Question

A constant torque of 1000 N-m turns a wheel of moment of inertia $ 200\text{ }kg-{{m}^{2}} $ about an axis through its centre. Its angular velocity after 3 second is, in rad/s:

• A. 1
• B. 5
• C. 15
• D. 10

Answer 1

Answer: (C)

The formula for torque is as $ \tau =I\alpha $ where $ I $ is moment of inertia and $ \alpha $ is the angular acceleration $ \alpha =\frac{\tau }{I}=\frac{1000}{200}=5\,\,rad/{{s}^{2}} $ $ \left( \begin{align} & \text{Given:}\,\tau \,=1000N \\ & I\text{ }=\text{ }2000\,kg-{{m}^{2}} \\ & \text{t = 3s } \\ \end{align} \right) $ Now, angular velocity $ \omega =\alpha \,t=5\times 3 $ $ =15\,rad/s $