A solid spherical ball of mass m is released from the topmost point of the shown semi – spherical shell. The track is  sufficiently rough to enable pure rolling motion. The normal force between the ball and the shell at the lowest position is $\frac{p}{7}mg$ . The value of p is _________
      

From the law of conservation of mechanical energy,
Final energy = Translational kinetic energy + Rotational kinetic energy
Hence, Final energy  $=\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2$
$=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{R}^2\times {\left(\frac{v}{R}\right)}^2=\frac{7}{10}m{v}^2$
Hence, we finally obtain,  $mg(R-r)=\frac{7}{10}m{v}^2\mathrm{.......}\left(1\right)$
and  $N-mg=\frac{m{v}^2}{R-r}\mathrm{.......}\left(2\right)$
Solving Eqs (1) and (2), we get
$N=\frac{17}{7}mg$