A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block is released from position A and leaves the track at B. The radius of curvature of its trajectory just after it leaves the track at B is ?
A
R
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B
R4
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C
R2
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D
None of these
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Solution
The correct option is CR2
x=R(1−cos53∘)=0.4R y=R(1−cos37∘)=0.2R
Apply conservation of energy at A and B, mgx=mgy+12mv2 mg(0.4R)=mg(0.2R)+12mv2 v2=0.4gR
Also as block leaves the block at B, thus Normal reaction from contact =0