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. 50 J
• B. 75 J
• C. 100 J
• D. 25 J

Hint:

For rotational motion problems, always use \( \tau = I\alpha \) and \( KE = \frac{1}{2}I\omega^2 \).

Answer 1

The Correct Option is A

Step 1: Mass from weight.
W = 500 N → m = 500/10 = 50 kg.

Step 2: Moment of inertia.
For a solid disc, I = ½ M R² = ½×50×1² = 25 kg m².

Step 3: Torque and angular acceleration.
τ = F R = 25×1 = 25 N m; α = τ/I = 25/25 = 1 rad s⁻².

Step 4: Angular velocity after 2 s.
ω = α t = 1×2 = 2 rad s⁻¹.

Step 5: Rotational kinetic energy.
KE = ½ I ω² = ½×25×4 = 50 J.