PC235 Winter 2013 Classical Mechanics
Midterm Exam Solutions #1 (5 points) The plot below shows the position vs. time for a weakly-damped oscillator. What is the approximat approximate e Q-factor Q-factor of this oscillator? oscillator? Explain Explain how you came up with this value. 1
0.5 ] m [ n o 0 i t i s o p
−0.5
−1 0
2
4
6
8
10
12
14
16
18
20
time [s]
Solution Equation (6.60) of the class notes tells us that with weak damping, Q is equal to π times the ratio of the 1/e 1 /e deca decay y time time to the period. From the plot, we see that the decay time is approximately 10 seconds while the period is 1.25 seconds (since there are almost exactly 8 cycles in the first 10 seconds). seconds). Therefore Therefore,, Q 25.. Your answers 25 answers may vary vary a bit dependin dependingg on how closely you read the plot.
≈
#2 (5 points) A mass m mass m is is subject to a two-dimensional, anisotropic Hooke’s-law potential 1 1 U (x, y) = kx x2 + ky y 2 . 2 2 The resulting motion of the mass traces out the path shown in the figure below. 1
What is the ratio kx /ky ? BONUS (2 points) Considering that the orbit of a 2D oscillator is a combination of sinusoidal x- and y-motion, it’s curious that this particular orbit looks like a parabola. In fact, it is a parabola. Prove this! 5
y
0
−5 −5
0
5
x
Solution Recall from the classroom demonstration and p. 172 of the text that the orbit of the mass can be decomposed into sinusoidal x and y functions of time, with time dependence of cos(ωx t) and cos(ωy t) respectively. The amplitudes and relative phase will contribute to the shape of the orbit, but are not relevant to this question. Starting at any point along the orbit, trace the path until you come back to the initial point again. You will see that the mass completes 1 full period in the x-direction for every 2 full periods that it completes in the y direction. Thus, the ratio of angular frequencies must be ω x /ωy = 1/2. Then, recalling that ω = k/m, we have
−
−
kx = ky
ωx ωy
2
=
1 2
2
1 = . 4
−
(1)
For the bonus marks, we now know that x(t) = Ax cos(ωt) and y(t) = Ay cos(2ωt) (we could read Ax and Ay off the graph, but it’s not relevant...since the orbit terminates in two “points” rather than taking the form of a complicated loop, we know that the phase difference between the x and y motion is either 0 or π, and is therefore irrelevant. We don’t know ωx or
2
ωy , but we do know their ratio, and this IS important). Thus, y = A y cos(2ωt) = A y 2cos2 ωt
1 = 2Ay cos2 ωt
−
− Ay =
2Ay A2x
x2
− Ay , (2)
which proves that y is parabolic in x. #3 (5 points) A 2-kilogram object is subjected to three forces. Measured in Newtons, they ˆ y ˆ , F2 = 2ˆ ˆ + 5ˆz, F3 = x ˆ + 2ˆ are: F1 = x x y y + 3ˆz.
− −
−
−
(a) What is the acceleration of the mass? (2 points) (b) Is the net force on the mass conservative? Clearly explain why or why not. (3 points) Solution
(a) From Newton’s second law, we know that F = ma. The net force on x + 0ˆ y + 8ˆz. the mass (measured in Newtons) is F = F 1 + F2 + F3 = 0ˆ Since the mass is 2 kilograms, the resulting acceleration is a =
8 zˆ N F = = 4 ˆz m/s 2 . 2 kg m
(3)
(b) The force is conservative if it passes two tests. First, does it depend only on position and nothing else? YES (in fact, it doesn’t even depend on position, it’s a constant). Second, does it satisfy F = 0? YES (we don’t even need to work out the curl explicitly - all of the derivatives are zero since F is constant). Together, we can claim that the net force is conservative.
∇×
#4 (5 points) A projectile is traveling horizontally and due north with speed v 0 , when all of a sudden it explodes into two pieces of equal mass. Just after the explosion, one of the pieces is traveling vertically upward with speed v 0 . What is the velocity of the other piece? Solution 3
Choose the x-axis pointing due north and y vertically upward, and let v be the velocity of the second fragment after the explosion. Conservation of momentum implies that 12 mvx = mv0 and 12 mvy = 12 mv0 . Therefore, v = (2v0 , v0 , 0). You can also put this in magnitude-angle terms, with v = 5v0 , with an angle of tan 1 (1/2) = 26.6 below the horizontal from due north.
