A body of mass m slides down an incline and r | vedclass

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(D) Case $1$: For a sliding body,the potential energy is converted into translational kinetic energy: $\frac{1}{2} m V^2 = mgh$ ... $(i)$ Case $2$: Fo

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$V \sqrt{\frac{2}{3}}$

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(D) Case $1$: For a sliding body,the potential energy is converted into translational kinetic energy: $\frac{1}{2} m V^2 = mgh$ ... $(i)$ Case $2$: For a rolling disc,the potential energy is converted into both translational and rotational kinetic energy: $\frac{1}{2} m (v')^2 + \frac{1}{2} I \omega^2 = mgh$ For a disc,the moment of inertia is $I = \frac{1}{2} mR^2$ and the rolling condition is $\omega = \frac{v'}{R}$. Substituting these into the energy equation: $\frac{1}{2} m (v')^2 + \frac{1}{2} (\frac{1}{2} mR^2) (\frac{v'}{R})^2 = mgh$ $\frac{1}{2} m (v')^2 + \frac{1}{4} m (v')^2 = mgh$ $\frac{3}{4} m (v')^2 = mgh$ ... $(ii)$ Equating $(i)$ and $(ii)$: $\frac{1}{2} m V^2 = \frac{3}{4} m (v')^2$ $V^2 = \frac{3}{2} (v')^2$ $(v')^2 = \frac{2}{3} V^2$ $v' = V \sqrt{\frac{2}{3}}$

$A$ body of mass $m$ slides down an incline and reaches the bottom with a velocity $V$. If the same mass were in the form of a disc which rolls down this same incline,the velocity of the disc at the bottom would be

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$Assertion$ : The velocity of a body at the bottom of an inclined plane of given height is more when it slides down the plane,compared to when it rolls down the same plane.
$Reason$ : In rolling down,a body acquires both kinetic energy of translation and rotation.

What is the acceleration parallel to the surface of an inclined plane for a rolling body? Also,write the equation for the friction force acting parallel to the surface of the slope for a rolling body.

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