A uniform disc of radius $$R$$, is resting on a table on its rim. The coefficient of friction between disc and table is $$\mu$$ (figure). Now, the disc is pulled with a force $$F$$ as shown in the figure. What is the maximum value of $$F$$ for which the disc rolls without slipping?
Consider the diagram below
Frictional force $$(f)$$ is acting in the opposite direction of $$F$$.
Let the acceleration of centre of mass of disc be a then
$$F-f=M a\quad \text{.... (i)}$$
where $$M$$ is mass of the disc
The angular acceleration of the disc is
$$\alpha=a / R\quad \text{(for pure rolling)}$$
from
$$\tau=l \alpha$$
$$f R=\left(\frac{1}{2} M R^2\right) \alpha \Rightarrow f R=\left(\frac{1}{2} M R^2\right)\left(\frac{a}{R}\right)$$
$$M a=2 f\quad \text{.... (ii)}$$
From Eqs. (i) and (ii), we get
$$\begin{array}{cc} & f=F / 3[\because N=M g] \\ \because & f \leq \mu N=\mu M g \\ \Rightarrow & \frac{F}{3} \leq \mu M g \Rightarrow F \leq 3 \mu M g \\ \Rightarrow & F_{\max }=3 \mu M g \end{array}$$