A round table is rotated with an angular velocity of 10 rad`cdot s^(-1)` about its axis. Two blocks of mass `m_(1) = 10`kg and `m_(2) = `5 kg , connected to each other by a weightless inextensible string of length 0.3 m, are placed along a diameter of the table. the coefficient of friction between the table and `m_(1)` is 0.5, while there is no frication between `m_(2)` and the table. Mass `m_(1)` is at a distance of 0.124 m from the centre of the table. The masses are at rest with respect to the table. (i) Calculate the frictional force on `m_(1)`. (ii) What should be the minimum angular speed of the table so that masses will slip from the table? (iii) how should these masses be kept so that the string remains taut but no frictional force acts on `m_(1)` ?

Explanation

The top view of the table is shown in
according to the problem,
AB = 0.3 m:
OA = 0.124 m
So, `" "` OB = 0.3 - 0.124
= 0.176 m
Centrifugal force acting on mass `m_(1)` along the direction OA,
`m_(1) omega^(2) r = 10 xx (10)^(2) xx 0.124 = 124` N
Again, centrifugal force acting on mass `m_(2)` along the direction OB,
`m_(2) omega^(2) r = 5 xx (10)^(2) xx 0.176 = 88` N
So, the resultant force along the direction OA
= 124 - 88 = 36 N
In spite of this resultant force, the two masses remain stationary with respect to the table. Hence, the frictional with respect to the table. Hence, the frictional force is equal and opposite to this force.
(ii) Reaction force of the table on mass `m_(1)`
= weight of the mass = `m_(1) g = 10 xx 9.8 = 98`N
As the coefficient of friction is 0.5 , the limiting frictional force = 0.5`xx`98 49 N.
If the minimum angular velocity of the table is `omega`, the resultant of the two opposite centrifugal forces acting on the two masses
= `m_(1) omega^(2) r_(1) - m_(2) omega^(2) r_(2) = omega^(2) (10 xx 0.124 - 5 xx 0.176) `
`= omega^(2)( 1.124 - 0.88) = 0.36 omega^(2)` N
According to the problem, resultant of the two centrifugal forces = limiting frictional force
`therefore" " 0.36 omega^(2) = 49 " " or, omega^(2) = (4900)/(36)`
or, `omega = (70)/(6) = 11.67 "rad"cdot s^(-1)`
If there is no frictional force acting on the masses, the resultant of the two centrifugal forces should be zero. In that case, if mass `m_(1)` is placed at a distance r from the centre, then the distance of `m_(2)` becomes (0.3 - r).
So, for any value of `omega`.
`m_(1)omega^(2) r = m_(2) omega^(2)(0.3 - r) " " or, 0.3 - r = (m_(1))/(m_(2))`r
or, 0.3 - r = 2r or, 3r = 0.3 or, r = 0.1 m