Question

Two heavy right circular rollers of radii $R$ and $r$ respectively rest on a rough horizontal plane as shown in figure. The larger roller has a string wound around it to which a horizontal force $P$ can be applied as shown. Assuming that the coefficient of friction $\mu$ has the same value for all surfaces of contact and the smaller cylinder should neither roll nor slide. The minimum coefficient of friction so that the larger roller can be pulled over the smaller one is

• A. $\frac{r}{R}$
• B. ${\left(\frac{r}{R}\right)}^2$
• C. $3\sqrt{\frac{r}{R}}$
• D. $\sqrt{\frac{r}{R}}$

Answer 1

Answer: (D)

Figure shows the forces acting on the smaller cylinder.
$N_1$ is reaction between the two cylinders and $N_2$ the normal reaction between the smaller cylinder and the horizontal surface.

The conditions of equilibrium are
$\sum F_x=N_1\cos \alpha -\mu N_2-\mu N_1\cos (90-\alpha )=0...(i)$
Taking torque about $O_2$, centre of smaller sphere,
$\sum \tau =\mu N_{1}r-\mu N_{2}r=0$
$N_1=N_2\dots (ii)$
On substituting equations (ii) in (i), we obtain
$\mu N_1+\mu N_1\sin \alpha =N_1\cos \alpha$
$\mu (1+\sin \alpha )=\cos \alpha ...(iii)$
From Figure (b), $\sin \alpha =\sqrt{\frac{R-r}{R+r}},\cos \alpha =\frac{2\sqrt{Rr}}{R+r}$
Now equation (iii) becomes; $\mu \left(1+\frac{R-r}{R+r}\right)=\frac{2\sqrt{Rr}}{R+r}\mu =\sqrt{\frac{r}{R}}$
Hence the required condition is $\mu \ge \sqrt{\frac{r}{R}}$.