A mass tied to a string is whirled in a horiz | vedclass

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(B) The correct option is $B$.Concept: Angular momentum $L$ of a particle of mass $m$ moving in a circle of radius $r$ with angular velocity $\omega$

Vedclass · Accepted Answer

$\frac{L}{4}$

Vedclass · Answer

(B) The correct option is $B$.Concept: Angular momentum $L$ of a particle of mass $m$ moving in a circle of radius $r$ with angular velocity $\omega$ is given by $L = I\omega = mr^2\omega$.Given that the mass $m$ and angular velocity $\omega$ remain constant.Initially,$L = m\omega r^2$.When the length of the string is halved,the new radius becomes $r' = \frac{r}{2}$.The new angular momentum $L'$ is given by:$L' = m\omega(r')^2 = m\omega\left(\frac{r}{2}\right)^2$$L' = m\omega\left(\frac{r^2}{4}\right) = \frac{1}{4}(mr^2\omega)$$L' = \frac{L}{4}$.

$A$ 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 the angular velocity the same,then the new angular momentum will be:

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