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(B) For pure rolling,the linear acceleration $a$ and angular acceleration $\alpha$ are related by $a = r\alpha$.Applying Newton's second law for trans

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$\frac{F}{3 M g}$

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(B) For pure rolling,the linear acceleration $a$ and angular acceleration $\alpha$ are related by $a = r\alpha$.Applying Newton's second law for translational motion:$F - f = M a \quad \dots (1)$Applying torque equation about the centre of mass:$	au = I \alpha$$f r = (\frac{1}{2} M r^2) (\frac{a}{r})$$f = \frac{1}{2} M a \quad \dots (2)$Substituting $(2)$ into $(1)$:$F - \frac{1}{2} M a = M a$$F = \frac{3}{2} M a \Rightarrow a = \frac{2 F}{3 M}$Now,substitute $a$ back into $(2)$ to find the friction force $f$:$f = \frac{1}{2} M (\frac{2 F}{3 M}) = \frac{F}{3}$For pure rolling to occur,the friction force must satisfy $f \le \mu N$,where $N = Mg$.$\frac{F}{3} \le \mu M g$$\mu \ge \frac{F}{3 M g}$Thus,the minimum value of the coefficient of friction is $\mu = \frac{F}{3 M g}$.

$A$ force $F$ is applied at the centre of a disc of mass $M$. The minimum value of the coefficient of friction of the surface for pure rolling is:

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