Question

A uniform disc of mass $M$ and radius $R$ rolls down an inclined (angle of inclination=$\theta$) plane without slipping then frictional force is given by $\frac{kMgsin\theta }{3}$,then the value of $k$,is

Answer 1

$f$ is the frictional force along the incline plane in upward direction

$g\sin \theta$ is the acceleration in the downward direction

$\alpha$ is the angular acceleration for pure rotation

$I=\frac{M{R}^2}{2}$ is the moment of inertia about center of mass.

For pure rolling motion,we have equation as

$Mg\sin \theta -f=Ma$ (1)

Torque about center of mass

$f.R=I\alpha$

$f.R=I\frac{a}{R}$ as $\alpha =\frac{a}{R}$

$f=\frac{M{R}^{2}a}{2R^2}$

$f=\frac{Ma}{2}$ (2)

Multiplying equation (2) by 2 and subtracting from (1) we get,

$2f+f-Mg\sin \theta =0$

$f=\frac{Mg\sin \theta }{3}$