Question

A solid cylinder of mass $10kg$ is rolling perfectly on a plane of inclination ${30}^0$. The force of friction between the cylinder and surface of the inclined plane is :

• A. $49N$
• B. $24.5N$
• C. $\frac{49}{3}N$
• D. $12.25N$

Answer 1

The correct option is C $\frac{49}{3}N$

Using Newton's 2nd Law of motion,
$Mg\sin \theta -f=Ma$ ......(1) along the inclined plane where f is the frictional force up the incline.
$Mg\cos \theta =N$ .........(2) perpendicular to the inclined plane.
Substituting (2) in (1) we get,
$Mg\sin \theta -\mu N=Ma$
$\therefore Mg\sin \theta -\mu Mg\cos \theta =Ma$
Only frictional force exerts torque since weight Mg and normal reaction N pass through the centre of the cylinder and do not exert any torque.
Torque due to frictional force is $\tau =fR$ where R is the radius of the cylinder.
Using equation for rotational motion,
$\tau =fR=I\alpha$ .....(3) where I is the MI of the cylinder about its center and $\alpha$ is the angular acceleration.
Since there is no slipping, linear acceleration $a=R\alpha$
Substituting $a=R\alpha$ in (3) we get,
$f=\frac{I\alpha }{R}=\frac{1}{2}M{R}^2\times \frac{a}{R^2}=\frac{Ma}{2}$ .....(4)
Substituting for Ma from eqn(1) in eqn(4)
$f=\frac{Mg\sin \theta -f}{2}$
$\Rightarrow 3f=Mg\sin \theta$
$\Rightarrow f=\frac{Mg\sin \theta }{3}=\frac{10\times 9.8\times \sin {30}^0}{3}=\frac{49}{3}N$