A solid sphere at rests rolls down an inclined plane of vertical height h without sliding. Its speed on reaching the bottom of plane is ( g = acceleration due to gravity)

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TestHub · Accepted Answer

By conservation of mechanical energy, the potential energy at the top is converted into translational and rotational kinetic energy at the bottom.

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$

For a solid sphere, the moment of inertia is $I = \frac{2}{5} m r^{2}$ .

For rolling without slipping, $ω = \frac{v}{r}$ .

Substituting these values into the energy equation:

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m r^{2}) (\frac{v}{r})^{2}$

$m g h = \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2}$

$m g h = \frac{7}{10} m v^{2}$

$v^{2} = \frac{10 g h}{7}$

$v = (\frac{10 g h}{7})^{\frac{1}{2}}$

Answer: $(\frac{10 g h}{7})^{\frac{1}{2}}$

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