Two blocks of mass `m_(1)=10kg` and `m_(2)=5kg` connected to each other by a massless inextensible string of length `0.3m` are placed along a diameter of the turntable. The coefficient of friction between the table and `m_(1)` is `0.5` while there is no friction between `m_(2)` and the table. the table is rotating with an angular velocity of `10rad//s`. about a vertical axis passing through its center `O`. the masses are placed along the diameter of the table on either side of the center `O` such that the mass `m_(1)` is at a distance of `0.124m` from `O`. the masses are observed to be at a rest with respect to an observed on the tuntable `(g=9.8m//s^(2))`.
(a) Calculate the friction on `m_(1)`
(b) What should be the minimum angular speed of the turntable so that the masses will slip from this position?
(c ) How should the masses be placed with the string remaining taut so that there is no friction on `m_(1)`.

Explanation

`r=0.3m,r_(1)=0.124m,r_(2)=r-r_(1)=0.176m`
`T=m_(2)r_(2)omega^(2)=5xx(0.176)xx(10)^(2)=88N`
`T+f=m_(1)omega^(2)r_(1)`
`88+f=10xx(10)^(2)xx0.124=124`
`f=36N`
Friction on `m_(1):36N`
(b) Considering `(m-(1)+m_(2))` as system
`=m_(2)r_(2)omega^(2)`
`T+f=m_(1)r_(1)omega^(2)`
`f=(m_(1)r_(1)-m_(2)r_(2))omega^(2)`
Mass will slip if
`(m_(1)r_(1)-m_(2)r_(2))omega^(2)gtf_(max)(=mum_(1)g)`
`omega_(min=sqrt((mum_(1)g)/(m_(1)r_(1)-m_(2)r_(2))=11.67 rad//s`
(c ) When `f=0impliesm_(1)r_(1)=m_(2)r_(2)implies10r_(1)=5r-(2)`
`r_(2)=2r_(1)`
`r_(1)+r_(2)=0.3impliesr_(1)=0.1m,r_(2)=0.2m`