=m_{2}/m_{1}=10$ , $e_{1}=0.6$ , and $e_{2}=0.9$ above, we get $h_{f}=3.11h$ .

Thus the tennis ball flies off to three times its original height.

Answer: $h_{f}=3.11h$

Note: From the expression obtained for $v_{1}^{+}$ , we see that if $M$ is very large then $v_{1}^{+}=\sqrt{2gh}(e_{1}+e_{2}+e_{1}e_{2})$ and $h_{f}=(e_{1}+e_{2}+e_{1}e_{2})^{2}h$ .

Problems for 10.5 1D Collisions

Preparatory Problems

10.5.1 Before a collision two particles, $m_{A}=1\,\mathrm{kg}$ and $m_{B}=2\,\mathrm{kg}$ , have velocities of $v_{A}^{-}=10\,\mathrm{m}/\mathrm{s}$ and $v_{B}^{-}=5\,\mathrm{m}/\mathrm{s}$ . After the collision the velocity of A is $v_{A}^{+}=6\,\mathrm{m}/\mathrm{s}$ .

(a)

What is the momentum of A before the collision?
(b)

What is the momentum of B before the collision?
(c)

What is the system momentum before the collision?
(d)

What is the momentum of A after the collision?
(e)

What is the system momentum after the collision?
(f)

What is the momentum of B after the collision?
(g)

What is the impulse that A applies to B during the collision?
(h)

What is the impulse that B applies to A during the collision?
(i)

What is the kinetic energy of the system before the collision?
(j)

What is the kinetic energy of the system after the collision?
(k)

What is the coefficient of restitution?

10.5.2 A ball is dropped from a height of $h_{0}=10\,\mathrm{m}$ onto a hard stationary surface. After the first bounce, it reaches a height of $h_{1}=6.4\,\mathrm{m}$ . What is the coefficient of restitution between the ball and ground? What is the height of the second bounce, $h_{2}$ ?

Filename:tfigure5-gen-rigid-body

Figure 10.79:

10.5.3 A 20 gram, 500 $\,\mathrm{m}/\mathrm{s}$ bullet embeds in an initially-stationary 50 kg rigid block. What is the coefficient of restitution? What is the velocity of the block after this collision? Answer: Note: the mass ratio is so high that there is no real loss of accuracy if you take the post-collision mass $m_{1}+m_{2}$ as 50 kg (the 0.004% error is surely much smaller than the inaccuracy in any of the other numbers in the problem).

10.5.4 A ball of mass $m$ is dropped vertically from a height $h$ . The only force acting on the ball in its flight is gravity. The ball strikes the ground with speed $v^{-}$ and after collision it rebounds vertically with reduced speed $v^{+}$ directly proportional to the incoming speed, $v^{+}=ev^{-}$ , where $0<e<1$ . What is the maximum height the ball reaches after one bounce, in terms of $h$ , $e$ , and $g$ . Answer: $h_{max}=e^{2}h$ .

10.5.5 Set up the following equations in matrix form and solve for $v_{A}$ and $v_{B}$ , if $v_{0}=2.6\,\mathrm{m}/\mathrm{s},\;e=0.8,\;m_{A}=2\,\mathrm{kg}$ , and $m_{B}=500\mbox{g}$ :

	$\displaystyle m_{A}v_{0}=m_{A}v_{A}+m_{B}v_{B}$
	$\displaystyle-ev_{0}=v_{A}-v_{B}.$

More-Involved Problems

10.5.6 Before a collision two particles, $m_{A}=7\,\mathrm{kg}$ and $m_{B}=9\,\mathrm{kg}$ , have velocities of $v_{A}^{-}=6\,\mathrm{m}/\mathrm{s}$ and $v_{B}^{-}=2\,\mathrm{m}/\mathrm{s}$ . The coefficient of restitution is $e=.5$ . Find the impulse of mass A on mass B and the velocities of the two masses after the collision.

10.5.7 Two frictionless masses $m_{A}=2\,\mathrm{kg}$ and $m_{B}=5\,\mathrm{kg}$ travel on straight collinear paths with speeds $V_{A}=5\,\mathrm{m}/\mathrm{s}$ and $V_{B}=1\,\mathrm{m}/\mathrm{s}$ , respectively. The masses collide since $V_{A}>V_{B}$ . Find the amount of energy lost in the collision. The coefficient of restitution is $e=0.5$ .

Filename:tfigure5-term1-a

Figure 10.80:

10.5.8 A ball of mass $m$ is dropped from height $h$ onto the solid hard ground where its coefficient of restitution is $e<1$ . The gravitational constant is $g$ .

