A disc of moment of inertia 5 kg- $m^{2}$ is acted upon by a constant torque of 40 Nm. Starting from rest the time taken by it to acquire an angular velocity of 24 ${rads}^{- 1}$ is

Question

Dr. Abubacker · Accepted Answer

(a) Torque is defined as rate of change of angular momentum
$	herefore$ $	au = \frac{dJ}{dt} = \frac{d(I\omega)}{dt}$
Given, $	au = 40\ Nm,\ I = 5\ kgm^{2},\ \omega = 24\ rads^{- 1}$
$dt = \frac{d(I\omega)}{	au} = \frac{5 	imes 24}{40} = 3\ $s

Q. A disc of moment of inertia 5 kg- $m^{2}$ is acted upon by a constant torque of 40 Nm. Starting from rest the time taken by it to acquire an angular velocity of 24 ${rads}^{- 1}$ is

Explanation

Reviewed by Dr. Abubacker

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