Question

Two blocks of mass m₁ = 10 kg and m₂ = 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 m₁ is 0.5 while there is no friction between m₂ and the table. The table is rotating with an angular velocity of 10 rad/s about the vertical axis passing through center O. The masses are placed along the diameter of the table on either side of center O such that the mass m₁ 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 of table.

(A) Calculate the frictional force on m₁ . 

(B) What should be the minimum angular speed of the turn table, 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 frictional force acting on the mass m₂ ?

Answer 1

Given,

m₁ = 10 kg, m₂ = 5 kg, ω = 10 rad/s

r = 0.3 m, r₁ = 0.124 m

∴ r₂ = r – r₁ = 0.176 m

(a) Masses m₁ and m₂ are at rest with respect to rotating table.

Let f be the friction between mass m₁ and table.

Free body diagram of m₁ and m₂ with respect to ground

(b) From eq. (iii)

f = (m₁ r₁ – m₂ r₂) ω²

Masses will start slipping when this force is greater than f_max or

i. e., mass m₂ should be placed at 0.2 m and m₁ at 0.1 m from the centre O.

Answer 2

Answer is :

(a) f = 36N inwards, 

(b) 11.67 rad/sec, 

(c) m₂ at 0.2m and m₁ at 0.1 m from O