Question

The rate of rotation of a solid disk with a radius of `2 m` and mass of `5 kg` constantly changes from `12 Hz` to `16 Hz`. If the change in rotational frequency occurs over `3 s`, what torque was applied to the disk?

Answer 1

Torque: `\qquad \tau = I\alpha = (80\pi)/3\quad kg.m^2. (rad)/s^2`

Explanation:

Given:
`\omega_i = (2\pi \quad rad)(12 Hz) = 24\pi \quad (rad)/s;`
`\omega_f = (2\pi\quad rad)(16 Hz) = 32\pi \quad (rad)/s;`

`\Delta\omega = 8\pi\quad (rad)/s; \qquad \Deltat = 3s; \qquad M = 5kg; \qquad R = 2m`

Moment-of-Inertia - Solid Disc: `\qquad I = 1/2 MR^2 = 10 \quad kg.m^2`

Angular Acceleration: `\qquad \alpha = (\Delta\omega)/(\Deltat) = (8\pi)/3\quad (rad)/s^2`

Torque: `\qquad \tau = I\alpha = (80\pi)/3\quad kg.m^2. (rad)/s^2`