Solutions for Problem Set 10

Last revised 1999/11/02

Chapter 10

27. (MathCad-based solution)

55. (MathCad-based solution)

67.
We are given that v_B2 is perpendicular to v_B1 and that v_B2 / v_B1 = 1/2 . We must find the direction that A moves after the collision. For any collision problem, we can use momentum conservation:

horizontal:	MBvB1 = mA vA2 cos(theta)
vertical:	0 = MB vB2 - MA vA2 sin(theta)

Rewriting the second equation as

M_B v_B2 = M_A v_A2 sin(theta)

We can divide the second equation by the first to get

tan(theta) = v_B2 / v_B1 = 1/2
theta = 27^o

We cannot, however, determine the speed of A , because we have not been given values for the masses.

68.
The figure for problem 67 applies to this problem as well. In this case, M_B is the mass of 1 nucleon, call it M, and M_A is then 2M. We are supposed to find the fraction of the initial kinetic energy carried away by the deuteron, in other words to evaluate f where

f = (1/2)(2 M) v_A2² / [(1/2) M v_B1².

We can again use momentum conservation, and this time energy conservation as well. After masses and factors of 1/2 are cancelled, the equations read

v_B1 = 2 v_A2 cos(theta)
0 = v_B2 - 2 v_A2 sin(theta)
v_B1² = v_B2² + 2 v_A2²

Moving the first term in the second equation to the left-hand side, then squaring each of the first two equations and adding, we can eliminate theta, getting

v_B2² + v_B1² = 4 v_A2²

We can use this equation to eliminate v_B2 from the third of the original equations, yielding

v_B1² = 3 v_A2²

Substituting this result into the definition of f gives

f = 2/3

as required.

Chapter 11

7.

We are given	alpha= 4 a t3 - 3 b t2
Integrating once gives	omega = omega0 + a t4 - b t3
and integrating again gives	theta = theta0 + omega0 t + (1/5) a t5 - (1/4) t 4

19. MathCad-based solution:

44.
We are given that the period T of the crab pulsar is given by

T = T0 + T1 T0 = 0.033 s T1 = 1.26 x 10 -5 s/yr = 3.99 x 10 -13 [using 1 yr = 3.16 x 107 s].	[Eq. 1]
We want to get (a) the angular acceleration alpha; (b) the time tf when the pulsar would quit spinning if its angular acceleration were to remain constant; (c) the period Ti of the pulsar when it was created in 1054 AD. We need the definition of omega in terms of the period, and the expression for the time-dependence of omega under constant angular acceleration:
omega = 2 pi / T	[Eq. 2]
omega = omega0 + alpha t	[Eq. 3]
Using Eqs. 1 and 2 we have omega = 2 pi / [ T0 + T1 t]           = [2 pi / T0 ] [1 + (t1 / T0) t ] -1 and using the binomial expansion, omega = (2 pi / T0 ) - [ (2 pi T1) / T0 2 ] t so that
omega0 = 2 pi / T0 = 190/s
alpha = - (2 pi T1) / T02 = -2.30 x 10 -9 / sec2	[Answer to (a)]
To get the time when omega = 0 we use eq. 3 0 = omega = omega0 + alpha t t = - omega0 / alpha
= 8.28 x 1010 sec = 2.62 x 103 yr3	[Answer to (b)]
We can substitute into the same equation to find what the value of omega was when the pulsar was formed in 1054 AD (943 years ago). The result is
omega = 259 / s	[Answer to (c)]

59.
We need to know something about the kinetic energy of a flywheel and about the relation between energy and power when the power is constant (or we are using average power). The relations are

Kinetic Energy: KE = (1/2) I omega²
Moment of Inertia of a wheel: I = (1/2) M R²
Power: P = dE/dt so that E = E₀ + P t      {P constant)

We have been given

omega = 200 pi / s
M = 500 kg
R = 1.0 m
P = - 8.0 kW

Substituting for the moment of inertia in the formula for the kinetic energy gives

KE = (1/4) M R² omega² = 49 x 10⁶ j.

The solving E = KE + P t gives

t = - KE/P = 6.1 x 10³ s = 100 min

Links to image files used in these solutions: 10-27, 27a, 27b, 27c, 27d; 10-55, 55a, 55b, 55c; 10-67 and 10-68; 11-19, 19a, 19b, 19c, 19d, 19e.