Question

A sanding disk with rotational inertia 3.8 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnitude 17 N·m about the central axis of the disk. About that axis and with torque applied for 79 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Answer 1

Answer:

(a) Angular momentum of disk is $1.343\, \frac{kg.m^{2}}{s}$

(b) Angular velocity of the disk is $353\frac{rad}{s}$

Explanation:

Given

Rotational inertia of the disk , $I=3.8\times 10^{-3}kg.m^{2}$

Torque delivered by the motor , $\tau =17N.m$

Torque is applied for duration , $\Delta t=79ms=0.079s$

(a)

Magnitude of angular momentum of the disk = Angular impulse produced by the torque

$\therefore L=\tau \Delta t=17\times 0.079\frac{kg.m^{2}}{s}$

=>$L=1.343\, \frac{kg.m^{2}}{s}$

Thus angular momentum of disk is $1.343\, \frac{kg.m^{2}}{s}$

(b)

Since Angular momentum , $L=I\omega$

where $\omega$= Angular velocity of the disk

$=>\omega =\frac{L}{I}=\frac{1.343}{3.8\times 10^{-3}}\frac{rad}{s}$

$\therefore \omega =353\frac{rad}{s}$

Thus angular velocity of the disk is $353\frac{rad}{s}$