Question

Consider two heavy right circular rollers of radii R and r, respectively, and rest on a rough horizontal plane as shown in fig. The larger roller has a string wound around it to which a horizontal force P can be applied as shown. Assuming that the coefficient, determine the necessary condition under which the larger roller can be pulled over the smaller one. Assume the smaller cylinder should neither roll nor slide

• A. $\mu$ $\ge$ $\sqrt{\frac{r}{R}}$
• B. $\mu$ $\le$ $\sqrt{\frac{r}{R}}$
• C. $\mu$ $\ge$ $\sqrt{\frac{r}{2R}}$
• D. None of these

Answer 1

The correct option is A $\mu$ $\ge$ $\sqrt{\frac{r}{R}}$

Sol:
As shown in fig the forces acting on the smaller cylinder. $N_1$ is reaction between the two cylindres and $N_2$ the normal reactoin between the smaller cylinder and the horizontal surface.
The conditions of equlibirium 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
$N_1$ = $N_2$ - (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 fig (b),
$\sin \alpha$ = $\sqrt{\frac{R-r}{R+r}}$
$\cos \alpha$ = $\frac{2\sqrt{Rr}}{R+r}$
Now equation (iii) becomes ;
$\mu (1+\frac{R-r}{R+r})$ = $\frac{2\sqrt{Rr}}{R+r}$
$\mu$ = $\sqrt{\frac{r}{R}}$
Hence the required condition is
$\mu$ $\ge$ $\sqrt{\frac{r}{R}}$