Question

A wheel is rotating with an angular speed of $20rad/sec$. It is stopped to rest by applying a constant torque in $4s$. If the moment of inertia of the wheel about its axis is 0.20 kg-m², then the work done by the torque in two seconds will be

• A. 10 J
• B. 20 J
• C. 30 J
• D. 40 J

Answer 1

Answer: (C)

30 J

Answer 2

$\omega_{1} = 20$ rad/sec, $\omega_{2} = 0,t = 4sec.$ So angular retardation $\alpha = \frac{\omega_{1} - \omega_{2}}{t} = \frac{20}{4} = 5rad/sec^{2}$Now angular speed after 2 sec $\omega_{2} = \omega_{1} - \alpha t = 20 - 5 \times 2$ = $10$rad/sec

Work done by torque in 2 sec = loss in kinetic energy

=$\frac{1}{2}I\left( \omega_{1}^{2} - \omega_{2}^{2} \right) = \frac{1}{2}(0.20)((20)^{2} - (10)^{2})$

$= \frac{1}{2} \times 0.2 \times 300$= 30 J.