Question

A small solid sphere of mass $m$ and radius $r$ rolls without slipping on the inside a large hemisphere of radius $R$ . The axis of symmetry of the hemisphere is vertical. Sphere starts at the top from rest. The sphere will exert a normal force on the hemisphere at its bottom equal to $\left(n\right)\frac{17}{14}mg$. Then the value of $n$ is

Answer 1

Potential energy of sphere at top = mgR.
Kinetic energy of sphere at top = 0.

At the bottom, the sphere has only kinetic energy.
K.E of sphere at the bottom will be
$1/2{mv}_{cm}^2+1/2I_{cm}{\omega }^2=1/2{mv}_{cm}^2+1/2^{\star }2/5{mr}^2{\left(v_{cm}/r\right)}^2$
$1/2{mv}_{cm}^2+1/5{mv}_{cm}^2=7/10{mv}_{cm}^2$

KE of sphere at bottom $=7/10{mv}_{cm}^2=mgR$

Centrifugal force $={mv}_{cm}^2/R=mg\ast 10/7$
Writing force equation at bottom in vertical direction,

$N$ - $F_{centrifugal}$ = mg
$⟹$ $mg+m\frac{v_{cm}^2}{R}$ = $N$
On solving we get $N$ = $\frac{17}{7}mg$
$\therefore$ $n$ = 2