Physics 16 Final Exam Take-Home

September 24, 2006

1a.

L = T −V , where L is the Langragian, T is the kinetic energy, and V is thepotential energy.

T =m (x′(t)2+y′(t)2)

2

V = k

(−d +

√(−` + x(t))2 + y(t)2

)2

+k

(−d +

√(` + x(t))2 + y(t)2

)2

+

2 k

(−d +

√x(t)2 + (−` + y(t))2

)2

+ 2 k

(−d +

√x(t)2 + (` + y(t))2

)2

L = −(

k

(−d +

√(−` + x(t))2 + y(t)2

)2)−k

(−d +

√(` + x(t))2 + y(t)2

)2

−

2 k

(−d +

√x(t)2 + (−` + y(t))2

)2

−2 k

(−d +

√x(t)2 + (` + y(t))2

)2

+m (x′(t)2+y′(t)2)

2

To find the energy, E, we can use

E =N∑

i=1

∂L

∂qiqi − L

where the qi’s are the variables in the Langrangian.E = (mx′(t)2 −my′(t)2)− m

2 (x′(t)2 + y′(t)2) + VE = T + V

b. If d = 0, then L simplifies to

L =12(x′(t)2 + y′(t)2)− 6k(x(t)2 + y(t)2 + `2)

This has symmetry under the infinitesimal transformation x → x +εy, y → y− εx to first order in ε. Thus, by Noether’s theorem, the followingquantity is conserved:

1

∑i

∂L

∂qiKi(q) = m(xy − yx)

c. This is easiest if we just write down the equations of motion using f =ma and then make some approximations using Taylor expansions. Summingthe force contributions from each spring, we get:

mx = 2k

[(`− x)

(1− d√

(`− x)2 + y2

)− (` + x)

(1− d√

(` + x)2 + y2

)]

−4k

[x

(1− d√

(`− y)2 + x2

)+ x

(1− d√

(` + y)2 + x2

)]+ F0 cos(ωt)

my = −4k

[(`− y)

(1− d√

(`− y)2 + x2

)− (` + y)

(1− d√

(` + y)2 + x2

)]

2k

[y

(1− d√

(`− x)2 + y2

)+ y

(1− d√

(` + x)2 + y2

)]+ F0 cos(ωt)

Since we are given that the oscillations about the equilibrium positionare small, we can Taylor expand both x and y about zero, using the followingidentities:

x√x2 + (`− y)2

≈ x

1− y≈ x

`+

xy

`2

x√x2 + (` + y)2

≈ x

1 + y≈ y

`+

xy

`2

`− x√y2 + (`− x)2

≈ 1

` + x√y2 + (` + x)2

≈ 1

Now, rewriting our equations of motion:

mx ≈ (2k) (−2x + (−d)(0))−4k

(2x− d

(x

`+

xy

`2+

x

`− xy

`2

))+F0 cos(ωt)

= −8kx + F0 cos(ωt)

2

my ≈ (4k) (−2y + (−d)(0))−2k

(2y − d

(y

`+

xy

`2+

y

`− xy

`2

))+F0 cos(ωt)

= −10kx + F0 cos(ωt)

Thus we just have driven (and not damped) simple harmonic motion.The specific equations of motion we want to solve are:

mx′′(t) = −8kx(t) + F0 cos(ωt)

my′′(t) = −10ky(t) + F0 cos(ωt)

These are inhomogeneous differential equations. Morin gives the solution tothis in eqn. 3.36. The solution is:

x(t) =F

Rcos(ωt− φ) + A cos(ω′t + θ0)

where ω′ =√

km , F = F0

m , R = ω′2 − ω2, sinφ = 2γωdR → φ = 0, and A and

ω0 are set by the initial conditions. Thus, the final solution is:

x(t) = A cos(√

8km t + θ0) + F0

mω2−8kcos(ωt)

y(t) = A′ cos(√

10km t + θ′0) + F0

mω2−10kcos(ωt)

Note that these do both give the correct units of distance. If F0 is small,and the motion is SHM, then we note that the frequency of the motion inthe y direction is higher than in the x direction, which is to be expected,since the springs with higher constants are in the y direction. Note thatif the driving force frequency is equal to the resonance frequency, then themotion will go to infinity, and R = 0 which is indeterminate and so ourequation does not work.

