Question

A sphere of mass \(5\,kg\) and radius \(4\,cm\) is rotating about a fixed axis along its diameter with \(1200\,\text{rpm}\). To stop it in \(10\,s\), a torque is applied. Find magnitude of torque required and number of revolutions made before it stops respectively :-

• A. \(0.08\,\text{N-m}\) and \(50\) rev.
• B. \(0.04\,\text{N-m}\) and \(100\) rev.
• C. \(0.016\,\text{N-m}\) and \(200\) rev.
• D. \(0.2\,\text{N-m}\) and \(100\) rev.

Answer 1

The Correct Option is B

Concept: Moment of inertia of a solid sphere about diameter: \[ I = \frac{2}{5}MR^2 \] Torque relation: \[ \tau = I\alpha \] Rotational kinematics: \[ \omega_f = \omega_i + \alpha t \] Step 1: Convert angular speed to rad/s.} \[ \omega_i = 1200\,rpm \] \[ \omega_i = 1200 \times \frac{2\pi}{60} \] \[ \omega_i = 40\pi \,rad/s \]
Step 2: Find angular acceleration.} Final angular velocity \[ \omega_f = 0 \] \[ 0 = 40\pi + \alpha(10) \] \[ \alpha = -4\pi \,rad/s^2 \]
Step 3: Find moment of inertia.} \[ R = 4\,cm = 0.04\,m \] \[ I = \frac{2}{5}(5)(0.04)^2 \] \[ I = 2 \times 0.0016 \] \[ I = 0.0032\,kg\,m^2 \]
Step 4: Calculate torque.} \[ \tau = I\alpha \] \[ \tau = 0.0032 \times 4\pi \] \[ \tau \approx 0.04\,N-m \]
Step 5: Find angular displacement.} \[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \] \[ 0 = (40\pi)^2 + 2(-4\pi)\theta \] \[ \theta = 200\pi \] Number of revolutions: \[ n = \frac{\theta}{2\pi} \] \[ n = \frac{200\pi}{2\pi} \] \[ n = 100 \] Final Result \[ \tau = 0.04\,N-m, \quad n = 100 \]