Physics 16 Final

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 the potential energy. m (x0 (t)2 +y 0 (t)2 ) T = 2 

V = k −d + 

2 k −d +

q

x(t)2 + (−` + y(t))2

L = − k −d + 2 k −d +

q

2

2

2



+k −d +

(−` + x(t)) + y(t)

q

 

2

2



+ 2 k −d +

q

2

2

2

2

x(t) + (−` + y(t))



−2 k −d +

2

2

(` + x(t)) + y(t)

q

x(t)2 + (` + y(t))2

2 !

(−` + x(t)) + y(t)

q



−k −d + q

2

+

2

q

(` + x(t))2 + y(t)2

2

2

x(t) + (` + y(t))

2

+

N X ∂L i=1

∂ q˙i

q˙i − L

where the qi ’s are the variables in the Langrangian. 0 2 0 2 E = (mx0 (t)2 − my 0 (t)2 ) − m 2 (x (t) + y (t) ) + V E =T +V b. If d = 0, then L simplifies to 1 L = (x0 (t)2 + y 0 (t)2 ) − 6k(x(t)2 + y(t)2 + `2 ) 2 This has symmetry under the infinitesimal transformation x → x + y, y → y − x to first order in . Thus, by Noether’s theorem, the following quantity is conserved: 1

−

m (x0 (t)2 +y 0 (t)2 ) 2

To find the energy, E, we can use E=

2

X ∂L i

Ki (q) = m(xy ˙ − yx) ˙

∂ q˙i

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

d m¨ x = 2k (` − x) 1 − p (` − x)2 + y 2 !

"

d −4k x 1 − p (` − y)2 + x2

!

d +x 1− p (` + y)2 + x2

"

d m¨ y = −4k (` − y) 1 − p (` − y)2 + x2 "

d 2k y 1 − p (` − x)2 + y 2

!

d − (` + x) 1 − p (` + x)2 + y 2

!

!#

!#

+ F0 cos(ωt)

d − (` + y) 1 − p (` + y)2 + x2

d +y 1− p (` + x)2 + y 2

!#

!#

+ F0 cos(ωt)

Since we are given that the oscillations about the equilibrium position are small, we can Taylor expand both x and y about zero, using the following identities: p

x2

p

x2

x x x xy ≈ ≈ + 2 2 1−y ` ` + (` − y) x x y xy ≈ ≈ + 2 2 1+y ` ` + (` + y) p

`−x ≈1 + (` − x)2

p

`+x ≈1 + (` + x)2

y2 y2

Now, rewriting our equations of motion: 



m¨ x ≈ (2k) (−2x + (−d)(0))−4k 2x − d

x xy x xy + 2 + − 2 ` ` ` `

= −8kx + F0 cos(ωt)

2



+F0 cos(ωt)

y xy y xy m¨ y ≈ (4k) (−2y + (−d)(0)) − 2k 2y − d + 2 + − 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: mx00 (t) = −8kx(t) + F0 cos(ωt) my 00 (t) = −10ky(t) + F0 cos(ωt) These are inhomogeneous differential equations. Morin gives the solution to this in eqn. 3.36. The solution is: x(t) =

F cos(ωt − φ) + A cos(ω 0 t + θ0 ) R

q

k d , F = Fm0 , R = ω 02 − ω 2 , sin φ = 2γω where ω 0 = m R → φ = 0, and A and ω0 are set by the initial conditions. Thus, the final solution is:

q

x(t) = A cos(

q

y(t) = A0 cos(

8k mt

+ θ0 ) +

F0 mω 2 −8k

10k m t

+ θ00 ) +

F0 mω 2 −10k

cos(ωt) cos(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 in the 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 that if the driving force frequency is equal to the resonance frequency, then the motion will go to infinity, and R = 0 which is indeterminate and so our equation does not work. 2. a. We can do this by converting the cylindrical coordinates into cartesian coordinates, and using the arc length formula for cartesian coordinates. A smooth curve with parametric equations for x, y, and z coordinates in terms of a variable θ that is traversed exactly once as θ ranges from θ1 to θ2 has length Z

