A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in the figure below. The established torque produces an angular acceleration of 2 rad/s$$^2$$. Moment of inertia of the wheel is _________ kg m$$^2$$.
(Acceleration due to gravity = 10 m/s$$^2$$)

Correct Answer: 1

The torque produced by the force acting tangentially at the rim is:

$$\tau = F \cdot R$$

$$\tau = 10 \text{ N} \times 0.2 \text{ m} = 2 \text{ N m}$$

$$I = \frac{\tau}{\alpha}$$

$$I = \frac{2}{2}$$

$$I = 1 \text{ kg m}^2$$

Create a FREE account and get:

Predict your JEE Main percentile, rank & performance in seconds

Check Your Score Now

Ask our AI anything

AI can make mistakes. Please verify important information.

Download resources