Question

A particle is moving in a circular path with constant angular velocity. Its initial angular momentum is L. If the radius of the circle is tripled by keeping angular velocity same, the new angular momentum is:

• A. \( 3L \)
• B. \( 6L \)
• C. \( 9L \)
• D. \( L/3 \)

Hint:

Angular momentum of a point particle in circular motion is proportional to the square of its radius if the angular velocity is constant.

Answer 1

The Correct Option is C

Step 1: Write angular momentum.
For a particle going in a circle, $L = I\omega$, where the moment of inertia is $I = mr^2$. So $L = mr^2\omega$.
Step 2: Note what changes.
The radius is tripled, $r \to 3r$, while the mass $m$ and the angular velocity $\omega$ are held fixed.
Step 3: New moment of inertia.
$I_{new} = m(3r)^2 = 9mr^2 = 9I$. The radius being squared turns the factor $3$ into $9$.
Step 4: New angular momentum.
$L_{new} = I_{new}\,\omega = 9mr^2\omega$.
Step 5: Compare with the original.
Since $L = mr^2\omega$, we have $L_{new} = 9\,(mr^2\omega) = 9L$.
Step 6: State the result.
Tripling the radius (at constant $\omega$) multiplies angular momentum by $9$.
\[ \boxed{9L} \]