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A body slides without friction from a he...

A body slides without friction from a height `H = 60 cm` and then loops the loop of radius `R = 20 cm` at the bottom of an incline. Find the ratio of forces exerted on the body by the track at the positions `A,B` and `C (g = 10 ms^(-2))`.
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Text Solution

Verified by Experts

From data `H = 3R`
Velocity ar `A, V_(A) = sqrt(2gH) = sqrt(2g(3R)) = sqrt(6 gR)`
Velocity ar `B, V_(B) = sqrt(4 gR)`
Velocity at `C, V_(C) = sqrt(2 gR)`
Reaction force at `A= R_(1) = (mV_(A)^(2))/( R) + mg cos (0^(@))`
=`(m xx 6gR)/( R) + mg = 7 mg`
Reaction force at `B = R_(2) = (mV_(B)^(2))/( R) + mg cos(90^(@))`
=`(mxx 4gR)/( R) + 0 = 4 mg`
Reaction force at `C = R_(3) = (mV_(C)^(2))/( R) + mg cos (180^@)`
=`(m xx 2gR)/( R) -mg = mg`
`:. R_(1) : R_(2): R_(3) = 7 : 4 : 1`.
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