A small particle of mass m starts sliding down from rest along the smooth surface of a fixed hollow hemisphere of mass M(4m). The distance of centre of mass of (particle+hemisphere) from centre O of hemisphere, when the particle separates from the surface of hemisphere is $\frac{\sqrt{69}}{5\theta }R.$ Find the value of  $\theta$.

Let the particle is separated when it is at the position $\theta$
Now, N=0
   $mg\cos \theta =\frac{m{v}^2}{R}$    $\Rightarrow v^2=Rg\cos \theta \mathrm{.......}\left(i\right)$
Decreases in PE=Increases in K.E  $\Rightarrow mgR(1-\cos \theta )=\frac{1}{2}m{v}^2\mathrm{......}\left(ii\right)$
From Eqs.(i) and (ii) , we get$\cos \theta =\frac{2}{3}$   
So,  $\sin \theta =\frac{\sqrt{5}}{3}$

Coordinates of particle,  $m=(x_1,y_1)=(\frac{\sqrt{5}}{3}R,\frac{2}{3}R)$
Coordinates of COM of hemisphere $=(x_2,y_2)=(0,\frac{R}{2})$
$\therefore$  X-coordinate of COM of (particle+hemisphere) is 
$X_{COM}=\frac{m\left(\frac{\sqrt{5}}{3}R\right)+4m\left(0\right)}{m+4m}=\frac{R}{3\sqrt{5}}$

$Y_{COM}=\frac{m\left(\frac{2}{3}R\right)+4m\left(\frac{R}{2}\right)}{m+4m}=\frac{8}{15}R$

Coordinate of point O=(0,0)
$\therefore$ Required distance= $\sqrt{{(0-\frac{R}{3\sqrt{5}})}^2+{(0-\frac{8R}{15})}^2}=\frac{\sqrt{69}}{15}R \Rightarrow \alpha =3$