The moment of inertia of a solid cylinder abo | vedclass

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(D) For a solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2}MR^2$,where $M$ is the mass and $R$ is the radius.Since the c

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$\frac{3}{2}I\omega^2$

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(D) For a solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2}MR^2$,where $M$ is the mass and $R$ is the radius.Since the cylinder rolls without slipping,the linear velocity $v$ is related to the angular velocity $\omega$ by $v = R\omega$.The total kinetic energy $(K.E.)$ of a rolling body is the sum of its translational kinetic energy and rotational kinetic energy:$K.E. = K.E._{trans} + K.E._{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$.Substituting $v = R\omega$:$K.E. = \frac{1}{2}M(R\omega)^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}MR^2\omega^2 + \frac{1}{2}I\omega^2$.Since $I = \frac{1}{2}MR^2$,we have $MR^2 = 2I$.Substituting this into the equation:$K.E. = \frac{1}{2}(2I)\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2 + \frac{1}{2}I\omega^2 = \frac{3}{2}I\omega^2$.

The moment of inertia of a solid cylinder about its axis is $I$. It is allowed to roll down an incline without slipping. If its angular velocity at the bottom is $\omega$,then the total kinetic energy $(K.E.)$ of the cylinder will be:

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