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Question

Two blocks of mass m1=10kgandm2=5kg connected to each other by a massless inextensible string of length 0.3m are placed along a diameter of turn table. The coefficient of friction between the table and m1is0.5 while there is no friction between m2 and the table. The table is rotating with an angular velocity 10 radian see about a vertical axis passing through its centre O. The masses are placed along a diameter of the table on either side of the centre O such that the mass m1 is at a distance of 0.124m from O. The masses are observed to be at rest with rest with respect to an observer on the turn table.
Calculate the frictional force on m1:

A
128N
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B
88N
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C
36N
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D
24N
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Solution

The correct option is C 36N
frictional force f

v=wr

v2=10x2

v1=10x1

m2v2x2=m1v21x1+f

m2w2x2=m1w2x1+f

5×100×0.176=10×100×0.124+f

f=88124=36N

-ve sign indicates direction of friction force acts opposite.


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