7. With Fapp  (10 N)iˆ , we solve the problem by applying Eq. 9-14 and Eq. 11-37.
(a) Newton’s second law in the x direction leads to


Fapp  f s  ma  f s  10N  10kg  0.60 m s  4.0 N.
2
In unit vector notation, we have f s  (4.0 N)iˆ which points leftward.
(b) With R = 0.30 m, we find the magnitude of the angular acceleration to be
|| = |acom| / R = 2.0 rad/s2,
from Eq. 11-6. The only force not directed towards (or away from) the center of mass

is f s , and the torque it produces is clockwise:
 I

 0.30 m  4.0 N   I  2.0 rad
s2 
which yields the wheel’s rotational inertia about its center of mass: I  0.60 kg  m2 .
13. (a) We choose clockwise as the negative rotational sense and rightwards as the
positive translational direction. Thus, since this is the moment when it begins to roll
b
g
. m .
smoothly, Eq. 11-2 becomes vcom   R  011
This velocity is positive-valued (rightward) since  is negative-valued (clockwise) as
shown in Fig. 11-57.
(b) The force of friction exerted on the ball of mass m is   k mg (negative since it
points left), and setting this equal to macom leads to
b gc
h
acom   g   0.21 9.8 m s2  2.1 m s2
where the minus sign indicates that the center of mass acceleration points left,
opposite to its velocity, so that the ball is decelerating.
(c) Measured about the center of mass, the torque exerted on the ball due to the
frictional force is given by    mgR . Using Table 10-2(f) for the rotational inertia,
the angular acceleration becomes (using Eq. 10-45)
2
  mgR 5 g 5  0.21  9.8 m/s 
 


 47 rad s 2
2
I 2m R 5
2R
2  0.11 m 

where the minus sign indicates that the angular acceleration is clockwise, the same
direction as  (so its angular motion is “speeding up’’).
(d) The center-of-mass of the sliding ball decelerates from vcom,0 to vcom during time t
according to Eq. 2-11: vcom  vcom,0  gt . During this time, the angular speed of the
ball increases (in magnitude) from zero to  according to Eq. 10-12:
 t
5gt vcom

2R
R
where we have made use of our part (a) result in the last equality. We have two
equations involving vcom, so we eliminate that variable and find
t
2vcom,0
7 g

2  8.5 m/s 
7  0.21  9.8 m/s 2 
 1.2 s.
(e) The skid length of the ball is (using Eq. 2-15)
x  vcom,0t 
1
1
2
  g  t 2  8.5 m/s 1.2 s    0.21  9.8 m/s 2  1.2 s   8.6 m.
2
2
(f) The center of mass velocity at the time found in part (d) is
vcom  vcom,0   gt  8.5 m/s   0.21  9.8 m/s 2  1.2 s   6.1 m/s.
8. (a) The amplitude is ym = 6.0 cm.
(b) We find  from 2/ = 0.020:  = 1.0×102 cm.
(c) Solving 2f =  = 4.0, we obtain f = 2.0 Hz.
(d) The wave speed is v = f = (100 cm) (2.0 Hz) = 2.0×102 cm/s.
(e) The wave propagates in the –x direction, since the argument of the trig function is
kx + t instead of kx – t (as in Eq. 16-2).
(f) The maximum transverse speed (found from the time derivative of y) is


umax  2 fym  4.0  s 1  6.0 cm   75cm s.
(g) y(3.5 cm, 0.26 s) = (6.0 cm) sin[0.020(3.5) + 4.0(0.26)] = –2.0 cm.
52. The nodes are located from vanishing of the spatial factor sin 5x = 0 for which
the solutions are
5x  0, , 2,3,
1 2 3
 x  0, , , ,
5 5 5
(a) The smallest value of x which corresponds to a node is x = 0.
(b) The second smallest value of x which corresponds to a node is x = 0.20 m.
(c) The third smallest value of x which corresponds to a node is x = 0.40 m.
(d) Every point (except at a node) is in simple harmonic motion of frequency f = /2
= 40/2 = 20 Hz. Therefore, the period of oscillation is T = 1/f = 0.050 s.
(e) Comparing the given function with Eq. 16–58 through Eq. 16–60, we obtain
y1  0.020sin(5x  40t ) and y2  0.020sin(5x  40t )
for the two traveling waves. Thus, we infer from these that the speed is v = /k =
40/5 = 8.0 m/s.
(f) And we see the amplitude is ym = 0.020 m.
(g) The derivative of the given function with respect to time is
u
y
 (0.040) (40)sin(5x )sin(40t )
t
which vanishes (for all x) at times such as sin(40t) = 0. Thus,
40t  0, , 2,3,

t  0,
1 2 3
, , ,
40 40 40
Thus, the first time in which all points on the string have zero transverse velocity is
when
t = 0 s.
(h) The second time in which all points on the string have zero transverse velocity is
when t = 1/40 s = 0.025 s.
(i) The third time in which all points on the string have zero transverse velocity is
when
t = 2/40 s = 0.050 s.
51. (a) The amplitude of each of the traveling waves is half the maximum
displacement of the string when the standing wave is present, or 0.25 cm.
(b) Each traveling wave has an angular frequency of  = 40 rad/s and an angular
wave number of k = /3 cm–1. The wave speed is
v = /k = (40 rad/s)/(/3 cm–1) = 1.2×102 cm/s.
(c) The distance between nodes is half a wavelength: d = /2 = /k = /(/3 cm–1) =
3.0 cm. Here 2/k was substituted for .
(d) The string speed is given by u(x, t) = y/t = –ymsin(kx)sin(t). For the given
coordinate and time,
 



9
u  (40 rad/s) (0.50cm) sin  cm 1  (1.5cm)  sin  40 s 1 

8
 3





s   0.
