A solid metallic cylinder of mass $m=1 \text{kg}$ and radius  $R=20 \text{cm}$ is free to roll (without sliding) over the inclined surface of a wooden wedge of mass $=0.28 \text{kg}$. Surface of wedge is inclined at ${37}^{\circ }$  with the horizontal and the wedge lies on a smooth horizontal floor.  $g\left(10{\text{\hspace{0.17em}m/s}}^2\right)$
When the system is released from rest, 

 For Cylinder 
(with respect to wedge)

$mg\sin \theta +m{a}_0\cos \theta -f=ma$   … (1) (Along incline)
$m{a}_0\cos \theta +{\rm N}=mg\cos \theta$   … (2)  (1 to incline)
$a=R\propto$         … (3)
$fR=I\propto$      … (4) (About C.M. of Cylinder)
For wedge

$N\sin \theta -f\cos \theta =M{a}_0$   .... (5)

$$\begin{gathered} a_0=37.5{\text{\hspace{0.17em}\hspace{0.17em}m/s}}^2 \\ \alpha =30{\text{\hspace{0.17em}rad/s}}^2 \\ N=5.75,f=3N \end{gathered}$$

Interaction force $=\sqrt{N^2+f^2}\approx 6.49$