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:

When angular velocity is constant, the angular momentum is directly proportional to the square of the radius of the circular path.

Answer 1

The Correct Option is C

Step 1: Recall the formula.
Angular momentum of a particle is $L = I\omega$ with $I = mr^2$, so $L = mr^2\omega$.
Step 2: Identify the change.
The radius triples, $r \to 3r$, while mass and angular velocity stay the same.
Step 3: New moment of inertia.
$I_{new} = m(3r)^2 = 9mr^2$, because the radius is squared.
Step 4: New angular momentum.
$L_{new} = I_{new}\omega = 9mr^2\omega$.
Step 5: Compare with the original.
Since $L = mr^2\omega$, this is $L_{new} = 9L$.
Step 6: Conclude.
The angular momentum becomes nine times the original.
\[ \boxed{9L} \]