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 incline, the velocity of the disc at bottom would h

Question

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 incline, the velocity of the disc at bottom would have been

TestHub · Accepted Answer

For a body sliding down a smooth incline of height $h$ , the conservation of energy gives:

$m g h = \frac{1}{2} m V^{2}$

$V^{2} = 2 g h$

For a disc rolling down the same incline, the total kinetic energy at the bottom is the sum of translational and rotational kinetic energies:

$m g h = \frac{1}{2} m v^{2} (1 + \frac{K ^{2}}{R ^{2}})$

For a disc, the radius of gyration $K$ is given by $\frac{K ^{2}}{R ^{2}} = \frac{1}{2}$ .

$m g h = \frac{1}{2} m v^{2} (1 + \frac{1}{2})$

$m g h = \frac{3}{4} m v^{2}$

$v^{2} = \frac{4 g h}{3}$

Substituting $2 g h = V^{2}$ into the above equation:

$v^{2} = \frac{2}{3} (2 g h) = \frac{2}{3} V^{2}$

$v = V \frac{2}{3}$

Answer: $v \frac{2}{3}$

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