s

θ2

dθ θ1

dx 2 dy 2 dz 2 + + dθ dθ dθ 3

In this case, x(θ) = r(θ) cos(θ), y(θ) = r(θ) sin(θ), and z(θ) = z(r(θ)). Thus, x0 (θ) = r0 cos(θ) cos(θ) − r(θ) sin(θ), y 0 (θ) = r0 (θ) sin(θ) + r(θ) cos(θ), and z 0 (θ) = z 0 (r(θ))r0 (θ). Squaring these and plugging them in yields a length: Z

θ2

q

dθ x0 (θ)2 + y 0 (θ)2 + z 0 (θ)2

θ1

Z

θ2

q

dθ (r0 cos(θ) cos(θ) − r(θ) sin(θ))2 + (r0 (θ) sin(θ) + r(θ) cos(θ))2 + (z 0 (r(θ))r0 (θ))2

θ1

Z

θ2

q

dθ r0 (θ)2 + r(θ)2 + z 0 (r(θ))2 r0 (θ)2

θ1

Z

θ2

q

dθ r(θ)2 + (1 + z 0 (r(θ))2 ) r0 (θ)2

θ1

b. The Euler-Lagrange equation for this problem is ∂L d ∂L = dθ ∂ r˙ ∂r p

where L = r(θ)2 + r0 (θ)2 + r0 (θ)2 z 0 (r(θ))2 . Thus, we must compute two partial derivatives and one full derivative (which we skip). ∂L 2 r(θ) + 2 r0 (θ)2 z 0 (r(θ)) z 00 (r(θ)) = q ∂r(θ) 2 r(θ)2 + r0 (θ)2 + r0 (θ)2 z 0 (r(θ))2 2 r0 (θ) + 2 r0 (θ) z 0 (r(θ))2 ∂L q = ∂r0 (θ) 2 r(θ)2 + r0 (θ)2 + r0 (θ)2 z 0 (r(θ))2 Thus, 2 d √ 2 r 0 (θ)+2 r 0 (θ) z 0 (r(θ)) dθ 2 r(θ)2 +r0 (θ)2 +r 0 (θ)2 z 0 (r(θ))2

0

r (θ) = 2√r(θ)+2 2 0 2

2

z 0 (r(θ)) z 00 (r(θ))

r(θ) +r (θ)2 +r 0 (θ)2 z 0 (r(θ))2

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

q

r(θ)2 + r0 (θ)2 + k 2 r0 (θ)2

∂L r(θ) =q ∂r r(θ)2 + r0 (θ)2 + k 2 r0 (θ)2 4

∂L r0 (θ) + k 2 r0 (θ) = q ∂ r˙ r(θ)2 + r0 (θ)2 + k 2 r0 (θ)2 Now we can plug in the possible solution r(θ) =

c cos( √

θ 1+k2

−φ)

, and check

it. In this case, we have: r0 (θ) =

θ θ − φ) tan( √1+k − φ) c sec( √1+k 2 2 √ 1 + k2

so, plugging into our expression for

∂L ∂r

above and simplifying,

θ ∂L = cos( √ − φ) ∂r 1 + k2 Similarly, after some algebraic simplification, θ ∂L p = 1 + k 2 sin( √ − φ) ∂ r˙ 1 + k2 But now the full derivative simplifies to: θ d ∂L = cos( √ − φ) dθ ∂ r˙ 1 + k2 Thus, equation.