2. a. We can do this by converting the cylindrical coordinates into carte-sian coordinates, and using the arc length formula for cartesian coordinates.A smooth curve with parametric equations for x, y, and z coordinates interms of a variable θ that is traversed exactly once as θ ranges from θ1 toθ2 has length

∫ θ2

θ1

dθ

√dx

dθ

2

+dy

dθ

2

+dz

dθ

2

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In this case, x(θ) = r(θ) cos(θ), y(θ) = r(θ) sin(θ), and z(θ) = z(r(θ)).Thus, x′(θ) = r′ cos(θ) cos(θ)− r(θ) sin(θ), y′(θ) = r′(θ) sin(θ) + r(θ) cos(θ),and z′(θ) = z′(r(θ))r′(θ). Squaring these and plugging them in yields alength: ∫ θ2

θ1

dθ√

x′(θ)2 + y′(θ)2 + z′(θ)2

∫ θ2

θ1

dθ√

(r′ cos(θ) cos(θ)− r(θ) sin(θ))2 + (r′(θ) sin(θ) + r(θ) cos(θ))2 + (z′(r(θ))r′(θ))2

∫ θ2

θ1

dθ√

r′(θ)2 + r(θ)2 + z′(r(θ))2r′(θ)2

∫ θ2

θ1

dθ√

r(θ)2 + (1 + z′(r(θ))2) r′(θ)2

b. The Euler-Lagrange equation for this problem is

d

dθ

∂L

∂r=

∂L

∂r

where L =√

r(θ)2 + r′(θ)2 + r′(θ)2z′(r(θ))2. Thus, we must computetwo partial derivatives and one full derivative (which we skip).

∂L

∂r(θ)=

2 r(θ) + 2 r′(θ)2 z′(r(θ)) z′′(r(θ))

2√

r(θ)2 + r′(θ)2 + r′(θ)2 z′(r(θ))2

∂L

∂r′(θ)=

2 r′(θ) + 2 r′(θ) z′(r(θ))2

2√

r(θ)2 + r′(θ)2 + r′(θ)2 z′(r(θ))2

Thus,ddθ

2 r′(θ)+2 r′(θ) z′(r(θ))2

2√

r(θ)2+r′(θ)2+r′(θ)2 z′(r(θ))2= 2 r(θ)+2 r′(θ)2 z′(r(θ)) z′′(r(θ))

2√

r(θ)2+r′(θ)2+r′(θ)2 z′(r(θ))2

c. We could just plug this into the above formula and obtain zero, but thealgebra is complicated enough that is simpler to start over. If z(r) = −kr,then

L =√

r(θ)2 + r′(θ)2 + k2 r′(θ)2

∂L

∂r=

r(θ)√r(θ)2 + r′(θ)2 + k2 r′(θ)2

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∂L

∂r=

r′(θ) + k2 r′(θ)√r(θ)2 + r′(θ)2 + k2 r′(θ)2

Now we can plug in the possible solution r(θ) = ccos( θ√

1+k2−φ)

, and check

it. In this case, we have:

r′(θ) =c sec( θ√

1+k2− φ) tan( θ√

1+k2− φ)

√1 + k2

so, plugging into our expression for ∂L∂r above and simplifying,

∂L

∂r= cos(

θ√1 + k2

− φ)

Similarly, after some algebraic simplification,

∂L

∂r=√

1 + k2 sin(θ√

1 + k2− φ)

But now the full derivative simplifies to:

d

dθ

∂L

∂r= cos(

θ√1 + k2

− φ)

Thus, ddθ

∂L∂r = ∂L

∂r , and we have found a solution to the Euler-Lagrangeequation.

