Question

The rate of rotation of a solid disk with a radius of `6 m` and mass of `5 kg` constantly changes from `5 Hz` to `3 Hz`. If the change in rotational frequency occurs over `1 s`, what torque was applied to the disk?

Answer 1

This is a lengthy calculation that requires we first find moment of inertia for the disk, and its angular acceleration, before finding the average torque to be 300 `pi` Newton-metres.

Explanation:

To answer this question, we must first determine the rotational inertia of the disk (also called the moment of inertia), I :
This will be given by `mr^2 /2`, where m is the mass and r is the radius of the disk. For the above disk, this is `6 (5^2) /2` = 75 kg `m^2`

Similar to Newton's second law F = ma, the corresponding relation for rotational motion is:

`tau` = I `alpha` where `alpha` is the angular acceleration of the disk.

To find `alpha` we take the change in angular velocity in radians and divide by the time taken (much like acceleration is the change in velocity divided by time)
The change in angular velocity is 2 Hz x 2`pi` = 4`pi` /s, so the angular acceleration is `alpha` = 4`pi` / 1 s = 4`pi` /`s^2`

Finally, `tau` = 75 kg `m^2` x 4`pi` /`s^2` = 300 `pi` kg`m^2/s^2` which is more commonly given as 300 `pi` N m (Newton-metres)