If L, M,and P are the angular momentum,mass,a | vedclass

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(C) The kinetic energy $(K.E.)$ of a particle is given by $K.E. = \frac{1}{2} M v^2$.We know that the linear momentum $P = Mv$,so $v = \frac{P}{M}$.Su

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

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(C) The kinetic energy $(K.E.)$ of a particle is given by $K.E. = \frac{1}{2} M v^2$.We know that the linear momentum $P = Mv$,so $v = \frac{P}{M}$.Substituting this into the kinetic energy formula: $K.E. = \frac{1}{2} M \left(\frac{P}{M}\right)^2 = \frac{P^2}{2M}$.For a particle rotating in a circle of radius $R$,the angular momentum $L$ is given by $L = MvR = PR$.Therefore,$P = \frac{L}{R}$.Substituting $P = \frac{L}{R}$ into the kinetic energy expression $K.E. = \frac{P^2}{2M}$,we get:$K.E. = \frac{(L/R)^2}{2M} = \frac{L^2}{2MR^2}$.

If $L, M$,and $P$ are the angular momentum,mass,and linear momentum of a particle respectively,which of the following represents the kinetic energy of the particle when the particle rotates in a circle of radius $R$?

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