Question

A particle executes uniform circular motion with angular momentum \(L\). Its rotational kinetic energy becomes half when the angular frequency is doubled. Its new angular momentum is

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

Hint:

Rotational kinetic energy varies as the square of angular momentum when moment of inertia is constant.

Answer 1

The Correct Option is D

Step 1: Expression for rotational kinetic energy.
Rotational kinetic energy is given by \[ K = \frac{L^2}{2I}. \]
Step 2: Condition given in the question.
When angular frequency is doubled, the rotational kinetic energy becomes half: \[ \frac{L'^2}{2I} = \frac{1}{2}\cdot\frac{L^2}{2I}. \]
Step 3: Solving for new angular momentum.
\[ L'^2 = \frac{L^2}{4} \Rightarrow L' = \frac{L}{4}. \]
Step 4: Conclusion.
The new angular momentum of the particle is \( \dfrac{L}{4} \).