(a)

How many times does the ball bounce before it comes to a stop?
(b)

How long does it take from first release until it comes to a stop?
(c)

What is the total distance the ball travels before coming to a stop (add up and down distances)?

10.5.9 A bullet of mass $m$ with initial speed $v_{0}$ is fired in the horizontal direction through block A of mass $m_{A}$ and becomes embedded in block B of mass $m_{B}$ . Each block is suspended by thin wires. The bullet causes A and B to start moving with speed of $v_{A}$ and $v_{B}$ respectively. Determine

(a)

the initial speed $v_{0}$ of the bullet in terms of $v_{A}$ and $v_{B}$ , Answer: $v_{0}=\frac{1}{m}(mv_{B}+m_{B}v_{B}+m_{A}v_{A})$ .
(b)

the velocity of the bullet as it travels from block A to block B,and Answer: $v_{1}=\frac{(m+m_{B})}{m}v_{B}$ .
(c)

the energy loss due to friction as the bullet (1) moves through block A and (2) penetrates block B. Answer: (1) $E_{loss}=\frac{1}{2}m\left[v_{0}^{2}-\left(\frac{m+m_{B}}{m}\right)^{2}v_{B}^{% 2}\right]-\frac{1}{2}m_{A}v_{A}^{2}$ .

(2) $E_{loss}=\frac{1}{2}\left[\frac{(m+m_{B})^{2}}{m}v_{B}^{2}-(m+m_{B})v_{B}^{2}\right]$ .

Filename:tfigure5-term1-b

Figure 10.81:

10.5.10 A basketball with mass $m_{b}$ is dropped from height $h$ onto the hard solid ground on which it has coefficient of restitution $e_{b}$ . Just on top of the basketball, falling with it and then bouncing against it after the basketball hits the ground, is a small rubber ball with mass $m_{r}$ that has a coefficient of restitution $e_{r}$ with the basketball.

(a)

In terms of some or all of $m_{b}$ , $m_{r}$ , $h$ , $g$ , $e_{b}$ and $e_{r}$ how high does the rubber ball bounce (measure height relative to the collision point)?
(b)

Assuming the coefficients of restitution are less than or equal to one, for given $h$ , what mass and restitution parameters maximize the height of the bounce of the rubber ball and what is that height?

10.5.11 Show that it is necessary that $|e|\leq 1$ for the net kinetic energy (sum of the two kinetic energies of the colliding particles) to not increase.

10.5.12 According to the problem above, unless energy is created in the collision (as in an explosion), $-1\leq e\leq 1$ . Show that, for given masses and given initial velocities, that the loss of system kinetic energy is maximized by $e=0$ .

$\displaystyle\makebox[54.2025pt][l]{If we let}\dot{x}$	$\displaystyle=$	$\displaystyle y,$
$\displaystyle\makebox[54.2025pt][l]{then}\dot{y}$	$\displaystyle=$	$\displaystyle\ddot{x}=-c\dot{x}-kx+a_{0}\sin\omega t$
	$\displaystyle=$	$\displaystyle-cy-kx+a_{0}\sin\omega t$

$\displaystyle\makebox[36.135pt][l]{If}\dot{x}$	$\displaystyle=$	$\displaystyle y$
$\displaystyle\makebox[36.135pt][l]{then}\dot{y}$	$\displaystyle=$	$\displaystyle\ddot{x}=c\dot{x}^{2}-kx^{3}=-cy^{2}-kx^{3}$
$\displaystyle\hbox to36.135pt{or\hfil}\left\{\begin{array}[]{c}\dot{x}\\ \dot{y}\end{array}\right\}$	$\displaystyle=$	$\displaystyle\left\{\begin{array}[]{c}y\\ -cy^{2}-kx^{3}\end{array}\right\}.$	(10.19)

$\displaystyle-2kx=ma_{x}$	$\displaystyle=$	$\displaystyle m\ddot{x}$	(10.39)
$\displaystyle\mbox{or}\quad m\ddot{x}+2kx$	$\displaystyle=$	$\displaystyle 0$	(10.40)
$\displaystyle\mbox{or}\quad\ddot{x}+\frac{2k}{m}x$	$\displaystyle=$	$\displaystyle 0.$	(10.41)

$\displaystyle\lambda^{2}$	$\displaystyle=$	$\displaystyle\frac{2k}{m}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\lambda$	$\displaystyle=$	$\displaystyle\sqrt{\frac{2k}{m}}$
	$\displaystyle=$	$\displaystyle\sqrt{\frac{2\mathord{\cdot}(1\,\mbox{kN/m})}{20\,\mathrm{kg}}}$
	$\displaystyle=$	$\displaystyle 10\,\mathrm{rad}/\mathrm{s}.$

