Question

A solid cylinder of mass 'm' rolls without slippling down an inclined plane making an angle $\theta$ with the horizontal. The frictional force between the cylinder and the incline is

• A. mg sin $\theta$
• B. $\frac{mgsin\theta }{3}$
• C. mg cos $\theta$
• D. $\frac{2mgsin\theta }{3}$

Answer 1

The correct option is B $\frac{mgsin\theta }{3}$

from newtons second law

incase of linear acceleration of rolling cylinder can be written as

$mgsin\theta -f_k=ma$

where $\theta$ is the angle of inclination

this frictional force is responsible for the rotation of the cylinder

$\tau =I\alpha$ where $\alpha$ is the angular acceleration

$R{f}_k=I\alpha ⟹\alpha =\frac{R{f}_k}{I}$

in general $\alpha =\frac{a}{R}$ now on substituting in the abovew we get $\frac{a}{R}=\frac{R{f}_k}{I}$

$f_k=\frac{aI}{R^2}$

$mgsin\theta -\frac{aI}{R^2}=ma$ [where $mgsin\theta -f_k=ma$]

$ma+\frac{aI}{R^2}=mgsin\theta$

$I=\frac{1}{2}m{R}^2$

$ma+\frac{a\left[\frac{1}{2}m{R}^2\right]}{R^2}=mgsin\theta$

$ma+\frac{1}{2}ma=mgsin\theta$

$\frac{3}{2}ma=mgsin\theta$

$a=\frac{2}{3}gsin\theta$

$\therefore mgsin\theta -f_k=m\left(\frac{2}{3}gsin\theta \right)$

$f_k=mgsin\theta -\frac{2}{3}mgsin\theta$ = $\frac{mgsin\theta }{3}$