Question

A particle executes uniform circular motion with an angular momentum $L$. If its kinetic energy is doubled and the angular frequency is halved, then its angular momentum becomes:

• A. $2\text{ L}$
• B. $4\text{ L}$
• C. $L/2$
• D. $L/4$

Hint:

Using the relationship $L = \frac{2K}{\omega}$, we can see how the variables scale: $$L \propto \frac{K}{\omega}$$ If $K$ is multiplied by 2 and $\omega$ is divided by 2, the scaling factor for $L$ is: $$\text{Factor} = \frac{2}{1/2} = 4$$ This allows you to find the answer without full algebraic substitution.

Answer 1

The Correct Option is B

Step 1: Write the two key formulas.
For circular motion, the kinetic energy and angular momentum are: \[ K = \frac{1}{2} I \omega^2, \qquad L = I \omega \] Here $I$ is the moment of inertia and $\omega$ is the angular frequency.

Step 2: Link energy, angular momentum and frequency.
From $L = I\omega$ we get $I = \dfrac{L}{\omega}$. Put this into the energy formula: \[ K = \frac{1}{2}\cdot \frac{L}{\omega}\cdot \omega^2 = \frac{1}{2} L \omega \] So we have the neat relation $K = \frac{1}{2} L \omega$.

Step 3: Solve for angular momentum.
Rearranging gives: \[ L = \frac{2K}{\omega} \] This tells us $L$ grows with energy and shrinks with frequency.

Step 4: Apply the changes.
The kinetic energy is doubled, so $K \to 2K$. The angular frequency is halved, so $\omega \to \frac{\omega}{2}$.

Step 5: Find the new angular momentum.
Put the new values in: \[ L_{new} = \frac{2(2K)}{\omega/2} = \frac{4K}{\omega/2} = \frac{8K}{\omega} \]

Step 6: Compare with the old value.
The old value was $L = \dfrac{2K}{\omega}$. So \[ L_{new} = 4 \times \frac{2K}{\omega} = 4L \] The angular momentum becomes four times larger. \[ \boxed{4\,L} \]