Question

A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straight down the slope. If the slope were frictionless, the cylinder would slide down the slope without rotating. What minimum coefficient of static
friction is needed for the cylinder to roll down without slipping?

Short answer

Coefficient of friction is $\frac{\tan \theta }{3}$

Step-by-step solution

Given Information

A cylinder of radius= R,

length = L, and

mass = M is released from rest on a slope

angle of inclination= θ

Explanation

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

$\tau =f_s\times R$

Torque is also expressed as

$\tau =I\alpha$

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

$R{f}_s=\frac{M{R}^2}{2}\alpha$

Upon simplification

$f_s=Ma/2$

Now substitute the value of f_s in equation (1)

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

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

So ${\mu }_s=\frac{f_s}{N}$

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

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