	$\displaystyle x(0)$	$\displaystyle=$	$\displaystyle A\cos(-\phi)=A\cos\phi$		(10.43)
	$\displaystyle\dot{x}(0)$	$\displaystyle=$	$\displaystyle-A\lambda\sin(-\phi)=A\lambda\sin\phi$		(10.44)

$\displaystyle A^{2}\cos^{2}\phi+\frac{A^{2}\lambda^{2}\sin^{2}\phi}{\lambda^{2}}$	$\displaystyle=$	$\displaystyle A^{2}=x^{2}(0)+\frac{\dot{x}^{2}(0)}{\lambda^{2}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad A$	$\displaystyle=$	$\displaystyle\sqrt{(25\,{\rm mm})^{2}+\frac{(160\,{\rm mm}/\,\mathrm{s})^{2}}{% (4\pi\,\mathrm{rad}/\mathrm{s})^{2}}}$
	$\displaystyle=$	$\displaystyle 28.06\,{\rm mm}.$

$\displaystyle\phi$	$\displaystyle=$	$\displaystyle\cos^{-1}\frac{x(0)}{A}$
	$\displaystyle=$	$\displaystyle\cos^{-1}\frac{25\,{\rm mm}}{28.06\,{\rm mm}}$
	$\displaystyle=$	$\displaystyle 0.471\,{\rm rad}\approx 27^{{}^{\circ}}.$

$\displaystyle\dot{y}_{1}$	$\displaystyle=$	$\displaystyle v_{1}$
$\displaystyle\dot{v}_{1}$	$\displaystyle=$	$\displaystyle-\frac{k}{m_{1}}(y_{1}-y_{2})+\frac{k\ell_{0}}{m_{1}}-g$
$\displaystyle\dot{y}_{2}$	$\displaystyle=$	$\displaystyle v_{2}$
$\displaystyle\dot{v}_{2}$	$\displaystyle=$	$\displaystyle\frac{k}{m_{2}}(y_{1}-y_{2})-\frac{k\ell_{0}}{m_{2}}-g.$

	$\displaystyle y_{1/{\rm cm}}$	$\displaystyle=$	$\displaystyle y_{1}-y_{\rm cm}=\frac{m_{2}}{m_{1}+m_{2}}(y_{1}-y_{2})$
	$\displaystyle y_{2/{\rm cm}}$	$\displaystyle=$	$\displaystyle y_{2}-y_{\rm cm}=-\frac{m_{1}}{m_{1}+m_{2}}(y_{1}-y_{2}).$

	$\displaystyle(m_{1}+m_{2})\ddot{y}_{\rm cm}$	$\displaystyle=$	$\displaystyle-m_{1}g-m_{2}g$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{y}_{\rm cm}$	$\displaystyle=$	$\displaystyle-g.$

Chapter 10 Dynamics in 1D

Download this chapter by section

The organization of this chapter

10.1 Force and motion in 1D

Position, velocity, and acceleration in one dimension

Differential equations

More differential equations.

The equations of dynamics

Linear momentum balance

Some special cases in 1D mechanics

The simplest ODEs

What do the terms in F=m⁢a mean?

Problems for 10.1 Force and motion in 1D

Preparatory Problems

More-Involved Problems

10.2 Energy methods in 1D

Power, work, kinetic energy and potential energy

Power and work

Power and rate-of-change of kinetic energy

Work is change of kinetic energy

Conservation of energy

Is energy balance a principle or a calculation trick?

Terrestrial locomotion: Trains, cars, bicycles and animals

Summary

Particle equations are exact for non-particle systems

More general energy balance equations

Problems for 10.2 Energy methods in 1D

Preparatory Problems

More-Involved Problems

10.3 A mass and spring: the harmonic oscillator

The harmonic oscillator

Solution of the harmonic oscillator differential equation

Checking the solution in detail

Angular frequency, period, and frequency

Initial conditions determine the constants A and B

Energy

Conservation of energy

Energy oscillations.

Using energy to derive the oscillator equation

Dealing with offsets

Numerical solution

Method I: separation of variables

Method II: complex variables

Problems for 10.3 Simple Harmonic Oscillator

Preparatory Problems

More-Involved Problems

10.4 Coupled motions in 1D

Relative motion in one dimension

Center of mass

Problems for 10.4 Coupled motion in 1D

Preparatory Problems

More-Involved Problems

10.5 Collisions in 1D

What happens during a collision

Extra assumptions are needed

One dimensional collisions

Sticking collisions

More general 1-D collisions

Elastic collisions

The coefficient of restitution

Problems for 10.5 1D Collisions

Preparatory Problems

More-Involved Problems

What do the terms in $F=ma$ mean?

Initial conditions determine the constants $A$ and $B$