Question

A force of 100 N is applied to the centre of a circular disc, of mass 10 kg and radius 1 m, resting on a floor as shown in the figure. If the disc rolls without slipping on the floor, the linear acceleration $in\frac{m}{s^2}$ of the centre of the disc is (correct to two decimal places).

Answer 1

m = 10 kg, r = 1 m
$100-f=m{a}_{CM}=mr\alpha$
(Newton's second law) ...(i)
$[Torque{]}_{CM}=I_{CM}\alpha$
$f\times r=\frac{m{r}^2}{2}\times \alpha$
$2f=mr\alpha$
$2f=m{a}_{CM}$ $[\because a_{CM}=r\alpha ]$
$f=\frac{m}{2}\times a_{CM}...\left(ii\right)$
By (i) and (ii):
$100-\frac{m{a}_{CM}}{2}=m{a}_{CM}$
$\frac{200}{3\times 10}=a_{CM}$
$a_{CM}=\frac{20}{3}=6.67\frac{m}{s^2}$