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Two blocks of mass m(1)=10kg and m(2)=5k...

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)`.

Text Solution

Verified by Experts

The correct Answer is:
(i) 36 N , (ii) `11.67` rad/s

(i) As for the circular motion of mass `m_(2)` , centripetal force is provided by tension alone .

`T = m_(2)r_(2) omega^(2)`
=`5 xx 0.176 xx (10)^(2) = 88 N`
And as for circular motion of mass `m_(1)` centripetal force is provided by both tension and friction , i.e.,
`T + f = m_(1) r_(1) omega^(2)`
So , `" " f = 10 xx 0.124 xx 10^(2) - 88 `
`= 124 - 88 = 36 N`
(ii) The masses will slip if
`T + f_(L) lt m_(1) r_(1) omega^(2) i.e., (m_(1) r_(1) - m_(2) r_(2))omega^(2) gt f_(L)`
[as `T = m_(2)r_(2) omega^(2)]`
i.e., `" " (10 xx 0.124 - 5 xx 0.176) omega^(2) gt "or" omega^(2) gt (49)/(0.36)`
i.e., ` " " omega gt 11.67 ` rad/s
So , for slipping `omega_("min") = 11.67` rad/s
(iii) In absence of friction , the masses will not slip if `m_(1) r_(1) omega^(2) = m_(2)r_(2)omega^(2) = T (-= CPF)`
i.e., `" " m_(1)r_(1) = m_(2) r_(2)` with `r_(1) + r_(2) = 0.3` m
i.e., `" " 10 r_(1) = 5(0.3 - r_(1))`
or `" " r_(1) = 0.1` m
And so , `r_(2) = 0.3 - 0.1 = 0.2` m
i.e. , if the centre of mass of `m_(1)` and `m_(2)` coincides with the centre of the table , the string , will remain taught and masses with not slip whatever be `omega`.
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