d ∂L dθ ∂ r˙

=

∂L ∂r ,

and we have found a solution to the Euler-Lagrange

3. By conservation of momentum, the two π 0 mesons must have equal mk and opposite velocity. Thus, mk = 2mπ γ, and γ = 2m . The energy π mk of this meson is Eπ0 = mπ γ = 2 . The momentum of the meson is pπ 0 =

mk 2 vγ

=

mk 2

r

1−

4m2π . m2k

To maximize the energy of the photons

coming from the decay of the π 0 meson, they must have velocities parallel to the velocity of the π 0 meson - any perpendicular velocity is ‘wasted’ (we discussed this in lecture). Let us call the momenta of the two photons (or their energies) p and p0 , with the p energy the one we want to maximize. By conservation of energy and momentum, p + p0 = pπ0 and kpk + kp0 k = Eπ0 . Since the p and p0 vectors are parallel, and pπ0 < Eπ0 , p and p0 must have opposite signs. Let us choose p0 to be the negative one. Then, p + p0 = pπ0 and p − p0 = Eπ0 . Solving these simultaneously yields 5

p=

Eπ0 +pπ0 2

=

mk 4



r

1−

1+

4m2π m2k



=

mk +

√

m2k −4m2π 4

.

This expression obviously has the right units of mass (energy). The energy is directly related to mk , as we would expect; it is also inversely related to mπ . This can be seen by considering the limiting cases. If mπ = 0, then Eπ0 = ppi0 , and the maximizing condition occurs when p0 = 0 and p is the whole m2k . If mπ = mk /2 (the highest for which the expression is physical), then the π 0 mesons have zero velocity, and p can be only m4k . 4. (Thanks to Edward) First let us calculate v(t) in the lab frame without using ζ. The acceleration in the ship frame is g, so by the transformation of acceleration equation, the acceleration in the lab frame is given by v˙ = γg3 = g(1 − v 2 )3 . Solving the differential equation yields dv (1−v 2 )3

v=

= gdt →

± √ gt 2 2 , 1+g t

v 1−v 2

+ c = gt.

Rearranging to solve for v, we get

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 get m0 dE that E = 1−v . We can relate all of this to ζ by noting that dp dt = dt = ζ(t)w(1 − v). The (1 − v) term can be explained in the following way. The beam is made up of photons, and the energy and momentum of a photon is given by E = hf . To someone on the spaceship, the beam of light will be red-shifted, and thus it will appear to have less energy in it than would appear to someone on earth. To calculate the apparent frequency, we use a standard, non-relativistic doppler effect (because we aren’t actually changing frames), which is just f 0 = f (c−v) = f (1−v). The energy is also transformed c by a factor of (1 − v). Thus we can calculate ζ(t) by: ζ(t) =

dE 1 dt w(1 − v)



=



d  m0  gt √ dt 1 − 2

 

1+g t2



=

gm0 g t + 1+

g 2 t2

p

−gt

1 + g 2 t2

p

1+



g 2 t2

6

1 w(1 − v) 1 

w 1−

√ gt 1+g 2 t2





=

gm0 g t + 

w −g t + = w

√

p

1 + g 2 t2

p

1 + g 2 t2



2

gm0 1+g 2 t2 −gt

3

Note that as t → ∞, ζ(t) → ∞, and since in the real world, ζ can’t exceed 1, any real spaceship will be able to keep up this motion for only so long before the laser beam is not powerful enough to maintain a ship frame acceleration of g. We can check the non-relativistic limit by seeing that ζ(0) = mg w , which is exactly what we would expect, because then wζ(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 starting position of particle 1. At the moment of the collision, particle 1 has energy E1 = 2m + 2T x and particle 2 has energy E2 = m + T (` − x). Using 2 2 2 the identity the momenta of the two particles: p E − P = m , we can findp 2 2 kp1 k = (2m + 2T x) − (2m) , kp2 k = (m + T (` − x))2 − m2 . But since R P = F dt, and the force on particle 1 acts for the same time as the force on particle 2 but is twice the magnitude, kp1 k = 2kp2 k. Expanding this, (2m + 2T x)2 − (2m)2 = (2m + 2T (` − x))2 − (2m)2 → 4T x = 2T ` → x = 2` . This is a sensible result, because the force on particle 1 is twice the force on particle 2, and the mass of particle 1 is twice the mass of particle 2d, so they should always have the same acceleration. Now we can solve for the energy and momentum of the new particle. E4 = E1 + E2 = 3m + 2T2 ` . The momenta √ of the two particles are in opposite directions, so kp4 k = kp1 k − ` T (4 m+` T )

