Consider two heavy right circular rollers of the radii `R` and `r` respectively and rest on a rough horizontal plane a 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 `m` has the same value for all surfaces of contact, 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.

Explanation

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 equlibrium are
`sumF_(x)=N_(1)cosalpha-muN_(2)=muN_(1)cos(90-alpha)=0`.....i
Taking torque about `O_(2)` centre of smaller sphere
`sumtau=muN_(1)r-muN_(2)r=0`
`N_(1)=N_(2)`..........ii
On substituting equatiuons ii in i, we obtain
`muN_(1)+muN_(1)sinalpha=N_(1)cosalpha`
`mu(1+sinalpha)=cosalpha`............iii
From figure b. `sin alpha=sqrt((R-r)/(R+r)), cosalpha=(2sqrt(Rr))/(R+r)`
Now equation iii becomes: `mu(1+(R-r)/(R+r))=(2sqrt(Rr))/(R+r)`
`mu=sqrt(r/R)`
Hence the required condition is `mugesqrt(r/R)`