Question

A particle performs rotational motion with an angular momentum \(L\). If frequency of rotation is doubled and its kinetic energy becomes one fourth, the angular momentum becomes.

• A. \(L\)
• B. \(\frac{L}{4}\)
• C. \(\frac{L}{8}\)
• D. \(\frac{L}{2}\)

Hint:

For rotational motion, \(K = \frac{L^2}{2I}\) and \(L = I\omega\). When \(\omega\) and \(K\) change, first find new \(I\), then new \(L\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Rotational kinetic energy \(K = \frac{L^2}{2I}\) and also \(K = \frac{1}{2}I\omega^2\). Angular momentum \(L = I\omega\).

Step 2: Key Formula or Approach:
Given frequency \(f\) is doubled \(\Rightarrow \omega\) is doubled. \(K\) becomes \(K/4\). Find new \(L\).

Step 3: Detailed Explanation:
\(K = \frac{1}{2}I\omega^2\). If \(\omega' = 2\omega\) and \(K' = \frac{K}{4}\), then \(\frac{1}{2}I'(2\omega)^2 = \frac{1}{4} \cdot \frac{1}{2}I\omega^2\).
\(\frac{1}{2}I' \cdot 4\omega^2 = \frac{1}{8}I\omega^2 \implies 2I'\omega^2 = \frac{1}{8}I\omega^2 \implies I' = \frac{I}{16}\).
Then \(L' = I'\omega' = \frac{I}{16} \times 2\omega = \frac{I\omega}{8} = \frac{L}{8}\).

Step 4: Final Answer:
Option (C) is correct.