Question

A particle attached to a string is rotating with a constant angular velocity and its momentum is $L$. If the string is halved and angular velocity is kept constant, the angular momentum will be

• A. $L/4$
• B. $L/2$
• C. $L$
• D. $2L$

Answer 1

The correct option is A $L/4$

As string length is halved,Let say $R$ is initial length, mass of the particle$=m$

Initial Angular Momentum$=L=m{R}^2{\omega }_0$

Final Angular Momentum$=m(\frac{R}{2}{)}^2{\omega }_0$

$=\frac{m{R}^2{\omega }_0}{4}$

$L_{final}=\frac{L}{4}$