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)`.

Answer

(i) 36 (ii) 11.67 (iii) 0.1

Explanation

The tension T of the string acting on `m_(2)` is responsible for rotation of the mass `m_(2)` .
`:. T = m_(2) r_(2) omega^(2) = 5 xx 0.176 xx 10 xx 10 - 88 N`
Since `m_(1)` is also rotating in a circular path, the required centripetal force on `m_(1)`
`:. F_(a) = m_(1) r_(1) omega^(2 = 10 xx 0.124 xx 10 xx 10 = 124 N`
Out of this 124 N required, 88 N will be provided by the tension T in the string and rest will be provided by frictional force between block and surface.
Therefore frictional force acting on `m_(1)`
(ii) Let `omega` be the maximum angular speed for which no slipping of masses occurs (or we may say that `omega` is the minimum angular speed for which slipping occurs)
`f_("max") = nu m_(1) g = 0.5 xx 10 xx 9.8 = 49 N`
The equation for `m_(1)` to move in circular motion is `T + f = m_(1) r_(1) omega^(2)`
When `f = f_("max")` then T = T' (say) and `T = m_(2) r_(2) omega^(2)` (for mass `m_(2)`) and `omega = omega'`
`:. m_(2) r_(2) omega^(2) + f_("max") = m_(1) r_(1) omega^(2) implies = (sqrt(f_("max"))))/(sqrt(m_(1)r_(1) - m_(2) r_(2)))`
`implies omega' = (sqrt(49))/(sqrt(10 xx 0.124 - 5 xx 0.176)) = 11.67` rad /s
(iii) For no friction force acting on mass `m_(1)`
The tension should be sufficient to provide centripetal force for both the masses. Then
`T = m_(1) r_(1) omega^(2)` and `T = m_(2) r_(2) omega^(2) implies (r_(1))/(r_(2)) = (m_(2))/(m_(1)) = (5)/(10) = (1)/(2)` ...(i)
But `r_(1) + r_(2)` = length of the read `implies r_(1) + r_(2) = 0.3 m`
Solving equation (i) and (ii) we get `r_(1) = 0.1 m` and `r_(2) = 0.2 m`