spring of constant k, as in the figure in Kleppner. Just as the mass M engages the
spring and starts to push it, a friction force acts with a coefficient of friction that.
Solutions to Problem Set 7 (in Phys 241X) John Hauptman Department of Physics and Astronomy, Iowa State University, Ames, Iowa 50011, USA (Dated: 12 October 2010) The solutions to Phys 241X Problem Set 7: 4.2, 4.8, 4.13, 4.22, 4.23, and 4.29.
I.
We just need to calculate E0 − Ex . I get
INTRODUCTION
This set of problems is extremely difficult, and you succeed just by trying them.
II.
Prob 4.2 A mass M and velocity v0 at x = 0 slides towards a spring of constant k, as in the figure in Kleppner. Just as the mass M engages the spring and starts to push it, a friction force acts with a coefficient of friction that depends on the position, µ = bx. Overview: there is an initial total energy in the system of 1 M v02 . 2
1 1 M v02 = kx2 + Q. 2 2
After a distance x, the heat generated by the friction force f = (bx)M g is x
Z
(bx0 )M gdx0 = bM g
Q= 0
x2 . 2
The spring is compressed by x, so energy conservation is 1 1 M v02 = kx2 + bM gx2 /2, 2 2 so the compression x is s x = v0
M . k + bM g
This is not the end of the problem. We want to know how much mechanical energy is lost to heat in this process. Since the compression is x, the mechanical energy stored in the spring after compression is Ex =
Prob 4.8 A mass M on a spring k oscillates back and forth with a small friction on the table so that only a small fraction of the oscillator energy is lost to frictional heat each cycle. The constant friction force is f (which presumably equals µM g, but just call it f ). The mass M is displaced by x0 and released. Let’s look at this problem. There is a starting total energy of E0 =
As the mass M pushes on the spring, there is potential energy stored in the spring, R and the friction force f generates heat given by Q = f dx, so conservation of energy is E0 =
bM g 1 M v02 [ ]. 2 k + bM g
Does this make sense? If we “turn off” the friction, b = 0, then ∆E = 0. If the friction is dominant, b → ∞, then ∆E = (1/2)M v02 , that is, all the energy is loss. Seems OK.
SOLUTIONS
E0 =
∆E = E0 − Ex =
1 2 1 M kx = k[v02 ( )]. 2 2 k + bM g
1 2 kx . 2 0
During the first cycle, there is heat energy, Q0 , generated by the friction force of Q0 = f · 4x0 , since a quarter cycle is x0 , and the whole round trip of one cycle is 4x0 . After one cycle, the energy left in the oscillator is E1 = E0 − Q0 . This oscillator energy will correspond to a spring energy (fully compressed on the next cycle) of E1 =
1 2 kx , 2 1
where x1 is the amplitude on the next cycle. a. The decrease in amplitude is x0 − x1 . Solving for x0 and x1 from the above relations will involve calculating the differences of square-roots, a likely algebraic mess. From my years of experience solving such problems, and taking the hint that the decrease is very small per cycle, let’s “linearize” this problem in the following way: define the decrease in amplitude as δx = x0 − x1 , then x1 = x0 − δx, and we write out energy conservation E0 = E1 + Q0 as 1 2 1 kx = k(x0 + δx)2 + 4f x0 . 2 0 2
2 The “linearization” part comes in squaring the binomial (x0 + δx), (x0 + δx)2 = x20 (1 +
δx δx 2 ) ≈ x20 (1 + 2 ). x0 x0
Solving for δx, I get δx ≈
4f . k
For the next cycle, I just repeated this calculation starting from an oscillator with energy E1 = (1/2)kx21 and a heat generated of Q = 4f x1 . In other words, I did not solve the general problem for the nth cycle. The decrease in amplitude is the same δx =
x0 x0 k ≈ . δx 4f
Prob 4.13 In this problem, an analytic formula for the interatomic potential energy function is given (frankly, I think it is a poor representation) and one wants to know the oscillation frequency of the molecule. This is important for O2 , N2 and H2 molecules, in addition to the sunlight absorbing molecules like CO2 and CH4 . You can plug the mass of Nitrogen into the formula and see where it lands in the electromagnetic spectrum. The method is actually given on pages 178-179 of Kleppner, including the derivation of the “reduced mass” of a two-body system. I did not mention this in class, but I should have. The inter-atomic potential energy function is U (r) = [(
r0 12 r0 ) − 2( )6 ]. r r
a. The minimum in U (r) is where the slope of U is zero, or where the derivative of U (r) is zero, 0=
d r0 1 r0 1 U (r) = [(−12)( )12 − 2(−6)( )6 ] = 0. dr r r r r
I solved for r by multiplying through by (r/r0 )6 , getting r = r0 . Evaluating U (r0 ), I get U (r0 ) = −. Since the potential U (r) goes to zero as r → ∞, − is the depth of the potential below zero. b. If the potential function U (r) is of the form U (r − r0 ) ≈
What is the form of U (r)? If you plot it, it looks like a parabola at the bottom. For small deviations from r0 , we can expand U (r) in a Taylor’s series expansion: (see Kleppner, page 42) U (r−r0 ) = U (r0 )+
1 k(r − r0 )2 2
dU (r) 1 d2 U (r) |r=r0 (r−r0 )2 +... |r=r0 (r−r0 )+ dr 2 dr2
Since the slope is zero at r = r0 and the depth is , this is
4f . k
Check: if the friction force is zero, then the amplitude never changes, δx = 0. b. The number of cycles before coming to rest is n=
at its minimum, then atoms of mass m will oscillate with a frequency r k ω= . m
1 U (r − r0 ) = − + 0 + k(r − r0 )2 2 if k=
d2 U (r) |r=r0 . dr2
This is the curvature of the function at it minimum. I took the next derivative and got 72 d2 U (r) |r=r0 = 2 . 2 dr r0 The molecule is like N2 or O2 , each atom of mass m. In this two-body system, the two “F=ma” equations actually result in a differential equation of an oscillator whose mass is what is called the “reduced mass” (label it µ), µ=
m1 m2 m → m1 + m2 2
(for equal masses).
