Question

A solid sphere rolls down from the top of an inclined plane. Its velocity on reaching the bottom of the plane is v. When the same sphere slides down from the top of the plane, its velocity on reaching the bottom is v'. The ratio v'/v is

Answer 1

Given:

Sold sphere rolls down from top of inclined plane and achieves velocity v at bottom. The same sphere slides down the plane and achieves velocity v' .

To find:

Ratio of v'/v

Calculation:

For any rolling body , the velocity at the bottom of an inclined plane is given by:

$\boxed{ \sf{ v = \sqrt{ \dfrac{2gh}{1 + \frac{ {k}^{2} }{ {r}^{2} } } } }}$

k - radius of gyration

r - radius of object

Rolling of sphere down the plane:

$\therefore \: v = \sqrt{ \dfrac{2gh}{1 + \frac{ {k}^{2} }{ {r}^{2} } } }$

$= > \: v = \sqrt{ \dfrac{2gh}{1 + { (\frac{k}{r}) }^{2} } }$

$= > \: v = \sqrt{ \dfrac{2gh}{1 + { ( \sqrt{\frac{2}{5}}) }^{2} } }$

$= > \: v = \sqrt{ \dfrac{2gh}{1 + \frac{2}{5} } }$

$= > \: v = \sqrt{ \dfrac{2gh}{ ( \frac{7}{5} )} }$

$= > \: v = \sqrt{ \dfrac{10gh}{7 } }$

Sliding of sphere:

Velocity at bottom of inclined plane is v' , let it be represented as v2

$\therefore \: \dfrac{1}{2} m {(v2)}^{2} = mgh$

$= > v2 = \sqrt{2gh}$

Required Ratio will be :

$\therefore \: \dfrac{v2}{v} = \dfrac{ \sqrt{2gh} }{ \sqrt{ \frac{10gh}{7} } }$

$= > \: \dfrac{v2}{v} = \sqrt{ \dfrac{2gh}{ \frac{10gh}{7} } }$

$= > \: \dfrac{v2}{v} = \sqrt{ \dfrac{7}{ 5} }$

$= > \: \dfrac{v2}{v} = \dfrac{ \sqrt{7}}{ \sqrt{5}}$

So , final answer is :

$\boxed{ \sf{ \bold{v2 : v = \sqrt{7} : \sqrt{5}}}}$