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

Step 1: Understanding the Concept:
We use the rotational analogues of the kinematic equations to find the angular acceleration and total angular displacement. Then, Newton's second law for rotation (\(\tau = I\alpha\)) yields the required torque.
Step 2: Key Formula or Approach:
1. \(\omega_i = \frac{2\pi N}{60}\)
2. \(\omega_f = \omega_i + \alpha t\)
3. \(\tau = I\alpha\) (where \(I = \frac{2}{5}MR^2\) for a solid sphere)
4. \(\theta = \left(\frac{\omega_i + \omega_f}{2}\right) t\) and \(n = \frac{\theta}{2\pi}\)
Step 3: Detailed Explanation:
Convert initial RPM to rad/s: \[ \omega_i = 1200 \times \frac{2\pi}{60} = 40\pi \, \text{rad/s} \] Since it comes to rest, \(\omega_f = 0\). Find \(\alpha\): \[ 0 = 40\pi + \alpha(10) \implies \alpha = -4\pi \, \text{rad/s}^2 \] Calculate the moment of inertia (\(I\)): \[ I = \frac{2}{5}MR^2 = \frac{2}{5} (5) (0.04)^2 = 2 \times 0.0016 = 0.0032 \, \text{kg}\cdot\text{m}^2 \] Magnitude of torque (\(\tau\)): \[ \tau = I|\alpha| = 0.0032 \times 4\pi = 0.0128\pi \, \text{N-m} \] Using \(\pi \approx 3.14\), \(\tau \approx 0.0402 \, \text{N-m} \approx 0.04 \, \text{N-m}\). Calculate the total angular displacement (\(\theta\)): \[ \theta = \frac{\omega_i t}{2} = \frac{40\pi \times 10}{2} = 200\pi \, \text{rad} \] Convert radians to revolutions: \[ \text{Revolutions} = \frac{\theta}{2\pi} = \frac{200\pi}{2\pi} = 100 \, \text{Rev.} \]
Step 4: Final Answer:
The torque is \(0.04 \, \text{N-m}\) and it makes \(100\) revolutions.