√ −
−
−
◦
#5 (5 points) By any means possible, find the moment of inertia of a flat, circular disk of mass M and radius R, which rotates about an axis which is perpendicular to the disk and which passes through a point on the rim of the disk, as in the figure below. Note: you might consider this a trick question, in that it is incredibly simple to answer - as long as you’ve been paying attention.
Solution The moment of inertia of a flat disk of mass M and radius R that rotates around an axis that is perpendicular to the disk and which passes through its center of mass is I CM = 12 M R2 (you can prove this, but it’s not required). When the axis of rotation is shifted, we apply the parallel axis theorem, which was provided in the solutions to Assignment 3, problem 10. Shifting the axis by a distance R from the center of mass produces a new moment of inertia 3 (4) I = I CM + M R2 = M R2 . 2
4
#6 (15 points) This question concerns a rocket that is launched vertically from the surface of a planet, where we assume drag to be negligible (this isn’t a realistic assumption, but...you’re welcome). The rocket is burning fuel at a rate of ˙ Since this represents a loss of the rocket’s total mass, we introduce m kg/s. the constant k = m, ˙ where k is a positive constant. The initial mass of the rocket is m 0 , and the spent fuel is ejected at a constant speed v ex .
−
(a) Prove that the rocket’s equation of motion is (3 points) ˙ mv = kv ex
− mg.
(b) Derive a formula for the rocket’s speed v(t), as a function of m0 , k , vex , and g. (6 points) (c) Derive and equation for the rocket’s height y(t), as a function of m0 , k , vex , and g. (6 points) Note: This will require an integral that you can find inside the front cover of the text.
Solution
(a) We know that the change in momentum (of the rocket plus the ejected fuel) in a time dt is dP = m dv + dm vex . But we also know from ˙ = F ext when there is an external force, Newton’s second law that P which means that dP = F ext dt (note that linear momentum is NOT conserved in this problem. Gravity is an external force). Equating these two expressions for dP gives us m dv + dm vex = F ext dt. Dividing by ˙ ˙ ex + F ext . Finally, with m = k ˙ dt and rearranging, we get mv = mv and F ext = mg, the equation of motion becomes
−
−
˙ mv = kv ex
−
− mg.
(5)
(b) Since dm = k, we can integrate with respect to t to find that m = dt m0 kt (the rocket’s mass decreases as a linear function of time, as expected). Therefore, the equation of motion separates to give
−
−
dv =
kvex m0 kt
−
5
−
g dt.
(6)
Integrating the LHS from 0 to v and the RHS from 0 to t, we get v(t) = v ex ln
m0 m0
− kt
−
(7)
gt.
This answer is physically reasonable. If the exhaust speed or the fuel burn rate is zero, then we simply have free-fall (the gt term).
(c) The height y(t) is found by integrating v(t) with respect to time. Using the identity ln(a/b) = ln(a) ln(b), we can split the integral into three parts:
−
t
y(t) =
t
′
′
v(t )dt = v ex
0
t
′
ln m0 dt
0
− vex
2
′
ln(m0
′
− kt )dt
0
−
′
′
gt dt .
0
(8) The first integrand on the RHS is independent of time, and evaluates to v ex t ln m0 . The third integral evaluates easily to 12 gt 2 . The middle integral requires a bit of work. We will change variables from t to m , integrating from m 0 to m. Since m = m 0 kt , we have dt = dm /k. The integral becomes
−
′
t
−vex
ln(m0
0
′
′
− kt )dt
′
′
−
′
′
′
0
′
′
−
−
0
−
where the last step uses the fact that m0 integrals together, we get
− m = kt.
Putting all the
− 12 gt2 + vkex (kt ln m0 − m0 ln m0 + m ln m) 1 mv ex m0 ln vex t − gt 2 − 2 k m 1 (m0 − kt)vex m0 ln vex t − gt 2 − . 2 k m − kt
y(t) = vex t
=
′
vex m = ln m dm (9) k m vex m = [m ln m m ]m (10) k vex = (m0 ln m0 m ln m) + vex t, (11) k ′
=
−
′
6
0
(12) (13) (14)
#7 (15 points) A mass m oscillates along the x-axis, on a spring with force constant k. The amplitude of the oscillations is d. At the moment when the the mass is at position x = d/2 and moving in the direction of increasing x (call this moment t = 0), it collides with and sticks to another equal mass m.