3. By conservation of momentum, the two π0 mesons must have equaland opposite velocity. Thus, mk = 2mπγ, and γ = mk

2mπ. The energy

of this meson is Eπ0 = mπγ = mk2 . The momentum of the meson is

pπ0 = mk2 vγ = mk

2

√1− 4m2

π

m2k

. To maximize the energy of the photons

coming from the decay of the π0 meson, they must have velocities paral-lel to the velocity of the π0 meson - any perpendicular velocity is ‘wasted’(we discussed this in lecture). Let us call the momenta of the two pho-tons (or their energies) p and p′, with the p energy the one we want tomaximize. By conservation of energy and momentum, p + p′ = pπ0 and‖p‖+ ‖p′‖ = Eπ0 . Since the p and p′ vectors are parallel, and pπ0 < Eπ0 , pand p′ must have opposite signs. Let us choose p′ to be the negative one.Then, p + p′ = pπ0 and p − p′ = Eπ0 . Solving these simultaneously yields

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p = Eπ0+pπ0

2 = mk4

(1 +

√1− 4m2

π

m2k

)= mk+

√m2

k−4m2

π

4 .

This expression obviously has the right units of mass (energy). The en-ergy is directly related to mk, as we would expect; it is also inversely relatedto mπ. This can be seen by considering the limiting cases. If mπ = 0, thenEπ0 = ppi0 , and the maximizing condition occurs when p′ = 0 and p is thewhole mk

2 . If mπ = mk/2 (the highest for which the expression is physical),then the π0 mesons have zero velocity, and p can be only mk

4 .

4. (Thanks to Edward) First let us calculate v(t) in the lab frame with-out using ζ. The acceleration in the ship frame is g, so by the trans-formation of acceleration equation, the acceleration in the lab frame isgiven by v = g

γ3 = g(1 − v2)3. Solving the differential equation yieldsdv

(1−v2)3= gdt → v

1−v2 + c = gt. Rearranging to solve for v, we get

v = ± gt√1+g2t2

, and we choose the positive solution so that v > 0.

Since the beam is made of light, in an infinitesimal time period, dp = dt.But p0 = 0, and E0 = m0, so we know that E = p + m0 at all times.We also know that p = Ev, and solving these two simultaneously, we getthat E = m0

1−v . We can relate all of this to ζ by noting that dpdt = dE

dt =ζ(t)w(1− v). The (1− v) term can be explained in the following way. Thebeam is made up of photons, and the energy and momentum of a photonis given by E = hf . To someone on the spaceship, the beam of light willbe red-shifted, and thus it will appear to have less energy in it than wouldappear to someone on earth. To calculate the apparent frequency, we use astandard, non-relativistic doppler effect (because we aren’t actually changingframes), which is just f ′ = f(c−v)

c = f(1−v). The energy is also transformedby a factor of (1− v). Thus we can calculate ζ(t) by:

ζ(t) =dE

dt

1w(1− v)

=d

dt

m0

1− g t√1+g2 t2

1w(1− v)

=gm0

(g t +

√1 + g2 t2

)1 + g2 t2 − g t

√1 + g2 t2

1

w

(1− g t√

1+g2 t2

)

6

=gm0

(g t +

√1 + g2 t2

)w(−g t +

√1 + g2 t2

)2

= gm0

w

(√1+g2t2−gt

)3

Note that as t → ∞, ζ(t) → ∞, and since in the real world, ζ can’texceed 1, any real spaceship will be able to keep up this motion for onlyso long before the laser beam is not powerful enough to maintain a shipframe acceleration of g. We can check the non-relativistic limit by see-ing that ζ(0) = mg

w , which is exactly what we would expect, because thenwζ(0) = mg = dp

dt = F , and mg is the naive non-relativistic force.

5. Let particles 1 and 2 collide at a distance x to the right of the start-ing position of particle 1. At the moment of the collision, particle 1 hasenergy E1 = 2m + 2Tx and particle 2 has energy E2 = m + T (`− x). Usingthe identity E2 − P 2 = m2, we can find the momenta of the two particles:‖p1‖ =

√(2m + 2Tx)2 − (2m)2, ‖p2‖ =

√(m + T (`− x))2 −m2. But since

P =∫

Fdt, and the force on particle 1 acts for the same time as the forceon particle 2 but is twice the magnitude, ‖p1‖ = 2‖p2‖. Expanding this,(2m + 2Tx)2− (2m)2 = (2m + 2T (`− x))2− (2m)2 → 4Tx = 2T` → x = `

