A wedge of mass M = 10 kg and inclination $\alpha ={30}^0$ is free to move on a smooth horizontal plane. A uniform solid cylinder of mass m = 20 kg is placed on the rough inclined face of the wedge. Find the acceleration $(inm/s^2)$ of the centre of the cylinder relative to the wedge down the face. There is no slipping between the cylinder and the wedge

When the wedge has moved through a distance y, let the point of contact of cylinder have moved down the plane through x (with respect to wedge). Let F be the friction between the cylinder and the plane. Since there is no horizontal force on (wedge + cylinder system)

$M\frac{d^{2}y}{d{t}^2}+m(\frac{d^{2}y}{d{t}^2}-\cos \alpha \frac{d^{2}x}{d{t}^2})=0 \dots \dots \dots \dots .\left(1\right)$

Also,  $m(\frac{d^{2}y}{d{t}^2}\cos \alpha -\frac{d^{2}x}{d{t}^2})=F-mg\sin \alpha$ and $\frac{ma}{2}\frac{d^{2}x}{d{t}^2}=\frac{m{a}^2}{2}\frac{d^2\theta }{d{t}^2}=Fa,$ so that 

$2\frac{d^{2}y}{d{t}^2}\cos \alpha -3\left(\frac{d^{2}x}{d{t}^2}\right)=-2g\sin \alpha \dots \dots \dots \dots \dots .\left(2\right)$

Equation (1) and (2) implies acceleration of the centre of the cylinder down the face, and relative to the wedge, is $\frac{2g\sin \alpha (M+m)}{3M+m+2m{\sin }^2\alpha }=5m/s^2.$