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A section of fixed smooth circular track...

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 `B` is ?

A

`R`

B

`(R)/(4)`

C

`(R)/(2)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
By energy conservation between `A & B`
`rArr Mg``(2R)/(5)+0=(MgR)/(5)+(1)/(2)MV^(2)`

`V=sqrt((2gR)/(5))`
Now, radius of curvature `r`
`=(V_(_|_)^(2))/(a_(f))=(2gR//5)/(gcos 37)=(R)/(2)`
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