q

q

kp2 k = . m4 = E42 − p24 = 94 (2m + T `)2 − 14 T `(4m + T `) = 2 √ 9m2 + 8mT ` + 2T 2 `2 . We note that each term in the radical does in fact have the correct units of mass (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 approaches 3m, the non-relativistic limit, which makes sense because if the tension is low the masses will be travelling slower. 6. a. Consider the axes ˆi = (−1, 0, 1), ˆj = (1, 0, 1), kˆ = (0, 1, 0) . We R will show that these are principal axes. ik = 0, because the following corners cancel each other out, having equal masses and ij coordinates of opR posite sign and equal magnitude: (2,5), (1,6), (7,4), (3,8). jk = 0 by the

7

same cancelations. Each corner has either zeroRi coordinate or zero j coordiR R 2 ij = 0. i + j 2 = 8m1 a2 + 8m2 a2 , j 2 + k 2 = 12m1 a2 + 4m2 a2 , nate, so R 2 2 2 i + k = 12m2 a + 4m1 a2 . Thus, Ii = a2 (12m1 + 4m2 ) Ij = a2 (4m1 + 12m2 ) Ik = a2 (8m1 + 8m2 ) b. We will show that principal axes are q       ˆi = − √1 , 0, √1 , ˆj = √1 , − 2 , √1 , kˆ = √1 , √1 , √1 . This can be done 3 2 2 6 6 3 3 3 by calculating Iij , Iik , and Ijk and showing them all to be zero, of course, but the following argument is a little more illuminating. A quick check shows that these are orthonormal, so at least they are a possibililty. Now, if the masses at corners 1 and 8 were present, then these would have to be principal axes. Both cornersR 1 and 8 have j coordinate zero and i coordinate zero. R R Thus, jk, ij, and ik are unchanged by removing the masses, q and must √ be zero. The new coordinates of the masses are: 2 : a 2, − 23 , √13 , 3 : q q q   √ q   √    a 0, 2 23 , √13 , 4 : a 2, 23 , − √13 , 5 : a − 2, − 23 , √13 , 6 : a 0, −2 23 , − √13 , 7 :  √ q  R R a − 2, 23 , − √13 . Simple computation of Ii = j 2 + k 2 , Ij = i2 + k 2 , and Ik = i2 + j 2 (by Mathematica ) gives Ii = 10ma2 , I2 = 10ma2 , I3 = 16ma2 . 7a. The total mass of the rod is ρπr2 + 2`κ, and the impulse is P x ˆ, so Px ˆ ˆ, the speed of the center of mass after the impact is vcm = ρπr2 +2`κ . The x yˆ, and zˆ axes as given in the diagram are principal axes for the top (at the moment of impact), so we can compute their principal moments (using the derived results in Morin’s table): R

1 2 Ix = Iy = I = πρr4 + κ`3 4 3 1 Iz = πρr4 2 z. The initial angular momentum of the top was L0 = I3 ωˆ z = 21 πρr4 ωˆ The angular momentum contributed by the hit was r × P = P `ˆ y . Thus, the total angular momentum (which is a constant after the impact, since there ~ = P `ˆ z . After the impact, the object can of are no torques) is L y + 12 πρr4 ωˆ course be described by the equations of a symmetric free top (in the center ~ at frequency ω = L . of mass frame), so the rod will just precess around L I

