Question:
A circular disc of weight 500 N and radius 1 m is started from rest by a constant horizontal force of 25 N applied tangentially to the disc. The kinetic energy of the disc after \(t = 2\,s\) is:
A circular disc of weight 500 N and radius 1 m is started from rest by a constant horizontal force of 25 N applied tangentially to the disc. The kinetic energy of the disc after \(t = 2\,s\) is:
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For rotational motion problems, always use \( \tau = I\alpha \) and \( KE = \frac{1}{2}I\omega^2 \).
Updated On: Jun 19, 2026
- 50 J
- 75 J
- 100 J
- 25 J
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The Correct Option is A
Solution and Explanation
Step 1: Find mass of disc.
Weight \(W = 500\,N = mg\). Taking \(g = 10\,m\,s^{-2}\): \[ m = \frac{500}{10} = 50\,kg \]
Step 2: Moment of inertia of disc.
For a solid disc: \[ I = \frac{1}{2}MR^2 = \frac{1}{2}(50)(1)^2 = 25\,kg\,m^2 \]
Step 3: Find torque due to tangential force.
\[ \tau = F R = 25 \times 1 = 25\,N\,m \]
Step 4: Find angular acceleration.
\[ \alpha = \frac{\tau}{I} = \frac{25}{25} = 1\,rad\,s^{-2} \]
Step 5: Find angular velocity after 2 s.
Starting from rest: \[ \omega = \alpha t = 1 \times 2 = 2\,rad\,s^{-1} \]
Step 6: Find kinetic energy.
\[ KE = \frac{1}{2}I\omega^2 = \frac{1}{2}(25)(4) = 50\,J \]
Final Answer: \[ \boxed{50\,J} \]
Weight \(W = 500\,N = mg\). Taking \(g = 10\,m\,s^{-2}\): \[ m = \frac{500}{10} = 50\,kg \]
Step 2: Moment of inertia of disc.
For a solid disc: \[ I = \frac{1}{2}MR^2 = \frac{1}{2}(50)(1)^2 = 25\,kg\,m^2 \]
Step 3: Find torque due to tangential force.
\[ \tau = F R = 25 \times 1 = 25\,N\,m \]
Step 4: Find angular acceleration.
\[ \alpha = \frac{\tau}{I} = \frac{25}{25} = 1\,rad\,s^{-2} \]
Step 5: Find angular velocity after 2 s.
Starting from rest: \[ \omega = \alpha t = 1 \times 2 = 2\,rad\,s^{-1} \]
Step 6: Find kinetic energy.
\[ KE = \frac{1}{2}I\omega^2 = \frac{1}{2}(25)(4) = 50\,J \]
Final Answer: \[ \boxed{50\,J} \]
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