Question

A sanding disk with rotational inertia 2.0 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnitude 11 N·m about the central axis of the disk. About that axis and with torque applied for 19 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Answer 1

Given,

Rotational inertia = 2.0 x 10-3 kg·m²

Torque = 11 N.m

time, t = 19 ms

a) Angular momentum

`\tau = (\Delta L)/(\Delta t)`

L is angular momentum

`\Delta L = \tau \Delta t`

`\Delta L = 11* 19 * 10^(-3)`

`\Delta L = 0.209\ Kg m^2/s`

b) Angular velocity

We know.

`L = I \omega`

`\omega = (L)/(I)`

`\omega = (0.209)/(2* 10^(-3))`

`\omega = 104.5\ rad/s`