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 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:
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Angular momentum of a point particle in circular motion is proportional to the square of its radius if the angular velocity is constant.
Updated On: Jun 9, 2026
- \( 3L \)
- \( 6L \)
- \( 9L \)
- \( L/3 \)
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The Correct Option is C
Solution and Explanation
Concept:
Angular momentum \( L \) is given by \( L = I\omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a particle of mass \( m \) in a circular path of radius \( r \), \( I = mr^2 \).
Step 1: Express initial angular momentum.
$$ L_{initial} = I_{initial} \omega = (mr^2) \omega = L $$
Step 2: Determine the new angular momentum after change.
The radius is tripled (\( r' = 3r \)), and \(\omega\) remains constant. $$ L_{new} = I_{new} \omega = (m(3r)^2) \omega $$ $$ L_{new} = (m \cdot 9r^2) \omega = 9 \cdot (mr^2 \omega) $$
Step 3: Relate to initial L.
$$ L_{new} = 9 \cdot L_{initial} = 9L $$ $$\boxed{9L}$$
Step 1: Express initial angular momentum.
$$ L_{initial} = I_{initial} \omega = (mr^2) \omega = L $$
Step 2: Determine the new angular momentum after change.
The radius is tripled (\( r' = 3r \)), and \(\omega\) remains constant. $$ L_{new} = I_{new} \omega = (m(3r)^2) \omega $$ $$ L_{new} = (m \cdot 9r^2) \omega = 9 \cdot (mr^2 \omega) $$
Step 3: Relate to initial L.
$$ L_{new} = 9 \cdot L_{initial} = 9L $$ $$\boxed{9L}$$
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