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

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Question

(A) The angular momentum $L$ of a particle of mass $m$ moving in a circular path of radius $r$ with angular velocity $\omega$ is given by the formula

Vedclass · Accepted Answer

$L/4$

Vedclass · Answer

(A) The angular momentum $L$ of a particle of mass $m$ moving in a circular path of radius $r$ with angular velocity $\omega$ is given by the formula $L = I\omega$, where $I$ is the moment of inertia.For a point mass $m$ at a distance $r$ from the center, the moment of inertia is $I = mr^2$.Therefore, the angular momentum is $L = mr^2\omega$.In the initial state, $L_1 = mr^2\omega$.In the final state, the length of the string (radius) is halved, so $r' = r/2$, and the angular velocity $\omega$ remains constant.The new angular momentum $L_2$ is given by $L_2 = m(r')^2\omega$.Substituting $r' = r/2$, we get $L_2 = m(r/2)^2\omega = m(r^2/4)\omega = (1/4)mr^2\omega$.Since $L_1 = mr^2\omega$, we have $L_2 = L_1/4 = L/4$.

$A$ 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 length of the string is now halved, keeping the angular velocity the same, then the angular momentum will be:

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