A heavy and a light cylindrical rollers of diameter D and d respectively rest on a horizontal plane as shown. The larger roller has a string wound around it to which a horizontal force 'P' can be applied as shown. Assuming 'mu' (coefficient of friction) is same at all surfaces of contact, find the critical value of 'mu' such that larger roller can be just pulled over smaller one.

Options

A. `mu=sqrt((D)/(d))`
B. `mu=2sqrt((D)/(d))`
C. `mu=sqrt((d)/(D))`
D. `mu=2sqrt((D)/(d))`

Answer

C

Explanation

`sin alpha=(D-d)/(D+d)`
`cos alpha=(2sqrt(Dd))/(D+d)`

If smaller cylinder neither rolls nor slides then larger cylinder can be pulled over it. Then sum of moments about `C_(2)` should be zero
`muN_(1)d=muN_(2)d`
`N_(1)=N_(2)`
for no sliding
`muN_(2)+muN_(1) cos(90-alpha)=N_(1) cos alpha`
`implies mu (1+sin alpha)=cos alpha`