elapsed time (Equation 8.2). Therefore, the angular
displacement is equal to the product of the average
angular velocity and the elapsed time The elapsed
time is given, so we need to determine the average
angular velocity. We can do this by using the graph of
angular velocity versus time that accompanies the
problem.
SOLUTION
The angular displacement
is
related to the average angular velocity
and the
3.0 s
8.0 s
+15 rad/ s
–9.0 rad/ s
0
Time (s)
Angular velocity
+3.0 rad/s

elapsed time
t
by Equation 8.2,
t
. The elapsed time is given as 8.0 s. To obtain
the average angular velocity, we need to extend the graph that accompanies this problem
from a time of 5.0 s to 8.0 s. It can be seen from the graph that the angular velocity
increases by +3.0 rad/s during each second. Therefore, when the time increases from 5.0
to 8.0 s, the angular velocity increases from +6.0 rad/s to 6 rad/s + 3
(3.0 rad/s) =
+15 rad/s. A graph of the angular velocity from 0 to 8.0 s is shown at the right. The
average angular velocity during this time is equal to one half the sum of the initial and
final angular velocities:
1
1
0
2
2
9.0 rad/s +15 rad/s
3.0 rad/s
The angular displacement of the wheel from 0 to 8.0 s is
3.0 rad/s
8.0 s
24 rad
t
31.
REASONING
According to Equation 3.5b, the time required for the diver to reach
the water, assuming free-fall conditions, is
2
/
y
t
y a
.
If we assume that the "ball"
formed by the diver is rotating at the instant that she begins falling vertically, we can use
Equation 8.2 to calculate the number of revolutions made on the way down.
SOLUTION
Taking upward as the positive direction, the time required for the
diver to reach the water is
t
2(–8.3 m)
–9.80 m/s
2
1.3 s
Solving Equation 8.2 for
, we find
(1.6 rev/s)(1.3 s)=
2.1 rev
t
32.
REASONING
The time required for the change in the angular velocity to occur can
be found by solving Equation 8.4 for
t
.
In order to use Equation 8.4, however, we must
know the initial angular velocity
0
.
Equation 8.6 can be used to find the initial angular
velocity.
SOLUTION
From Equation 8.6 we have
1
2
(
0
)
t
Solving for
0
gives
0
2
t

Since the angular displacement
is zero,
0
= –
.
Solving
0
t
(Equation 8.4) for
t
and using the fact that
0
= –
give
t
2
2(
25.0
rad/s)
4.00
rad/s
2
12.5 s
33.
REASONING
The angular displacement of the child when he catches the horse is,
from Equation 8.2,
c
c
t
. In the same time, the angular displacement of the horse is,
from Equation 8.7 with
0
0
rad/s,
h
1
2
t
2
.
If the child is to catch the horse
c
h
(
/ 2)
.
SOLUTION
Using the above conditions yields
2
1
1
c
2
2
0
t
t
or
2
2
1
1
2
2
(0.0100 rad/s
)
0.250 rad/s
rad
0
t
t
The quadratic formula yields
t
= 7.37 s and 42.6 s; therefore, the shortest time
needed to catch the horse is
t
= 7.37 s
.

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- Fall '16
- Velocity