Problem 36P

A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from 0.00 to 4.00 rev/s in 3.00 s. What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is (a) 2.00 rev/s and (b) 3.00 rev/s?

Solution 36P

a.)

Step 1 of 2

We have to find the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00 rev/s

The tangential acceleration is given by the expression

where,

radius of disk = 6 cm = 0.06 m

angular acceleration in rad/s2

The angular accelerationis given by

=

= 8.38 rad/s2

Hence, the tangential acceleration is

=

= 0.503 rad/s2

Therefore, the tangential acceleration of a point on the outer rim of the disk is 0.503 rad/s2.

b.)