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Question

Two blocks of mass m1=10 kg and m2=5 kg connected to each other by a massless inextensible string of length 0.3m are placed along a diameter of a 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 10rad/sec about a vertical axis passing through its centre. The masses are placed along the diameter of the table on either side of the centre O such that m1 is at a distance of 0.124m from O. The masses are observed to be at rest with respect to an observer on the turn table.
(i) Calculate the frictional force on m1
(ii) What should be the minimum angular speed of the turn table so that the masses will slip from this position.

A
36, 11.7
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B
45, 13
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C
55, 16
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D
78, 19
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Solution

The correct option is A 36, 11.7
Given mass m1andm2 10 kg and 5 kg respectively length of the string l=0.3 , the coefficient of friction μ1 = 0.5 and angular velocity ω= 10 rad/sec.

(1) Frictional force on m1=Centrifugal force on m1
- centrifugal force on m2
Centrifugal force on m1
F1=m1ω2
centrifugal force on m2
F1=m2ω2
Thus frictional force on m1
Ff=F2F1=ω2{m2(Ll)m1l}=102{5(0.30.124)10×0.124}=36N

(2) For the mass to slip from the position it must satisfy the equation
F kf= F fm1gμ1=ω2{m2(Ll)m1l}ω=m1gμ1{m2(Ll)m1l}=10×10×0.5{5(0.30.124)10(0.124)=11.78rad/sec


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