A stone of mass m,tied to the end of a string | vedclass

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(D) The angular momentum $L$ of the stone is given by $L = mvr = mr^2\omega$,where $m$ is mass,$v$ is linear velocity,$r$ is the radius,and $\omega$ i

Vedclass · Accepted Answer

$-3$

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(D) The angular momentum $L$ of the stone is given by $L = mvr = mr^2\omega$,where $m$ is mass,$v$ is linear velocity,$r$ is the radius,and $\omega$ is angular velocity.Since the angular momentum is constant,we have $mr^2\omega = C$ (where $C$ is a constant).This implies $\omega = \frac{C}{mr^2}$.The tension $T$ in the string provides the necessary centripetal force for circular motion:$T = m\omega^2r$.Substituting the expression for $\omega$ into the tension formula:$T = m \left( \frac{C}{mr^2} \right)^2 r = m \left( \frac{C^2}{m^2r^4} \right) r = \frac{C^2}{mr^3} = \left( \frac{C^2}{m} \right) r^{-3}$.Comparing this with $T = Ar^n$,we identify $A = \frac{C^2}{m}$ and $n = -3$.

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