PHYSICS 311: Classical Mechanics – Midterm Exam Solution Key (2019)

1. [20 points] Short Answers (5 points each)

(a) Consider the complex number A = 4− 3i. Please compute AA∗.

Sol.

First, note A∗ = 4 + 3i, so:

AA∗ = 16 + 9 = 25

(b) As best you are able (in a sentence or a few words and an equation),please describe a “conservative force.”

Sol.

A conservative force has a number of important properties:

• The work done by the force around a closed loop is zero:∮~F ·

d~r = 0

• It is generated by a potential: ~F = −∇U• It is curl-free: ∇× ~F = 0.

Any of the three (even descriptively) are acceptable, as all are iden-tical mathematically. A simple statement that conservative forcesdon’t include friction will get you partial credit.

(c) Using scaling analysis (and recognizing that air resistance behavesquadratically), how much faster will a 200 pound parachuter fall thana 100 pound parachuter, assuming their parachutes are the same size?

E.C. (2 points) How much larger (in diameter) will the larger jumper’sparachute need to be so that the two fall at the same race?

Sol.

First, note that:mg = cv2

can be used for terminal velocity, so:

v =

√mg

c

vbig =√

2vsmall

As for the extra credit, c ∝ A, so Abig = 2Asmall produces the sameterminal velocity, or:

dbig =√

2dsmall

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(d) Consider a potential expressed in spherical coordinates:

U = mgr cos θ

What is the force arising from this potential? Be sure to give youranswer in spherical coordinates.

Sol.

Note:dU

dr= mg cosφ

anddU

dθ= −mgr sin θ

so

~F = −∇U= −mg cos θr +mg sin θθ

= mg(− cos θr + sin θθ)

I should note that the direction vector is simply a spherical coordinatedescription of −k.

2. [15 points] Consider the vector field:

~v =1

2x2i+ xyj

and a scalar field:f = x2y

(a) Compute the divergence of the vector field: ∇ · ~vSol.

In Cartesian coordinates:

∇ · ~v =∂vx∂x

+∂vy∂y

+∂vz∂z

= 2x

(b) Compute the gradient of the scalar field: ∇fSol.

∇f =∂f

∂xi+

∂f

∂yj

= 2xyi+ x2j

2

(c) Now compute the curl of the result from the previous part: ∇×(∇f)

Sol.

0

We proved this in general, since the curl of a gradient is always zero,but to be sure:

x : (∂uy∂z − ∂uz∂y) = (0− 0) = 0

y : (∂uz∂x− ∂ux∂z) = (0− 0) = 0

z : (∂ux∂y − ∂uy∂x) = 2x− 2x = 0

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3. [15 points] Consider a piece of wood in roughly the shape of an oar. Ithas a length, L, which is which has a density per unit Length, of:

λ(x) = λ0x

L

(a) What is the mass of the rod?

Sol.

As a reminder, the 1-d version of the integral is

m =

∫ L

0

λ(x)dx

= λ0

∫ L/2

0

x

Ldx

= λ0

(L2

2

)1

L

=Lλ0

2

(b) What is the center of mass?

Sol.

Again

xCM =1

M

∫xλ(x)dx

=2

Lλ0

λ0L

x3

3

∣∣∣∣L0

=2

3L

(c) What is the moment of inertia if rotated around the left end?

Please give your answer in terms of M and L.

Sol.

4

Same basic idea.

I =

∫ L

0

x2λ(x)dx

=λ0L

L4

4

=λ0L

3

4

=ML2

2

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4. [15 points] Consider the central force:

~F = −krr

which acts like a bungee cord pulling a “planet” in to the center of a solarsystem.

At some instant, the planet is a distance, R from the center, and movingwith a purely tangential velocity, v0.

(a) In terms of k, m, and R, what angular velocity ω0 = φ is required tomaintain the planet in a circular orbit?

Hint: A circular orbit requires r to be constant.

Sol.

As a reminder, the radial acceleration is given by:

ar = r − rφ2 = −krm

So if r = 0

φ2 =k

mor

ω0 =

√k

m

Surprise!

(b) What is the angular momentum, `, of the planet in terms of k, m,and R?

