Basic Rotational Motion problem

Force Diagram First we draw all the forces acting on the cylinder. Drawing mg and N is easy but to draw friction we need to decide whether friction is acting upwards or downwards. We can imagine that the cylinder will roll clockwise with increasing angular acceleration. Therefore friction must act upwards to provide cylinder clockwise angular acceleration. The magnitude of friction will be equal to $\mu$ N as this is the limiting case.

Now we have to draw the acceleration of the cylinder. The translational part of acceleration of center of mass will be downwards as Mgsin60 is the only force downwards along the incline plane. We can easily imagine the rotational part of acceleration of the cylinder which is clockwise.

Now applying F=ma we get, N=Mgcos60

Mgsin60- $\mu$ N =M(Acm)

Applying $\tau =I \alpha$ we get,

$\mu Nr={ I }_{ CM }\alpha$

Also if an object rolls without slipping on a stationary rigid surface then the point of contact must be momentarily at rest.

${ A }_{ CM }=r\alpha$

As the object is cylinder its moment of inertia is given by,

${ I }_{ CM }=\frac { M{ r }^{ 2 } }{ 2 }$

Solving all the equations we get,

$\mu =\frac { 1 }{ \sqrt { 3 } } =0.577$