Question

The figure shows a hemisphere of radius 4R. A solid sphere of radius R is released from position ‘P’. It rolls without slipping along the inner surface of the hemisphere. Linear speed of its centre of mass when the ball is at lowest position is:
a) \sqrt{\frac{30gR}{7}}
b) \sqrt{\frac{24gR}{5}}
c) \sqrt{\frac{40gR}{9}}
d) \sqrt{6gR}

Answer 1

Given:

The figure shows a hemisphere of radius 4R. A solid sphere of radius R is released from position ‘P’. It rolls without slipping along the inner surface of the hemisphere.

To find:

Linear speed of ball at lowest position ?

Calculation:

In this type of questions , you need to apply the principle of CONSERVATION OF MECHANICAL ENERGY , as the whole Potential Energy (at the height) will be converted to rolling Kinetic Energy.

Now , the net change in height will be (4R - R) = 3R:

$\therefore \: \Delta KE = - \Delta PE$

$\implies \: \dfrac{1}{2} m {v}^{2} \bigg(1 + \dfrac{ {k}^{2} }{ {R}^{2} } \bigg) - 0 = - \bigg \{ 0 - mg(3R) \bigg \}$

$\implies \: \dfrac{1}{2} m {v}^{2} \bigg(1 + \dfrac{ {k}^{2} }{ {R}^{2} } \bigg) =mg(3R)$

$\implies \: \dfrac{1}{2} {v}^{2} \bigg(1 + \dfrac{ {k}^{2} }{ {R}^{2} } \bigg) =g(3R)$

$\implies \: {v}^{2} \bigg(1 + \dfrac{ {k}^{2} }{ {R}^{2} } \bigg) =6gR$

$\implies \: {v}^{2} \bigg(1 + \dfrac{ 2 }{ 5} \bigg) =6gR$

$\implies \: {v}^{2} \bigg( \dfrac{ 7 }{ 5} \bigg) =6gR$

$\implies \: {v}^{2} = \dfrac{30gR}{7}$

$\implies \: v = \sqrt{\dfrac{30gR}{7}}$

So, final answer is:

$\boxed{ \bf\: v = \sqrt{\dfrac{30gR}{7}}}$