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I am facing trouble with the following problem.

A solid cylinder of mass M=5kg rests on two supports of same height as shown.One support is rough and stationary while other is am smooth plank of same mass M=5kg placed on a smooth horizontal surface.Initially $\theta=37 degree and there is no slipping between the cylinder and stationary support.The support is released from rest.

The question is to find out the acceleration of cylinder just after release.

enter image description here

I drew a free body diagram of the body. enter image description here

Since friction force is the only tangential force acting on the sphere .So $fR=(1/2)mR^2( a/R)$ where $a$ is the acceleration of the cylinder.This gives $f=ma/2$.Also since one of the block is stationary.The component of forces along that direction should reduce to zero.So resolving forces in the perpendicular direction we get $$mgcos(\theta)-N_1sin(2\theta)-f=ma$$.All things are known except for $ N_1$.I am facing trouble in finding out $N_1$ in terms of the acceleration of the cylinder.Any hint to go ahead will be highly appreciated.Thanks.

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closed as off-topic by John Rennie, Bill N, sammy gerbil, Jon Custer, AccidentalFourierTransform Jan 20 '17 at 9:32

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I think you chose the pivot point at the center of the sphere??? I would choose it at the corner of the "rough" plank, since it's stationary. Then write a torque equation from that.

Then, that would give you the instantaneous rotation of the sphere about the corner of the rough edge. But knowing this you can break it into an x and y component.

After that, you can write Newton's equations for the x and y component. I think that should give you 3 unknowns and 3 equations.

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