2 .This is a sensible result, because the force on particle 1 is twice the forceon particle 2, and the mass of particle 1 is twice the mass of particle 2d, sothey should always have the same acceleration. Now we can solve for theenergy and momentum of the new particle. E4 = E1 + E2 = 3m + 2T`

2 . Themomenta of the two particles are in opposite directions, so ‖p4‖ = ‖p1‖ −‖p2‖ =

√` T (4 m+` T )

2 . m4 =√

E24 − p2

4 =√

94(2m + T`)2 − 1

4T`(4m + T`) =√

9m2 + 8mT` + 2T 2`2 .We note that each term in the radical does in fact have the correct units ofmass (in units where distance = time). The result increases with each of m,T , and `, as we would expect. Note that if T → 0, this answer approaches3m, the non-relativistic limit, which makes sense because if the tension islow the masses will be travelling slower.

6. a. Consider the axes i = (−1, 0, 1), j = (1, 0, 1), k = (0, 1, 0) . Wewill show that these are principal axes.

∫ik = 0, because the following

corners cancel each other out, having equal masses and ij coordinates of op-posite sign and equal magnitude: (2,5), (1,6), (7,4), (3,8).

∫jk = 0 by the

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same cancelations. Each corner has either zero i coordinate or zero j coordi-nate, so

∫ij = 0.

∫i2 + j2 = 8m1a

2 + 8m2a2,∫

j2 + k2 = 12m1a2 + 4m2a

2,∫i2 + k2 = 12m2a

2 + 4m1a2. Thus,

Ii = a2(12m1 + 4m2)

Ij = a2(4m1 + 12m2)

Ik = a2(8m1 + 8m2)

b. We will show that principal axes are

i =(− 1√

2, 0, 1√

2

), j =

(1√6,−√

23 , 1√

6

), k =

(1√3, 1√

3, 1√

3

). This can be done

by calculating Iij , Iik, and Ijk and showing them all to be zero, of course, butthe following argument is a little more illuminating. A quick check showsthat these are orthonormal, so at least they are a possibililty. Now, if themasses at corners 1 and 8 were present, then these would have to be princi-pal axes. Both corners 1 and 8 have j coordinate zero and i coordinate zero.Thus,

∫jk,

∫ij, and

∫ik are unchanged by removing the masses, and must

be zero. The new coordinates of the masses are: 2 : a(√

2,−√

23 , 1√

3

), 3 :

a(0, 2√

23 , 1√

3

), 4 : a

(√2,√

23 ,− 1√

3

), 5 : a

(−√

2,−√

23 , 1√

3

), 6 : a

(0,−2

√23 ,− 1√

3

), 7 :

a(−√

2,√

23 ,− 1√

3

). Simple computation of Ii =

∫j2 + k2, Ij =

∫i2 + k2,

and Ik =∫

i2 + j2 (by Mathematica ) gives Ii = 10ma2, I2 = 10ma2, I3 = 16ma2 .7a. The total mass of the rod is ρπr2 + 2`κ, and the impulse is Px, so

the speed of the center of mass after the impact is vcm = P xρπr2+2`κ

. The x,y, and z axes as given in the diagram are principal axes for the top (at themoment of impact), so we can compute their principal moments (using thederived results in Morin’s table):

Ix = Iy = I =14πρr4 +

23κ`3

Iz =12πρr4

The initial angular momentum of the top was L0 = I3ωz = 12πρr4ωz.

The angular momentum contributed by the hit was r×P = P`y. Thus, thetotal angular momentum (which is a constant after the impact, since thereare no torques) is ~L = P`y + 1

2πρr4ωz. After the impact, the object can ofcourse be described by the equations of a symmetric free top (in the centerof mass frame), so the rod will just precess around ~L at frequency ω = L

I .

