Two hemispheres of radii `R` and `r(lt R)` are fixed on a horizontal table touching each other (see figure). A uniform rod rests on to spheres as shown. The coefficient of friction between the rod and two spherers is `mu`. Then the maximum value of the ratio `(r)/(R)` for which the rod will not slide is :

Options

A. `(sqrt(1 + mu^(2))-mu)/(sqrt(1 + mu^(2))+mu)`
B. `(sqrt(1 + 2mu^(2))-mu)/(sqrt(1 + mu^(2))+mu)`
C. `(sqrt(2 + mu^(2))-mu)/(sqrt(1 + mu^(2))+mu)`
D. None of these

Answer

A

Explanation

In figure `(b)`
`sintheta = (R - r)/(R + r) = (1 - eta)/(1 + eta)`[where`eta = (r)/(R)`]………(1)
The equilibrium of rod gives
`N_(1) + N_(2) = Mg cos theta`…..(2)
and `mu(N_(1) + N_(2)) = mg sintheta` .......(3)
Assuming the friction to be at its limiting value.
`(3)//(2)` gives
`tan theta = mu`
`:. sin theta = (mu)/(sqrt(1+mu^(2))`
Put in `(1)`
`(1 - mu)/(1 + mu) = (mu)/(sqrt(1 + mu^(2)))`
`sqrt(1)+mu^(2) - etasqrt(1 + mu^(2)) = mu + mueta`
`:. (sqrt(1 + mu^(2))-mu)/(sqrt(1 + mu^(2))+ -mu)=eta`
If the ratio `eta` is decreased, `theta` wil increase and the rod will begin to slide.