Question

A mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its angular momentum is L. If the string is now halved, keeping angular velocity same, then the angular momentum will be

• A. L
• B. \(\frac {L}{4}\)
• C. 2L
• D. \(\frac {L}{2}\)

Answer 1

The Correct Option is B

The angular momentum (L) of an object rotating about an axis is given by the equation: 
L = Iω 
When the string is halved while keeping the angular velocity constant, the moment of inertia changes. The moment of inertia for a mass rotating in a circular path is given by: 
I = mr² 
Since the radius is halved, the new moment of inertia (I') can be calculated as: 
I' = m\((\frac {r}{2})^2\)= m\((\frac {r^2}{4})\) = \((\frac {1}{4})\)mr² 
Substituting the new moment of inertia into the angular momentum equation: 
L' = I'ω = \(\frac {1}{4}\)mr²ω = \(\frac {1}{4}\)L 
Therefore, the angular momentum is reduced to one-fourth of its original value. 
The answer is (B) \(\frac {L}{4}\).