91影视

A cylinder of radius= R,

length = L, and

mass = M is released from rest on a slope

angle of inclination= 胃

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 cos胃 ............................(1)

and

mg sin胃 - 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

$蟿=f_s脳R$

Torque is also expressed as

$蟿=I伪$

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

$R{f}_s=\frac{M{R}^2}{2}伪$

Upon simplification

$f_s=Ma/2$

Now substitute the value of f_s in equation (1)

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

We know frictional force is $f_s={渭}_{s}N$

So ${渭}_s=\frac{f_s}{N}$

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

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