8

7b. Since the rod is travelling in the x ˆ direction, this motion will not affect the position relative to the y = 0 plane, and the next time the rod is in the y = 0 plane will just be after one full rotation of the top, which will occur at a time 



2π 14 πρr4 + 23 κ`3 2π 2πI 8κ`3 + 3πρr4 = 2 2 1 2 2 8 2 = p 2 2 = ~ ω P ` + 4π ρ r ω 6 4` P + ρ2 ω 2 π 2 r8 kLk The top is moving at a speed vcm found above, so at this time, the position of the upper tip is: √ 8κ` 6

3 +3πρr 4

4`2 P 2 +ρ2 ω 2 π 2 r8

P x ˆ ρπr 2 +2`κ

+ `ˆ z

8. Let kˆ be the symmetry axis of the top, and let ˆj point perpendicular to kˆ from zˆ to the top, and ˆi be orthogonal to them. Now since torque is ~r × F~ and the centrifugal force is −m~ ω × (~ ω × r), we can evaluate the torque due to the centrifugal force by evaluating this integral over the body of the top: −

Z

~r × (~ ω × (~ ω × ~r))dm

But ω ~ = ωp x ˆ, and we can express zˆ as zˆ = kˆ cos θ − ˆj sin θ. We can let rˆ (for each point of mass on the top) be rˆ = (a, b, c). Then, ω ~ × ~r = ωp zˆ × ~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)) = ωp2 (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, and then by the terms in each of the sums.R Since ˆi, ˆj, kˆRare (instantaneous) prinR ciple axes of the object, we know thatR (ab)dm = (bc)dm = (ac)dm = 0. R 2 R (a + b2 )dm = I3 , b2 + c2 = I, a2 + c2 = I. Thus, since θ is a constant, most of the terms in the integrand go to zero, and we are left with R (ωp2 cos θ sin θ(b2 − c2 ), 0, 0)dm, which is (ωp2 cos θ sin θ(I − I3 ), 0, 0) . We could use a similar method to calculate the torque to the coriolis force directly, 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 the torques 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 more fictitious forces), so the torque due the coriolis force is: (mgl sin θ − ωp2 cos θ sin θ(I − I3 ), 0, 0) . 9. Using just f = ma and gravity, we can derive the equations of motion of the system: m1 r~¨1 =

m1 m2 (r~2 − r~1 ) m1 m3 (r~3 − r~1 ) + kr~2 − r~1 k3 kr~3 − r~1 k3

m2 m3 (r~3 − r~2 ) m2 m1 (r~1 − r~2 ) m2 r~¨2 = + kr~3 − r~2 k3 kr~1 − r~2 k3 m3 r~¨3 =

m3 m1 (r~1 − r~3 ) m3 m2 (r~2 − r~3 ) + kr~1 − r~3 k3 kr~2 − r~3 k3

Let us choose our origin at the center of mass so that m1 r~1 + m2 r~2 + m3 r~3 = 0. Then, letting ` designate the side of our equilateral triangle, we can simplify the equations of motion: Gm1 m1 r~¨1 = 3 (−r~1 (m2 + m3 ) + m2 r~2 + m3 r~3 ) ` G r~¨1 = − 3 (m1 + m2 + m3 ) r~1 ` And similarly, G r~¨2 = − 3 (m1 + m2 + m3 ) r~2 ` G r~¨3 = − 3 (m1 + m2 + m3 ) r~3 ` Now what we hope that is that there exists a ω such that r~˙j = ω ~ × r~j for j ∈ {1, 2, 3}. Taking the derivative of both sides, we see that this is equivalent to the existence of some ω such that r~¨j = ω ~ × r~˙j + ω ~˙ × r~j = ω×(ω× 2 r~j ) = −~ rj k~ ω k (because ω ~˙ = 0). Note that this is just the standard equation for centripetal acceleration. Thus, if we let ω = |~ ω| =

q

G (m1 `3

+ m2 + m3 ) ,

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

10