Question

A wheel is rotating freely at some angular speed. A second wheel initially at rest and with thrice the rotational inertia of the first, is suddenly coupled to the first wheel. The fraction of the original rotational kinetic energy lost is

• A. 0.50
• B. 0.25
• C. 0.66
• D. 0.75

Hint:

When rotating bodies are coupled together, angular momentum is conserved but rotational kinetic energy is generally not conserved. The loss occurs due to internal friction during coupling.

Answer 1

The Correct Option is D

Step 1: Find final angular speed using angular momentum conservation.
Let first wheel: inertia \( I \), angular speed \( \omega \). Second wheel: inertia \( 3I \), initially at rest. After coupling: \( I\omega = (I+3I)\omega_f \Rightarrow \omega_f = \omega/4 \).

Step 2: Compute the fractional KE lost.
Initial KE \( = \frac{1}{2}I\omega^2 \). Final KE \( = \frac{1}{2}(4I)(\omega/4)^2 = \frac{I\omega^2}{8} \). Fraction lost \( = 1 - \frac{1/8}{1/2} = 1 - \frac{1}{4} \) \[ \boxed{0.75} \]