Solution: Question 17
Diagram: Pulling the String
Step 1: Conservation of Angular Momentum
The tension force passes through the center hole, so it exerts zero torque on the disc. Angular momentum \( L \) is conserved.
Kinetic Energy can be expressed as \( K = \frac{L^2}{2I} = \frac{L^2}{2mR^2} \).
Step 2: Final Kinetic Energy
Initial KE: \( K_0 \propto \frac{1}{R^2} \).
Final Radius: \( R’ = R/\eta \).
Final KE: \( K_f \propto \frac{1}{(R/\eta)^2} = \eta^2 \left( \frac{1}{R^2} \right) = \eta^2 K_0 \).
Step 3: Work Done
By the Work-Energy Theorem, the work done by the pulling force equals the change in kinetic energy: \[ W = K_f – K_i = \eta^2 K_0 – K_0 = (\eta^2 – 1)K_0 \]
Answer: (c) \( W = (\eta^2 – 1)K_0 \)