A constant torque of $100 ext{ N m}$ turns a wheel of moment of inertia $300 ext{ kg m}^2$ about an axis pas

Question

A constant torque of $100	ext{ N m}$ turns a wheel of moment of inertia $300	ext{ kg m}^2$ about an axis passing through its centre. Starting from rest, its angular velocity after $3	ext{ s}$ is:

Accepted Answer

Given:
Torque $	au = 100	ext{ N m}$
Moment of inertia $I = 300	ext{ kg m}^2$
Initial angular velocity $\omega_0 = 0$ (since it starts from rest)
Time $t = 3	ext{ s}$

From Newton's second law for rotational motion, the angular acceleration $\alpha$ is given by:
$	au = I\alpha \implies \alpha = \frac{	au}{I} = \frac{100}{300} = \frac{1}{3}	ext{ rad/s}^2$

Using the first equation of rotational kinematics, the final angular velocity $\omega$ is:
$\omega = \omega_0 + \alpha t$
$\omega = 0 + \left(\frac{1}{3}ight) 	imes 3 = 1	ext{ rad/s}$.

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