So the final oscillation frequency is s s r 72 144 ω= = = 12 . (m/2)r02 mr02 mr02 Prob 4.22 The dropped ball has velocity v0 just as it hits the floor the first time. It rebounds slightly inelastically with velocity v1 = ev0 . On the second bounce, it rebounds with v2 = ev1 = e2 v0 . In other words, the ball loses a fraction e in velocity each bounce. The question is: how long until it comes to rest. There may be many ways to go after this problem, but one way is to calculate the time for each bounce, then add them up. If the ball hits the floor with velocity v0 , it must have been accelerated for a time t0 such that v0 = gt0 , so that t0 = v0 /g. Since it has to go up and back down again for the first bounce, the time is twice this: t1 = 2v1 /g = e2v0 /g.
3 Similarly, the time for the second bounce is t2 = 2v2 /g = e2 2v0 /g, and so on. The total time T is 2v0 T = [e + e2 + e3 + ...]. g This is a converging series, which you can Google, or look up in a math book. You can even guess the answer: if e = 12 , then 21 + 14 + 18 + ... seems to fill a square of area 1. The mathematical series 1 + r + r2 + r3 + ... (where r is less than 1) sums to 1 , 1−r which equals 2 for r = 1/2. Our series does not have the 1 as the first term, so ours sums to 1 for e = 1/2. Then, the total time for v0 = 5 m/s and e = 1/2 is T =
2 × 5m/s 2
10m/s
1. the momentum transfer to the wall is larger since the wall is approaching the particle. The velocity gain is 2V per reflection. 2. the box is getting smaller, so the round trip travel time is less. 3. the power being delivered to the particle in the box is F · V (work per second), and so the particle must gain this much energy per second. The only energy term in the problem is the particle’s kinetic energy. Collision with the wall: give the wall a very large mass M moving to the right with a small velocity V . The momentum before the collision is M V − mv0 , and after the collision is M V 0 + mv, so V 0 = V − (m/M )(v0 + v). Energy conservation is 1 1 1 1 M V 2 + mv02 = M V 02 + mv 2 . 2 2 2 2
× 1 = 1s.
Prob 4.23 The large ball √ of mass M is dropped from height h so that v0 = mgh when it hits the floor and rebounds upwards. The small ball m falls the same distance and has the same v0 just as the mass M collides with it. Momentum conservation with positive momentum pointing up is
Let r = m/M , solve for V 0 from momentum, and substitute into the energy equation. Factor (v02 − v 2 ) = (v0 − v)(v0 + v), and cancel all (v0 + v) terms. Neglect r2 compared to r, and you get v = v0 + 2V. The velocity gain happens every 2`/v0 seconds, so dv 2V = , dt 2`/v0
+M v0 − mv0 = M vM + mv. Conservation of energy is 1 1 1 1 2 M v02 + mv02 = M vM + mv 2 , 2 2 2 2 where v is the vertical velocity of the small ball. I divided everywhere through by M to get small terms in r = m/M . Then, solved for vM from the momentum equation and substituted it into the energy equation. About a half-page of algebra and neglecting r2 compared to r gets you to
and inserting F = mv02 /`, FV dv = . dt mv0 This same result can be gotten from directly setting the power (work per second), F V , equal to the time derivative of the particle kinetic energy, FV =
v = 3v0 .
d 1 m dv mv 2 = 2v , dt 2 2 dt
so that Prob 4.29 a. The round trip time is ∆t = 2`/v0 and the momentum transfer per wall is ∆p = 2mv0 , so the average force is F =
∆p 2mv0 = = ∆t 2`/v0
mv02 `
dv FV = . dt mv From dv/dt given above, the velocity of the particle as a function of time t is
.
This is just like the result for the pressure in a gas volume from the kinetic theory. b. As the wall is pushed in at velocity V , which is much smaller than v, the particle speeds up (the “temperature” rises). There are three things going on:
v = v0 +
2V · t, 2x/v
and V t = ` − x, so v = v0 +
v0 vV t = v0 + (` − x), x x
4 so `−x ` v =1+ = . v0 x x Finally, the force as a function of x is F (x) =
mv02 ` 3 mv 2 mv02 `2 = = ( ) . x x · x2 ` x
This was an extremely difficult problem. I had a lot of trouble finding my way to this solution. I tried taking a small ∆t and expanding the changes in v and x, then linearizing, but without success. You having tried these problems are already a success. Never are the problems in a first semester physics course this difficult.
