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Question

Two blocks of mass m1 = 10kg and m2 = 5kg connected to each other by a massless inextensible string of length 0.3m are placed along a diameter of a turntable. 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 10rad/sec about a vertical axis passing through its center. The masses are placed along the diameter of the table on either side of the center O such that 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 turntable.
(i) Calculate the frictional force on m1
(ii) What should be the minimum angular speed of the turntable so that the masses will slip from this position?

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Solution

m1=10kg,m2=5kg,l=0.3m,l=l+l1+l2,μ1= coefficient of friction between table and m1=0.5,w=10rad/s,l1=0.124m
(i) Centrifugal force acting on m1:
F1=m1w2l1
Centrifugal force acting on m2
F2=m2w2(ll1)
The frictional force on m1 is equal to the difference of these:
f=|F2F1|=w2|m2(ll1)m1l1|=(10)2|5×(0.30.124)10×0.124|=36N

(ii) If the masses start to slip, f should be equal to kinetic friction:
f=|F2F1|=w2|m2(ll1)m1l1|=(10)2|5×(0.30.124)10×0.124|=36N

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