PHYSICS 311: Classical Mechanics – Midterm Soluion Key

2015

1. [15 points] A particle of mass, m, is in orbit in a central potential givenby:

U = − κr2

where κ is a positive constant.

(a) What is the force on the particle? Is the force attractive or repulsive?Sol.

The force, in general, can be computed by:

~F = −∇U

In this particular case, we only have a radial component since deriva-tives in θ and φ vanish.

But:

~F = −∂U∂r

r̂ = −2κr3r̂

As this is in the negative r̂ direction, the force is attractive.

(b) Please compute ∇× ~F . Is this force conservative?Sol.

∇× ~F = 0

I can say this with certainty because the force was generated by agradient and

∇× (∇U) = 0

according to Stoke’s theorem. Also according to Stoke’s theorem:∮~F · d~l = 0

if the curl vanishes. This is the definition of a conservative force.

(c) The particle is in a circular orbit at radius, r. What is the velocityfor a circular orbit?

Sol.

For a circular orbit, r̈ = ṙ = 0. Thus:

ar = −rφ̇2

Thus:

−2κr3

= −mrφ̇2

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or

φ̇ =

√2κ

mr2

and thus:

vφ = rφ̇ =

√2κ

mr2

2. [15 points] Let’s do a bit of math.

Consider a vector:~v = 2x3î− 4xĵ

and a complex number:A = 2− 3i

Compute

(a) ∇ · ~vSol.

Noting:

∂vx∂x

= 6x2

∂vy∂y

= 0

∂vz∂z

= 0

we get:

∇ · ~v = 6x2

(b) ∇× ~vSol.

By a similar argument as above, I note that the only non-zero “off-diagonal” derivative is:

∂vy∂x

= −4

so

∇× ~v = −4k̂

(c) AA∗

Sol.

First, we have:

A∗ = 2 + 3i

soAA∗ = (2− 3i)(2 + 3i) = 4− 6i+ 6i+ 9 = 13

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3. [20 points] Let’s consider a slight variation of a classic. A mass, m, isreleased from rest, on an inclined plane, as shown:

Instead of contact friction, the mass is moving through a viscous oil witha linear coefficient of friction, b.

(a) What is the net force on the mass the instant it is released?

Force is a vector, so be sure to define your directional vectors clearly.

Sol.

I won’t draw a free-body diagram, but I’ll simply note that x refersto the direction down the plane. Since the particle starts at rest, thefriction doesn’t matter (at first). The net force is simply:

~F = mg sin θî

But since this problem is manifestly 1-d (only up or down the planematters), from now on, I’ll exclude the î.

(b) Write down a differential equation describing the velocity as a func-tion of time.

Sol.

Obviously, after a while, friction kicks in. We get:

~F = mg sin θ − bv

or

v̇ +b

mv = g sin θ

(c) Assume the plane is arbitrarily long. What is the terminal velocityof the mass?

Sol.

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At terminal velocity, ẍ = 0. So:

vT =mg

bsin θ

whch simply solves the relation above.

(d) What is v(t)? (This is a toughie, so think carefully).

Sol.

We now need to solve the inhomogeneous differential equaiton above.The trick, as you’ll recall, is solving the homogeneous equation:

v̇ +b

mv = 0

which is solved for:v = v0e

− bm t

To solve the inhomogeneous equation, we assume the form:

v(t) = v0e− bm t + C

where we know two things:

1) v(0) = 0

which gives:

v(t) = v0

(1− e− bm t

)and

2) v(t =∞) = vTwhich gives:

v(t) =mg

b

(1− e− bm t

)(e) E.C. What is x(t)? Show that in the limit of short times (quadratic

in time) that this reduces to the non-friction result.

Sol.

This is fairly straightforward. We simply integrate:

x(t) =mg

bt+ g

m2

b2e−

bm t + C

undet the limit of x(0) = 0. This yields:

x(t) = vT t+ vT τ[e−t/τ − 1

]where I’ve defined:

τ =m

b

4

as the rough amount of time for friction to kick in.

At small times:

e−t/τ ' 1− tτ

+1

2

t2

τ2

and thus the position becomes:

x(t) ' vT t+ vT τ[− tτ

+1

2

t2

τ2

]=

1

2vTt2

τ

=1

2g sin θt2

as required.

4. [35 points] Let’s do a single system, 2 ways.

Consider a double pendulum with two equal masses, one a distance L andone a distance L/2 from the pendulum pivot, as shown:

The pendulum operates in an ordinary earth gravitational field of strength,g with no friction. Do not assume small oscillations unless you are askedto.

(a) Approach 1: As a rotational system.

What is the moment of inertia of the pendulum?

Sol.

5

As there are two masses, a distance, L and L/2, the moment of inertiais:

I =5

4mL2

(b) What is the torque on the pendulum?

Sol.

There is a force on each mass of mg, producing a torque of:

Γ1 = −mgL sin θ

Γ2 = −1

2mgL sin θ

So

Γ = L̇ =3

2mgL sin θ

(c) Based on your results above, what is φ̈ as a function of φ?

Sol.

We have:

Iφ̈ = −32mgL sin θ

5

4mL2φ̈ = −3

2mgL sin θ

so

φ̈ = −65

g

Lsin θ

(d) Approach 2: Based on Energy.

What is the kinetic energy of the pendulum as a function of φ̇?

Sol.

The kinetic energies combine:

T =1

2mL2φ̇2 +

1

2m

(L

2

)2φ̇2 =

5

8mL2φ̇2

(e) What is the potential energy of the pendulum as a function of φ?

Sol.

Again, we simply combine the results for a single particle:

U =3

2mgL(1− cosφ)

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(f) Write the Euler-Lagrange equation.

Sol.

Start with the full Lagrangian:

L = 58mL2φ̇2 − 3

2mgL(1− cosφ)

so:∂L∂φ̇

=5

4mL2φ̇

and∂L∂φ

= −32mgL sinφ

so5

4mL2φ̈ = −3

2mgL sinφ

or

φ̈ = −65

g

Lsinφ

Exactly as we found using torques.

(g) For small oscillations, what is the frequency of oscillations?

Given the nature of the problem (and the fact that I’m spoon-feedingyou sanity checks), it is especially important that you show yourwork.

Sol.

In the smal angle limit, sinφ ' φ, so:

φ̈ = −65

g

Lφ

and thus:

ω =

√6g

5L

5. [15 points] A particle of mass, m, moves around an inverted cone wherethe half opening angle is α as shown. There is no friction.

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(a) What is the Lagrangian of the system? You should express both thisand subsequent answers in terms of the orbital angle, φ, and r, thephysical distance to the tip of the cone.

Hint: You may need to do a bit of trig to figure out the tangentialcomponent of the velocity.

Sol.

I’m going to define a coordinate:

ρ = r sinα

Sovφ = ρφ̇ = r sinαφ̇

and thus, the total kinetic energy is:

T =1

2m(ṙ2 + r2 sin2 αφ̇2

)and similarly:

U = mgr cosα

so

L = 12m(ṙ2 + r2 sin2 αφ̇2

)−mgr cosα

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

Sol.

There are two.

For r:

mr̈ = mr sin2 αφ̇2 −mg cosα

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For φ:

md

dt

(r2 sin2 αφ̇

)= 0

(c) What velocity, if any, produces a stable orbit around the cone?

Sol.

We’re basically looking for ṙ = ṙ = 0, which is satisfied for:

φ̇2 =g

r

cosα

sin2 α

Noting:vφ = r sinαφ̇

we get:

vφ =√gr cosα

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