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7b. Since the rod is travelling in the x direction, this motion will notaffect the position relative to the y = 0 plane, and the next time the rod isin the y = 0 plane will just be after one full rotation of the top, which willoccur at a time

2π

ω=

2πI

‖~L‖=

2π(

14πρr4 + 2

3κ`3)

P 2`2 + 14π2ρ2r8ω2

=8κ`3 + 3πρr4

6√

4`2P 2 + ρ2ω2π2r8

The top is moving at a speed vcm found above, so at this time, theposition of the upper tip is:

8κ`3+3πρr4

6√

4`2P 2+ρ2ω2π2r8

Pρπr2+2`κ

x + `z

8. Let k be the symmetry axis of the top, and let j point perpendicularto k from z to the top, and i be orthogonal to them. Now since torque is~r× ~F and the centrifugal force is −m~ω× (~ω×r), we can evaluate the torquedue to the centrifugal force by evaluating this integral over the body of thetop:

−∫

~r × (~ω × (~ω × ~r))dm

But ~ω = ωpx, and we can express z as z = k cos θ− j sin θ. We can let r(for each point of mass on the top) be r = (a, b, c). Then,

~ω × ~r = ωpz × ~r = ωp(b cos θ − c sin θ, a cos θ, a sin θ)

Similarly, the whole integrand is given by:

~r × (~ω × (~ω × ~r)) =

ω2p(bc cos2 θ+b2 cos θ sin θ−bc sin2 θ−c2 cos θ sin θ,−ab sin θ cos θ−ac+ac sin2 θ,

ab + ac sin θ cos θ − ab cos2 θ)

We can integrate this by splitting it up by components of the vector, andthen by the terms in each of the sums. Since i, j, k are (instantaneous) prin-ciple axes of the object, we know that

∫(ab)dm =

∫(bc)dm =

∫(ac)dm = 0.∫

(a2 + b2)dm = I3,∫

b2 + c2 = I,∫

a2 + c2 = I. Thus, since θ is a con-stant, most of the terms in the integrand go to zero, and we are left with∫

(ω2p cos θ sin θ(b2 − c2), 0, 0)dm, which is (ω2

p cos θ sin θ(I − I3), 0, 0) . Wecould use a similar method to calculate the torque to the coriolis force di-rectly, but the following is easier. Since in this frame I and ~ω are fixed,

9

there can be no azimuthal force (since dω/dt = 0), and dL/dt = 0. Thus thetorques must sum to zero. The torque due the translation force (gravity)is (mgl sin θ, 0, 0), and there can be no others (because there are no morefictitious forces), so the torque due the coriolis force is:(mgl sin θ − ω2

p cos θ sin θ(I − I3), 0, 0) .

9. Using just f = ma and gravity, we can derive the equations of motionof the system:

m1 ~r1 =m1m2(~r2 − ~r1)‖~r2 − ~r1‖3

+m1m3(~r3 − ~r1)‖~r3 − ~r1‖3

m2 ~r2 =m2m3(~r3 − ~r2)‖~r3 − ~r2‖3

+m2m1(~r1 − ~r2)‖~r1 − ~r2‖3

m3 ~r3 =m3m1(~r1 − ~r3)‖~r1 − ~r3‖3

+m3m2(~r2 − ~r3)‖~r2 − ~r3‖3

Let us choose our origin at the center of mass so that m1 ~r1 + m2 ~r2 +m3 ~r3 = 0. Then, letting ` designate the side of our equilateral triangle, wecan simplify the equations of motion:

m1 ~r1 =Gm1

`3(−~r1(m2 + m3) + m2 ~r2 + m3 ~r3)

~r1 = −G

`3(m1 + m2 + m3) ~r1

And similarly,

~r2 = −G

`3(m1 + m2 + m3) ~r2

~r3 = −G

`3(m1 + m2 + m3) ~r3

Now what we hope that is that there exists a ω such that ~rj = ~ω × ~rj

for j ∈ {1, 2, 3}. Taking the derivative of both sides, we see that this isequivalent to the existence of some ω such that ~rj = ~ω× ~rj+~ω× ~rj = ω×(ω×~rj) = −~rj‖~ω‖2 (because ~ω = 0). Note that this is just the standard equation

for centripetal acceleration. Thus, if we let ω = |~ω| =√

G`3

(m1 + m2 + m3) ,

then we see that we have indeed met the condition, and the masses orbit inan equilateral triangle.

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