1. Write the basic formulas for uniform circular motion

For a particle of mass $m$ moving in a circle of radius $r$ with angular frequency $\omega$:

$v = r\omega$

Kinetic energy,

$K = \frac{1}{2}mv^2 = \frac{1}{2}m r^2 \omega^2$

Angular momentum about the center,

$L = mvr = mr^2\omega$

2. Relate angular momentum to kinetic energy and angular frequency

From

$K = \frac{1}{2}m r^2 \omega^2$

we get

$mr^2 = \frac{2K}{\omega^2}$

Substitute into

$L = mr^2\omega$

So,

$L = \frac{2K}{\omega^2}\cdot \omega = \frac{2K}{\omega}$

Hence,

$L \propto \frac{K}{\omega}$

3. Apply the changed conditions

Given:

• angular frequency is doubled: $\omega' = 2\omega$
• kinetic energy is halved: $K' = \frac{K}{2}$

Now,

$L' = \frac{2K'}{\omega'}$

Substitute the new values:

$L' = \frac{2\left(\frac{K}{2}\right)}{2\omega}$

$L' = \frac{K}{2\omega}$

But original angular momentum was

$L = \frac{2K}{\omega}$

Therefore,

$\frac{L'}{L} = \frac{K/(2\omega)}{2K/\omega} = \frac{1}{4}$

So,

$L' = \frac{L}{4}$

4. Check options

• A: $\dfrac{L}{4}$ ✅
• B: $2L$ ❌
• C: $4L$ ❌
• D: $\dfrac{L}{2}$ ❌

Therefore, the correct option is A.