Question

A constant force $F$ acts tangentially at the highest point of a uniform disc of mass $m$ kept on a rough horizontal surface as shown in figure. If the disc rolls without slipping, calculate the acceleration of the centre $C$ and points $A$ and $B$ on the disc.

• A. $a_C=\frac{4F}{3m},a_A=0,a_B=\frac{8F}{3m}$
• B. $a_C=\frac{2F}{3m},a_A=0,a_B=\frac{4F}{3m}$
• C. $a_C=\frac{F}{m},a_A=0,a_B=\frac{2F}{3m}$
• D. $a_C=\frac{F}{2m},a_A=0,a_B=\frac{F}{2m}$

Answer 1

The correct option is A $a_C=\frac{4F}{3m},a_A=0,a_B=\frac{8F}{3m}$


As shown in figure, the force $F$ rotates the disc, so the point of contact has a tendency to slip towards left and the static friction on the disc will act towards right.
As there is no slipping, aceeleration of point $A,a_A=0$,
and angular acceleration of disc $\alpha =\frac{a}{r}$
(Here, $a$=Linear acceleration, $\alpha$=angular acceleration)

For the linear motion of the centre,$\sum F_x=ma$
$F+f=ma$ ... (i)
For the rotational motion about the centre,
${\tau }_{com}=I\alpha$
$-Fr+fr=I\alpha$
$Fr-fr=\left(\frac{1}{2}m{r}^2\right)\left(\frac{a}{r}\right)$ $F-f=\frac{1}{2}ma$ ... (ii)
From (i) and (ii),
$2F=\frac{3}{2}ma$ or $a=\frac{4F}{3m}$
$\Rightarrow a_C=a=\frac{4F}{3m}$

Acceleration of point $B$, $a_B=a_C+\alpha \times r$
$\Rightarrow a_B=a+\frac{a}{r}\times r$
$\Rightarrow a_B=2a=2\left(\frac{4F}{3m}\right)=\left(\frac{8F}{3m}\right)$