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

Answer: (B)

$\frac {L}{4}$

Answer 2

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) $\frac {L}{4}$.