Question

The angular momentum of a rotating body is '\(L\)'. When frequency is tripled and kinetic energy is made one-third, the new angular momentum becomes

• A. \(\frac{1}{9} L\)
• B. \(\frac{1}{3} L\)
• C. \(6 L\)
• D. \(9 L\)

Hint:

Remember $K.E. = \frac{1}{2} L \omega$, which is analogous to $K.E. = \frac{1}{2} p v$ in linear motion.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The angular momentum (\(L\)) and kinetic energy (\(K\)) of a rotating body are related to its frequency (\(f\)). We need to find the new \(L\) after changes to \(f\) and \(K\).
Step 2: Key Formula or Approach:
Rotational Kinetic Energy: \(K = \frac{1}{2} I \omega^2\)
Angular Momentum: \(L = I \omega\)
Combining these: \(K = \frac{1}{2} L \omega\) or \(L = \frac{2K}{\omega}\)
Angular frequency: \(\omega = 2\pi f\)
Step 3: Detailed Explanation:
Initial state: Angular momentum \(L\), Kinetic Energy \(K\), Frequency \(f\).
We have \(L = \frac{2K}{2\pi f} = \frac{K}{\pi f}\).
Final state:
New kinetic energy \(K' = \frac{1}{3} K\)
New frequency \(f' = 3f\)
The new angular momentum \(L'\) is:
\[ L' = \frac{K'}{\pi f'} = \frac{K/3}{\pi (3f)} \]
\[ L' = \frac{1}{9} \left( \frac{K}{\pi f} \right) \]
Substituting \(L = \frac{K}{\pi f}\):
\[ L' = \frac{1}{9} L \]
Step 4: Final Answer:
The new angular momentum is \(\frac{1}{9} L\).