A solid sphere rolls down from the top of an inclined plane. On reaching the bottom of the plane, its velocity is ' V 1 '. When the same sphere slides down from the top of the same plane o

Question

A solid sphere rolls down from the top of an inclined plane. On reaching the bottom of the plane, its velocity is ' V 1 ​ '. When the same sphere slides down from the top of the same plane of same height, its velocity on reaching the bottom is ' V 2 ​ '. The ratio V 1 ​ : V 2 ​ is (neglect friction)

TestHub · Accepted Answer

When a solid sphere rolls down an inclined plane of height $h$ , by conservation of mechanical energy:

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

For a solid sphere, the moment of inertia is $I = \frac{2}{5} m r^{2}$ and for pure rolling, $ω = \frac{V _{1}}{r}$ .

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

$m g h = \frac{1}{2} m V_{1}^{2} + \frac{1}{5} m V_{1}^{2} = \frac{7}{10} m V_{1}^{2}$

$V_{1} = \frac{10 g h}{7}$

When the sphere slides down the same plane without friction, it only acquires translational kinetic energy. By conservation of mechanical energy:

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

$V_{2} = 2 g h$

The ratio of the velocities is:

$\frac{V _{1}}{V _{2}} = \frac{\frac{10 g h}{7}}{2 g h} = \frac{10}{7 \times 2} = \frac{5}{7}$

Thus, $V_{1} : V_{2} = 5 : 7$ .

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