A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane 5 x where the

Question

A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane 5 x ​ ​ where the value of x is

TestHub · Accepted Answer

By conservation of mechanical energy for a body rolling down an incline from rest:

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

Since $v = ω R$ , we have $m g h = \frac{1}{2} m v^{2} (1 + \frac{I}{m R ^{2}})$

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

For the circular ring, $I = m R^{2}$ , so $1 + \frac{I}{m R ^{2}} = 2$ .

$v_{r in g} = \frac{2 g h}{2} = g h$

For the solid sphere, $I = \frac{2}{5} m R^{2}$ , so $1 + \frac{I}{m R ^{2}} = \frac{7}{5}$ .

$v_{s p h ere} = \frac{2 g h}{7/5} = \frac{10 g h}{7}$

The ratio of their velocities is:

$\frac{v _{r in g}}{v _{s p h ere}} = \frac{g h}{\frac{10 g h}{7}} = \frac{7}{10} = \frac{3.5}{5}$

Comparing with $\frac{x}{5}$ , we get $x = 3.5$ .

Answer: $3.5$

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