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12.4 Motion with Time-Dependent Angular Acceleration 377

In the special case of constant angular acceleration , Eq. (12.22) gives us
t, which agrees with our previous result, Eq. (12.17). If we insert this into
0 t
1
2
Eq. (12.23), we obtain f f ∫ 0 ( t)dt t t , which agrees with
2
0 0 0
our previous Eq. (12.18).
In the general case of a time-dependent angular acceleration , we proceed in the
same way: first, use Eq. (12.22) to find as a function of time, and then insert this
function into Eq. (12.23) to find the angular position as a function of time, as in the
following example.

When turned on, a motor rotates a circular saw wheel, begin-
EXAMPLE 7
ning from rest, with an angular acceleration that has an initial
2
value 60 radians/s at t 0 and decreases to zero acceler-
0
ation during the interval 0 t 3.0 s according to
t
a 1 b
0
3.0s
After t 3.0 s, the motor maintains the wheel’s angular velocity at a constant
value. What is this final angular velocity? In the process of “getting up to speed,”
how many revolutions occur?
SOLUTION: The angular acceleration is given as an explicit function of time. Since
we are beginning from rest, the initial angular velocity is 0, so Eq. (12.22) gives
0
as a function of t:
0 t t t
œ
dt
0 a 1 b dt
0

0 0 3.0 s
0 t 1 t t 1 t
2 t
¢ dt
t
dt
b a t
` ` b
0
0 3.0 s 0 0 3.0 s 2 0
2
t
a t b (12.24)
0
6.0s
where we have used the property that the integral of the sum is the sum of the
n n+1
integrals, and that ∫t dt t (n 1). At t 3.0 s, this angular velocity reaches
its final value of
(3.0 s) 2
2
60 radians>s a 3.0 s b 90 radians>s
6.0 s
To obtain the number of revolutions during the time of acceleration, we can cal-
culate the change in angular position and divide by 2 . To do so, we must insert
the time-dependent angular velocity obtained in Eq. (12.24) into Eq. (12.23):

0 t t t
2
f f dt
a t
b dt
0
0 0 6.0 s
0 t 1 t 2 t
2 t 1 t
3 t
¢ t
dt
t
dt
b a ` ` b
0
0 6.0s 0 2 0 6.0 s 3 0
2 3
t t
a b
0
2 18 s