A CD with a diameter of 12.0 cm starts from rest and with a constant angular acceleration of 1.00 rad/sec2 acquires an angular velocity of 5.00 rad/sec. The CD continues rotating at 5.00 rad/sec for 15.0 seconds and then slows to a stop in 12.0 second with a constant angular acceleration. What is the angular distance traveled by a point 4.00 cm from the center, at the time 25.0 seconds from the start?

Explanation:

Given:

Now the time taken to get to the maximum angular velocity:

using equation of motion,

Now we observe the time to be in the deceleration phase, by:

Now the angular acceleration during the phase:

using eq. of motion,

where:

final angular velocity of the disk during the phase

acceleration during the deceleration phase

Now the speed at the instant of 5 seconds of the deceleration phase:

The angular distance travelled by the disk during this 5 seconds:

....................................(1)

Now the angular distance travelled in the phase of acceleration:

.....................................(2)

Now the angular acceleration during the constant angular speed phase:

........................(3)

Now the total angular distance travelled by any point on the disk will be same:

so, from (1), (2) and (3)

the answer is b ) square root of 5