Question

Two blocks of masses m and M are connected by a chord passing around a frictionless pulley which is attached to a rotating frame, which rotates about a vertical axis with an angular velocity $\omega$. If the coefficient of friction between the two masses and the surface are ${\mu }_1$ and ${\mu }_2$, respectively, determine the value of $\omega$ at which the block starts sliding radially (M >m).

• A. $\omega =[\frac{g(2{\mu }_{1}m+{\mu }_{2}M)}{M{r}_2-m{r}_1}{]}^{1/2}$
• B. $\omega =[\frac{g({\mu }_{1}m+{\mu }_{2}M)}{M{r}_2-m{r}_1}{]}^{1/3}$
• C. $\omega =[\frac{g({\mu }_{1}m+{\mu }_{2}M)}{M{r}_2-m{r}_1}{]}^{1/2}$
• D. $\omega =[\frac{g({\mu }_{1}m+{\mu }_{2}M)}{M{r}_2-m{r}_1}{]}^{1/4}$

Answer 1

Answer: (C)

$\omega =[\frac{g({\mu }_{1}m+{\mu }_{2}M)}{M{r}_2-m{r}_1}{]}^{1/2}$

Owing to the large force experienced by block of mass $M$ is tends in fly off redally. In the situation of limiting equilibriun we have,

$T=m{\omega }^2{r}_1+f_1\Rightarrow T+f_2=M{\omega }^2{r}_2$

$f_1={\mu }_{1}mg\Rightarrow f_2={\mu }_{2}mg-\left(i\right)$

The above two equations get reduced to

$T=m{\omega }^2{r}_1=mi{\mu }_{1}mg$

$T+{\mu }_{2}mg+M{\omega }^2{r}_2$

Subtracting eq (i) from eq (ii)

${\mu }_{2}Mg=M{\omega }^2{r}_2-m{\omega }^2{r}_1-{\mu }_1\mathrm{mg}$

${\omega }^2=\frac{g[{\mu }_{1}m+{\mu }_{2}M]}{M{r}_2-w{r}_1}$ $\Rightarrow \omega ={\left[\frac{g[{\mu }_{1}m+{\mu }_{2}M]}{M{r}_2-m{r}_1}\right]}^{1/2}$