Question

A circular disc of mass $2kg$ is placed over a rough horizontal surface. Coefficient of friction is $0.5$. A force F is applied on the periphery of the disc, tangentially in vertically downward direction. The disc will undergo pure rolling only if $F\le F_0.$ When $F=F_0$, the value of $F_0$ in $N$ is (Take $g=10m/s^2$) (Answer upto two digit after decimal place)

Answer 1

When the disc is on the verge of slipping
$f_{max}=\mu N=\mu (F_0+Mg)$
$a_{cm}=\frac{f_{max}}{M}=\frac{\mu (F_0+Mg)}{M}$
Applying torque equation by point touching the ground,
$\alpha =\frac{\tau }{I}=\frac{F_0.R}{\frac{3}{2}M{R}^2}$
For pure rolling : $a_{cm}=R\alpha$
$\frac{\mu (Mg+F_0)}{M}=\frac{F_{0}R}{\frac{3}{2}M{R}^2}$
$\Rightarrow F_0=60N$