A solid cylinder rolls down an inclined plane | vedclass

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(B) The rotational kinetic energy of the cylinder is $K_{rot} = \frac{1}{2} I \omega^2$.For a solid cylinder,the moment of inertia about its central a

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$1:3$

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(B) The rotational kinetic energy of the cylinder is $K_{rot} = \frac{1}{2} I \omega^2$.For a solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2} MR^2$ and the angular velocity is $\omega = \frac{v}{R}$.Substituting these,we get $K_{rot} = \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2 = \frac{1}{4} Mv^2$.The translational kinetic energy is $K_{trans} = \frac{1}{2} Mv^2$.The total kinetic energy is $K_{total} = K_{rot} + K_{trans} = \frac{1}{4} Mv^2 + \frac{1}{2} Mv^2 = \frac{3}{4} Mv^2$.The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{4} Mv^2}{\frac{3}{4} Mv^2} = \frac{1}{3}$.

$A$ solid cylinder rolls down an inclined plane from a height $h$. At any moment,the ratio of rotational kinetic energy to the total kinetic energy is:

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