Question

Two blocks of mass m, 10 Kg and m,- 5 Kg connected to each other by a massless inextensible string
of length 0.3 m are placed along a diameter of turn table. The coefficient of friction between the table and m,
is 0.5 while there is no friction between m, and the table. The table is rotating with an angular velocity of 10
rad/s about a vertical axis passing through its centre O. The masses are placed along the diameter of the
table on either side of the centre Osuch that the mass m, is at a distance of 0.124 m from O. The masses
are observed to be at rest with respect to an observer on the turn table.
Calculate the frictional force on m,.
What should be the minimum angular speed of the turn table so that the masses will slip

Answer 1

If masses start to slip, the force F = F1 - F2 must be equal to the force of kinetic friction Fk = m1gu1 acting on the mass m1. So, using F1 = m1 * (omega square) * L1 and F2 = m2 * (omega square) * L2 we can get below answer.
Note : L1 = 0.124m and L1+L2 = 0.3m, m1 = 10Kg, m2 = 5Kg, u1=0.5, omega = 10rad/s
Answer is : 11.7 rad s