(a) What is the speed of the combined mass immediately after the collision (3 points)? (b) What is x(t) for the combined mass? (4 points)? (c) What is the amplitude of oscillation for the combined mass? (4 points)? (d) What fraction of the initial total mechanical energy is lost during the collision? (4 points)? Solution
(a) This is a momentum conservation problem, for a perfectly inelastic collision between two equal masses in which the initial velocity of the second mass is zero. We need to find the velocity of the first mass immediately before the collision. For SHM, we know that x(t) = d cos(ωt δ ), where ω = k/m. The collision occurs at t = 0 and at x = d/2, and therefore x(0) = d cos( δ ) = d cos(δ ) = d/2, which gives δ = π/3. The velocity immediately before the collision is therefore
v(0) =
− ±
−
−ωd sin(−δ ) = −ωd sin(±π/3) = ∓(
√
3/2)ωd. (15)
We want the positive solution, because we know that the mass is moving in the direction of increasing x. Now, we know all of the masses and speeds before the collision, and we apply conservation of momentum:
→ vf = v2i ,
m1 vi + 0 = (m1 + m2 )vf
(16)
Which tells us that the speed of the combined mass immediately after the collision is vf = ( 3/4)ωd.
√
(b) From chapter 6, we know that there are 4 different ways to express SHM. However, since we know the initial position (x0 = d/2) and velocity (v0 = ( 3/4)ωd), the most useful expression is x(t) = B 1 cos ω t +
√
′
7
B2 sin ω t, where B1 and B2 are x0 and v0 /ω , respectively. ω = k/2m is the new angular frequency, corresponding to the doubled mass. Therefore, ′
′
√
′
6d d sin ω t. x(t) = cos ω t + 2 4 ′
′
(17)
(c) The amplitude of the oscillations is A =
B12 + B22 =
d2 4
+
6d2 16
=
5 d. 8
(18)
(d) For SHM, E = 12 kA2 . Before the collision, A = d, and E = 12 kd2 . After
the collision, A = 58 d, and E = 12 k(5/8)d2 . Comparing these values, we see that 3/8 of the initial total mechanical energy is lost. Compare this with the case where two equal masses (one initially stationary), without any springs collide in a perfectly inelastic fashion. There, 1/2 of the energy is lost. The difference in our case is that there is potential energy stored in the spring, and that isn’t affected by the collision. #7 (15 points) The potential energy of two atoms in a molecule can sometimes be approximated by the Morse function, U (r) = A
(S −r)/R
e
2
− − 1
1 ,
where r is the distance between the atoms and A,R, and S are positive constants with S R.
≫
(a) Find the equilibrium separation r0 at which U (r) is a minimum; use any logical reasoning to prove that r0 is in fact a minimum and not a maximum (4 points ). (b) Write r = r 0 + x so that x is the displacement from equilibrium. Then, show that for small displacements, U has the approximate form U = const + 12 kx2 (8 points ). (c) What is the force constant k, in terms of A, R, and S ? (3 points ) 8
Solution
(a) r0 is found by setting dU/dr to zero. However, in the case of this particular U (r), it should be clear that we just want to minimize the magnitude of e(S r)/R 1. This simply gives r0 = S . We know that it must be a minimum because U (r) is large and positive for r 0 (since R , and S ), U (r) is negative and approaching zero for r U (S ) = A. −
−
→ →∞
≪ −
(b) If we set r = r 0 +x = S +x, then a Taylor expansion of the exponential term about x = 0 gives e(S
r)/R
−
= e
x/R
−
2
x ≈ 1 − Rx + 2R − · · · 2
(19)
and therefore
−
x x2 + R 2R2
− · · · − − ≈ − 2
1 = const+ kx2 , 2 (20) where we retained only enough terms in the series to express U up to second order in x. U = A
1
1
(c) From part (b), we see that k = 2A/R2 .
9
1
A+A
x R
2
