469
Chapter 11
1. The velocity of the car is a constant
?
80 km/h (1000 m/km)(1 h/3600 s) i
( 22m s)i,
v
and the radius of the wheel is
r
= 0.66/2 = 0.33 m.
(a) In the car’s reference frame (where the lady perceives herself to be at rest) the road is
moving toward the rear at
v
v
road
m s
22
, and the motion of the tire is purely
rotational. In this frame, the center of the tire is “fixed” so
v
center
= 0.
(b) Since the tire’s motion is only rotational (not translational) in this frame, Eq. 10-18
gives
top
ˆ
( 22m/s)i.
v
(c) The bottom-most point of the tire is (momentarily) in firm contact with the road (not
skidding) and has the same velocity as the road:
bottom
ˆ
( 22m s)i.
v
This also follows
from Eq. 10-18.
(d) This frame of reference is not accelerating, so “fixed” points within it have zero
acceleration; thus,
a
center
= 0.
(e) Not only is the motion purely rotational in this frame, but we also have
= constant,
which means the only acceleration for points on the rim is radial (centripetal). Therefore,
the magnitude of the acceleration is
2
2
2
3
top
(22 m/s)
1.5 10 m s .
0.33 m
v
a
r
(f) The magnitude of the acceleration is the same as in part (d):
a
bottom
= 1.5
10
3
m/s
2
.
(g) Now we examine the situation in the road’s frame of reference (where the road is
“fixed” and it is the car that appears to be moving). The center of the tire undergoes
purely translational motion while points at the rim undergo a combination of translational
and rotational motions. The velocity of the center of the tire is
ˆ
v
(h) In part (b), we found
v
v
top,car
and we use Eq. 4-39:
top, ground
top, car
car, ground
?
?
i
i
2 i
v
v
v
v
v
v

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*Sign up*CHAPTER 11
470
which yields 2
v
= +44 m/s.
(i) We can proceed as in part (h) or simply recall that the bottom-most point is in firm
contact with the (zero-velocity) road. Either way, the answer is zero.
(j) The translational motion of the center is constant; it does not accelerate.
(k) Since we are transforming between constant-velocity frames of reference, the
accelerations are unaffected. The answer is as it was in part (e): 1.5
10
3
m/s
2
.
(1) As explained in part (k),
a
= 1.5
10
3
m/s
2
.
2. The initial speed of the car is
80 km/h (1000 m/km)(1 h/3600 s)
22.2 m/s
v
.
The tire radius is
R
= 0.700/2 = 0.350 m.
(a) The initial speed of the car is the initial speed of the center of mass of the tire, so Eq.
11-2 leads to
com0
0
22.2 m/s
63.4 rad/s.
0.350 m
v
R
(b) With
= (30.0)(2
) = 188 rad and
= 0, Eq. 10-14 leads to
2
2
2
2
0
(63.4 rad/s)
2
10.7 rad/s .
2 188 rad
(c) Equation 11-1 gives
R
= 65.8 m for the distance traveled.
3. By Eq. 10-52, the work required to stop the hoop is the negative of the initial kinetic
energy of the hoop. The initial kinetic energy is
K
I
mv
1
2
2
1
2
2
(Eq. 11-5), where
I =
mR
2
is its rotational inertia about the center of mass,
m
= 140 kg, and
v
= 0.150 m/s is the
speed of its center of mass. Equation 11-2 relates the angular speed to the speed of the
center of mass:
=
v
/
R
. Thus,
2
2
2
2
2
2
1
1
140 kg
0.150 m/s
3.15 J

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