Sol.

The speed of the orbit is Rφ, so:

` = mR2ω0 = R2√km

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(c) The planet is then given a kick pointed purely in the radial direction(maintaining the same angular momentum, in case you missed it).Express φ as a function of r and `.

Hint: There is definitely a conservation law at work.

Sol.

It’s simply:

φ =`

mr2

(d) E.C. (3 points) Following this, write, but do not solve, a differentialequation describing the evolution of r, incorporating only `, r, andm.

Sol.

Very importantly:` = mr2φ

is conserved.

But our radial equation yields:

r = rφ2 − kr

m

Plugging in our result from the previous part:

r =`2

m2r3− kr

m

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5. [20 points] Consider two blocks of equal mass m connected by a springof constant k and confined to move in one dimension. The entire systemmoves without friction. At equilibrium, the spring has a length, `.

(a) Write down the Lagrangian of this system as a function of x1 and x2and their time derivatives. Assume x2 > x1.

Sol.

The Lagrangian is simply the sum of two free Lagrangians plus acoupling term:

L =1

2m2x21 +

1

2m2x22 −

1

2k(x2 − x1 − `)2

(b) Write the Euler-Lagrange equations for this system.

For x1:

mx1 = k(x2 − x1 − `)

and, for x2:

mx2 = −k(x2 − x1 − `)

Note that the sum of these two expressions guarantees overall con-servation of linear momentum.

(c) Make the change of variables:

∆ ≡ x2 − x1 − `

X ≡ 1

2(x1 + x2)

Write the Euler-Lagrange equations in these new variables.

Sol.

First note that:∆ = x2 − x1

which simply gets rid of the `. So:

x2 = X +1

2∆

and

x1 = X − 1

2∆

8

so

x21 + x22 = 2X2 +1

2∆2

and thus:

L = mX2 +1

4m∆2 − 1

2k∆2

and the Euler-Lagrange equations:

X = 0

∆ = −2k

m∆

(d) What are the frequencies of oscillation if the two masses are out ofphase with one another?

Sol.

Reading off from the previous equation for ∆:

ω =

√2k

m

Note that this is the exact solution that you’d get by computing thefrequency for the reduced mass, µ = m

2 .

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6. [15 points] Okay. This is the toughest problem, even though it looksinnocuous. Be careful, and make sure you show your work. I know youcan do it.

I throw a marble of mass, m, downward through a fluid with a linearviscous coefficient, b very, very fast.

You don’t need to compute the terminal velocity, vT , explicitly (but Iwould suggest doing so for your own sake). I initially throw the marble atv0 = 8vT

(a) Write, but do not solve, a differential equation describing the velocity,v, of the marble.

As a suggestion, you should define v as positive if it is going down-ward.

Sol.

There is a simple force:

F = mg − bv

which is canceled if:vT ≡

mg

b

But writing the first version as a differential equation:

v = g − b

mx

or, cleaning up a little bit:

v = g

(1− v

vT

)(b) I care about you all deeply, so I will tell you that the solution to the

motion has the form:v(t) = A+Bert

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Solve explicitly for A,B, and r, and determine the motion of themarble at all times.

Hint: As a strong suggestion, you should make careful note of termswhich vary over times, and which ones don’t.

Sol.

First, note that:v = rBert

So plugging into the cleaned up version above:

rBert = g

(1− A

vT− Bert

vT

)Since the term on the left has no time dependance, we can say:

1− A

vT= 0 ⇒ A = vT

and similarly:

rB = −gBvT

so:

r = − g

vT

Finally, we can look at the initial conditions, t = 0:

v0 = A+B

= vT +B

soB = v0 − vT

Just to summarize:

v = (v0 − vT )e−gt/vT + vT

(c) Sketch the velocity of the marble. How long will it take before themarble is moving at only v = 2vT ?

Sol.

First, for convenience, define:

τ =vTg

so the exponent is just:e−t/τ

We’re solving for:2vT = 7vT e

−t/τ + vT

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or

e−t/τ =1

7

Thus:

t = ln(7)τ = ln(7)vTg' 1.95

vTg

I don’t need exact labels, just a trendline.

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