Question

A solid uniform disc of mass $m$ rolls without slipping down a fixed inclined plane with an acceleration $a$. The frictional force on the disc due to surface of the plane is :

• A. $2ma$
• B. $\frac{3}{2ma}$
• C. $ma$
• D. $\frac{1}{2}ma$

Answer 1

The correct option is D $\frac{1}{2}ma$

We first draw the free body diagram.
We get the equation as
$f_s=mgsin\theta -ma$
here $f_s$ is the frictional force and $ma$ is the inertia force.
Now using the relation $a=\frac{gsin\theta }{1+\frac{k^2}{r^2}}$ for a rigid body descending down an inclined plane we substitute $gsin\theta$. Thus we get
$f_s=ma(1+\frac{k^2}{r^2})-ma$
Now for a solid uniform disc we have k as $\frac{r}{\sqrt{2}}$
Thus we get
$f_s=ma(1+\frac{1}{2}-1)$
or
$f_s=\frac{1}{2}ma$