Question

A wheel is rotating with angular speed \( \omega \). Another identical wheel with moment of inertia three times the first is initially at rest. It is suddenly coupled to the first. What fraction of the original kinetic energy is lost?

• A. \( 0.50 \)
• B. \( 0.25 \)
• C. \( 0.66 \)
• D. \( 0.75 \)

Hint:

In sudden coupling (no external torque), angular momentum is conserved, but energy is not.
Use \( I_1 \omega_1 = (I_1 + I_2) \omega_f \) to find final speed.
Compare energies to find the fraction lost.

Answer 1

The Correct Option is D

Step 1: Initial kinetic energy: \[ E_i = \frac{1}{2} I \omega^2 \] Step 2: Angular momentum conserved: \[ I \omega = (I + 3I) \omega_f \Rightarrow \omega_f = \frac{\omega}{4} \] Step 3: Final energy: \[ E_f = \frac{1}{2} (4I) \cdot \left( \frac{\omega}{4} \right)^2 = 2I \cdot \frac{\omega^2}{16} = \frac{I \omega^2}{8} \] So, fraction of energy lost: \[ \frac{E_i - E_f}{E_i} = \frac{\frac{1}{2} I \omega^2 - \frac{1}{8} I \omega^2}{\frac{1}{2} I \omega^2} = \frac{3}{4} = 0.75 \]