Question
Show that the kinetic energy $K$ of a particle of mass $m$, moving in a circular path, is $K=L^{2} / 2 I,$ where $L$ is its angular momentum and $I$ is its moment of inertia about the center of the circle.
Step 1
Step 1: First, we define the kinetic energy of a particle of mass $m$ moving with velocity $v$ as $K = \frac{1}{2}mv^2$. Show more…
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Show that the kinetic energy of an object rotating about a fixed axis with angular momentum $L=I \omega$ can be written as $K=L^{2} / 2 I .$
A particle of mass $m$ is describing a circular path of radius $r$ with uniform speed. If $L$ is the angular momentum of the particle about the axis of the circle, the kinetic energy of the particle is given by [CPMT 1995] (a) $L^{2} / m r^{2}$ (b) $L^{2} / 2 m r^{2}$ (c) $2 L^{2} / m r^{2}$ (d) $m r^{2} L$
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