A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm . By applying an external torque of 25 π Nm for 40 s , the speed increases to 2100 rpm . The diameter of the disk is ___________ m.

Solution

Physics sir · 4 steps

Torque times time is the change in angular momentum, so first find I, then read off the radius.

1. Change in speed. The speed goes from 1800 rpm to 2100 rpm, a rise of 300 rpm. In rad/s that is Δω = 300 × 2π/60 = 10π rad s⁻¹.

   τ t = I Δω

2. Find I. The angular impulse is 25π × 40 = 1000π N m s. So I = 1000π/10π = 100 kg m².
3. Diameter axis. For a thin disk about a diameter, I = 1/4MR², not 1/2MR². With M = 1 kg, R²/4 = 100, so R² = 400 and R = 20 m.
4. So. The diameter is 2R = 40 m.

The common slip: 20. Stopping at the radius gives 20, not 40. The question asks for the diameter, so double R.

The idea to keep. A constant torque for a time t changes L by τ t. Convert rpm to rad/s, and check the axis: a disk about its diameter has I = 1/4MR².

Answer from the NTA final answer key. Solution by PrepWiser.