c.
The following drawing shows the relevant geometry:
r
s
celestial body
person on earth
pe rson on e arth
moon
r
moon
r
sun
s
sun
R
sun
sun
moon
s
b
R
b

388
ROTATIONAL KINEMATICS
The apparent circular area of the sun as measured by a person standing on the earth is given
by:
2
sun
sun
A
R
, where
R
sun
is the radius of the sun.
The apparent circular area of the sun
that is blocked by the moon is
2
blocked
b
A
R
, where
R
b
is shown in the figure above. Also
from the figure above, it follows that
R
sun
= (1/2)
s
sun
and
R
b
= (1/2)
s
b
Therefore, the fraction of the apparent circular area of the sun that is blocked by the moon is
2
2
2
2
blocked
b
b
b
moon
sun
2
2
sun
sun
sun
sun
sun
sun
2
2
3
moon
3
sun
(
/ 2)
(
/ 2)
9.04
10
rad
0.951
9.27
10
rad
A
R
s
s
r
A
s
r
R
s
The moon blocks out
95.1 percent
of the apparent circular area of the sun.
19.
REASONING
The golf ball must travel a distance equal to its diameter in a maximum
time equal to the time required for one blade to move into the position of the previous blade.
SOLUTION
The time required for the golf ball to pass through the opening between two
blades is given by
/
t
, with
1.25 rad/s
and
(2
rad)/16 = 0.393 rad
.
Therefore, the ball must pass between two blades in a maximum time of
t
0.393 rad
1.25 rad/s
0.314 s
The minimum linear speed of the ball is
v
x
t
4.50
10
–2
m
0.314 s
1.43
10
–1
m/s
20.
REASONING
We know that the skater’s final angular velocity is
0.0 rad /s
, since she
comes to a stop.
We also that her initial angular velocity is
0
15 rad / s
and that her
angular displacement while coming to a stop is
5.1 rad
.
With these three values, we
can use
2
2
0
2
(Equation 8.8) to calculate her angular acceleration
α
.
Then,
knowing
α
, we can use
0
t
(Equation 8.4) to determine the time
t
during which she
comes to a halt.
SOLUTION
a.
With
0.0 rad /s
for the skater’s final angular velocity, Equation 8.8 becomes
2
0
0
2
.
Solving for
, we obtain

Chapter 8
Problems
389
2
2
2
0
(15 rad/s)
–22 rad/s
2
2(5.1 rad)
b.
With
0.0 rad /s
for the skater’s final angular velocity, Equation 8.4 becomes
0
0
t
.
Solving for
t
, we obtain
0
2
15 rad/s
0.68 s
–22 rad/s
t
____________________________________________________________________________________________
21.
SSM
REASONING
The angular displacement is given as
θ
= 0.500 rev, while the initial
angular velocity is given as
0
= 3.00 rev/s and the final angular velocity as
= 5.00 rev/s.
Since we seek the time
t
, we can use Equation 8.6
1
2
0
t
from the equations of
rotational kinematics to obtain it.

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