Question

Four identical rods of mass $M=6\text{kg}$ each are welded at their ends to form a square and then welded to a massive ring having mass $M=4\text{kg}$ and radius $R=1\text{m}$. If the system is allowed to roll down the incline of inclination $\theta ={30}^{\circ }$.


The minimum value of friction coefficient to prevent slipping is

• A. $\frac{5}{7}$
• B. $\frac{5}{12\sqrt{3}}$
• C. $\frac{5\sqrt{3}}{7}$
• D. $\frac{7}{5\sqrt{3}}$

Answer 1

The correct option is B $\frac{5}{12\sqrt{3}}$

Friction force, $f=\mu N$ ....(1)
where, $N=Mgcos\theta 28\times g\times \frac{\sqrt{3}}{2}$

As we know
$fR=\frac{I{a}_{cm}}{R}f=\frac{20\times 7g}{24}$

using eq. (1)
$\frac{27}{24}\times 7g=\mu \times 28g\times \frac{\sqrt{3}}{2}\mu =\frac{5}{12\sqrt{3}}$