Question

A rigid body of mass 'M' and radius R is rolling without sliding, along with bock of radius of gyration K, and mass $m_1$, then find the minimum value of friction coefficient $\mu$?

• A. $\frac{M{k}^2}{R^2\{M+m(1+\frac{k^2}{R^2}\left)\right\}}$
• B. $\frac{M{k}^2}{R^2\{M+m(1+\frac{k^2}{R^2}\left)\right\}}$
• C. $\frac{M{R}^2}{k^2\{M+m(1+\frac{k^2}{R^2}\left)\right\}}$
• D. $\frac{M{R}^2}{k^2\{M+m(1+\frac{k^2}{R^2}\left)\right\}}$

Answer 1

The correct option is B

$\frac{M{k}^2}{R^2\{M+m(1+\frac{k^2}{R^2}\left)\right\}}$

Let us assume that friction acts in backward direction {anyways the sign of final equation for friction will give the right idea even if assumed wrong initially}

Free body diagram

Linear acceleration of Both will be same by constraint equations:

$T-f_r=ma$ - - - - - - (1)

Torque about $C\Rightarrow f_r\times R=I\alpha$

Where $a=R\alpha andI=m{k}^2$

$\Rightarrow f_r\times R=\frac{m{k}^{2}a}{R}$

$f_r=\frac{m{k}^{2}a}{R^2}$

put in (1)

$T=\frac{m{k}^{2}a}{R^2}+ma$ ....(3)

Mg-T=Ma ....(2)

From(2) and (3)

$Mg=\frac{m{k}^{2}a}{R^2}-ma=Ma$

$Mg=a\{M+m+\frac{m{k}^2}{R^2}\}$

$a=\frac{Mg}{M+m+\frac{k^2}{R^2}m}$

$f_r=\frac{Mmg{k}^2}{R^2\{M+m+\frac{k^2}{R^2}m\}}\le \mu mg$

$\mu \ge \frac{M{k}^2}{R^2\{M+m+\frac{k^2}{R^2}m\}}$