Question

A solid cylinder rolls without slipping down a ${30}^0$ slope. The minimum coefficient of friction needed to prevent slipping, will be:

• A. 0.192
• B. 0.18
• C. 0.15
• D. 0.2

Answer 1

The correct option is A 0.192

Lets consider, $R=$ radius of the cylinder

From free body diagram,

$mgsin{30}^0-f=ma$. . . . . .(1)

$W=N=mgcos{30}^0$. . . . . .(2)

Torque, $\tau =I\alpha =fR$

$I\alpha =fR$

$fR=\frac{m{R}^2}{2}\alpha$

$f=\frac{mR}{2}\alpha$. . . . . . . . .(3)

As we know that, $\alpha =\frac{a}{R}$. . . . . .(4)

Substitute equation (3) and (4) in equation (1), we get

$mgsin{30}^0-\frac{mR}{2}\frac{a}{R}=ma$

$gsin{30}^0=\frac{3a}{2}$

$a=\frac{2}{3}gsin{30}^0$. . . . . . .(5)

From equation (1),

$f=mgsin{30}^0-ma$

$f=mgsin{30}^0-\frac{2}{3}mgsin{30}^0$

$f=\frac{mgsin{30}^0}{3}$

$\mu N=\mu mgcos{30}^0=\frac{mgsin{30}^0}{3}$ (from equation 2)

$\mu =\frac{sin{30}^0}{3cos{30}^0}=\frac{tan{30}^0}{3}$

$\mu =0.192$

The correct option is A.