The portion AB of the wedge shown in the figu | vedclass

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(B) For a solid cylinder rolling without slipping,the total energy is conserved. Let the vertical height difference between $A$ and $B$ be $H$. At poi

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$5$

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(B) For a solid cylinder rolling without slipping,the total energy is conserved. Let the vertical height difference between $A$ and $B$ be $H$. At point $B$,the cylinder has reached the bottom of the rough incline. The translational kinetic energy $K_T$ and rotational kinetic energy $K_R$ are related by $K_T = 2K_R$ for a solid cylinder $(I = \frac{1}{2}mR^2)$. From energy conservation,$mgH = K_T + K_R = 2K_R + K_R = 3K_R$. Thus,$K_R = \frac{1}{3}mgH$ and $K_T = \frac{2}{3}mgH$. From $B$ to $C$,the surface is smooth,so there is no friction to exert a torque. Thus,the angular velocity $\omega$ and rotational kinetic energy $K_R$ remain constant. The potential energy lost from $B$ to $C$ is converted entirely into translational kinetic energy. If the vertical height difference between $B$ and $C$ is also $H$,the additional translational kinetic energy gained is $mgH$. Total translational kinetic energy at $C$ is $K_{T,C} = K_{T,B} + mgH = \frac{2}{3}mgH + mgH = \frac{5}{3}mgH$. The rotational kinetic energy at $C$ is $K_{R,C} = K_{R,B} = \frac{1}{3}mgH$. The ratio is $\frac{K_{T,C}}{K_{R,C}} = \frac{\frac{5}{3}mgH}{\frac{1}{3}mgH} = 5$.

The portion $AB$ of the wedge shown in the figure is rough and $BC$ is smooth. $A$ solid cylinder rolls without slipping from $A$ to $B$. The ratio of translational kinetic energy to rotational kinetic energy,when the cylinder reaches point $C$,is:

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