Question

A solid sphere of radius $4\text{ cm}$ and mass $5\text{ kg}$ is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of $1200\text{ rpm}$. It is brought to rest in $10\text{ s}$ by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are ______ and ______ respectively.

• A. $0.0128 \pi\text{ Nm}, 100$
• B. $0.0128 \pi\text{ Nm}, 200$
• C. $0.128 \pi\text{ Nm}, 100$
• D. $0.128 \pi\text{ Nm}, 200$

Hint:

Convert rpm to rad/s. Calculate I using (2/5)mR^2. Use angular kinematic equations to find alpha and theta. Then find torque using tau = I*alpha.

Answer 1

The Correct Option is A

Alternatively, we can use the Work-Energy Theorem for rotation. The work done by the torque is equal to the change in rotational kinetic energy.

Initial angular velocity $\omega_0 = 40\pi\text{ rad/s}$.
Final angular velocity $\omega = 0$.
Moment of inertia $I = \frac{2}{5} m R^2 = \frac{2}{5} \times 5 \times (0.04)^2 = 0.0032\text{ kg}\cdot\text{m}^2$.

Change in Kinetic Energy $\Delta K = \frac{1}{2} I \omega^2 - \frac{1}{2} I \omega_0^2$:
$$\Delta K = 0 - \frac{1}{2}(0.0032)(40\pi)^2 = -0.0016 \times 1600\pi^2 = -2.56\pi^2\text{ J}$$

Now, find the angular displacement $\theta$ from the average velocity:
$$\theta = \frac{\omega_0 + \omega}{2} \times t = 20\pi \times 10 = 200\pi\text{ rad}$$
Number of rotations $N = \frac{\theta}{2\pi} = \frac{200\pi}{2\pi} = 100$.

Work done by torque is $W = -\tau \times \theta$ (since torque opposes motion):
$$-2.56\pi^2 = -\tau \times (200\pi)$$
$$\tau = \frac{2.56\pi^2}{200\pi} = \frac{2.56\pi}{200} = 0.0128\pi\text{ Nm}$$
Both methods give a torque of $0.0128\pi\text{ Nm}$ and $100$ rotations.