Question

A sanding disk with rotational inertia 6.5 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnitude 21 N·m about the central axis of the disk. About that axis and with torque applied for 54 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Answer 1

Answer:

The angular momentum and angular velocity are 1.134 kg.m²/s and 174.5 rad/s.

Explanation:

Given that,

Moment of inertia $I= 6.5\times10^{-3}\ kg.m^2$

Torque = 21 N.m

Time dt = 54 ms

(a). We need to calculate the angular momentum

Using formula of torque

$\tau=\dfrac{dL}{dt}$

$dL =\tau\times t$

Where, dL = angular momentum

t = time

$\tau$ = torque

Put the value into the formula

$dL=21\times0.054$

$dL=1.134\ kg.m^2/s$

(b). We need to calculate the angular velocity of the disk

Using formula of angular velocity

$dL=I\omega$

$\omega=\dfrac{dL}{I}$

$\omega=\dfrac{1.134}{6.5\times10^{-3}}$

$\omega=174.5\ rad/s$

Hence, The angular momentum and angular velocity are 1.134 kg.m²/s and 174.5 rad/s.