If the system shown in the figure is rotated in a horizontal circle with angular velocity `omega` . Find `(g=10m//s^(2))`
(a) the minimum value of `omega` to start relative motion between the two blocks.
(b) tension in the string connecting `m_(1)` and `m_(2)` when slipping just starts between the blocks
The coefficient of frition between the two masses is `0.5` and there is no friction between `m_(2)` and ground. The dimensions of the masses can be neglected. (Take `R=0.5m,m_(1)=2Kg,m_(2)=1Kg)`

Answer

A, B, C, D

Explanation

(a) Force diagrams of `m_(1)` and `m_(2)` are as show below

(Only horizontal forces have been shown)
Equations of motion are
`T+mum_(1)g=m_(1)Romega^(2)`
`T-mum_(1)g=m+_(2)Romega^(2)` ,brgt Solving Eqs. `(i)` and `(ii)` we have
`omega=sqrt((2m_(1)mug)/((m_(1)-m_(2))^(R))`
Substituting the values, we have
`omega_(min)=6.32rad//s`
(b) `T=m_(2)Romega^(2)+mu_(1)g`
`=(1)(0.5)(6.32)^(2)+(0.5)(2)(10)`
`~~30N`