91Ó°ÊÓ

A cylinder of radius= R,

length = L, and

mass = M is released from rest on a slope

angle of inclination= θ

Lets first draw the free body diagram as below

Equate horizontal and vertical force

As net force in perpendicular direction is zero so

N = mg cosθ ............................(1)

and

mg sinθ - f_s = ma ........................(2)

Where f_sis frictional force

Find the torque acting on cylinder.

As mass passes through axis of center so only frictional force will cause the torque

$\mathrm{τ}=f_s×R$

Torque is also expressed as

$\mathrm{Ï„}=I\mathrm{Î\pm }$

Substitute $I=\frac{M{R}^2}{2}\mathrm{and}\mathrm{Î\pm }=ar$

$R{f}_s=\frac{M{R}^2}{2}\mathrm{Î\pm }$

Upon simplification

$f_s=Ma/2$

Now substitute the value of f_s in equation (1)

$$\begin{gathered} Mg\sin \mathrm{θ}-M\frac{a}{2}=Ma \\ Mg\sin \mathrm{θ}=\frac{3}{2}Ma........................\left(3\right) \end{gathered}$$

We know frictional force is $f_s={\mathrm{μ}}_{s}N$

So ${\mathrm{μ}}_s=\frac{f_s}{N}$

Substitute the value from equation(1) and (3)

$$\begin{gathered} \left({\mathrm{μ}}_s\right)=\frac{\left(\frac{M_g\sin \mathrm{θ}}{3}\right)}{Mg\cos \mathrm{θ}} \\ {\mathrm{μ}}_s=\frac{\tan \mathrm{θ}}{3} \end{gathered}$$