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(A) The kinetic energy $E$ of a particle in rotational motion is given by $E = \frac{1}{2} I \omega^2$.Angular momentum $L$ is defined as $L = I \omeg

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$\frac{L^2}{2 mr^2}$

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(A) The kinetic energy $E$ of a particle in rotational motion is given by $E = \frac{1}{2} I \omega^2$.Angular momentum $L$ is defined as $L = I \omega$,which implies $L^2 = I^2 \omega^2$.Substituting $\omega^2 = \frac{L^2}{I^2}$ into the kinetic energy equation,we get $E = \frac{1}{2} I \left( \frac{L^2}{I^2} \right) = \frac{L^2}{2I}$.For a particle of mass '$m$' moving in a circle of radius '$r$',the moment of inertia about the axis passing through the center is $I = mr^2$.Substituting $I = mr^2$ into the expression for kinetic energy,we get $E = \frac{L^2}{2(mr^2)} = \frac{L^2}{2 mr^2}$.

$A$ particle of mass '$m$' is performing uniform circular motion along a circular path of radius '$r$'. Its angular momentum about the axis passing through the centre and perpendicular to the plane is '$L$'. The kinetic energy of the particle is

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