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A bead of mass m is atached to a spring of natural length `(3R)/(4)` with other end of it fixed at O.The bead moves in a track of which part ABC is semicircular of radius R and part CDA varies from R to `(R)/(2)` and then again to R. The bead is in equilibrium at position D and starts moving downward. Find the ratio of normal reaction on the bead to the cetrifugal force at the bottom most position. A. 2B. `(1)/(2)`C. `1,0`D. `(1)/(3)` |
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Answer» In position D spring is in equilibrium. ltbr. `N+mg-F=0`. . . (1) in bottom most position. `N+f-mg=(mv^(2))/(2)`. . . .(2) [since spring is elongated] where F=kx and `x=(3R)/(4)-(R)/(2)=(R)/(4)`in position D and `x=R-(3R)/(4)=(R)/(4)` in bottom position From 1 and 2 `N+(N+mg)-mg=(mv^(2))/(r)` `2N=(mv^(2))/(r)` `(N)/(mv^(2)//r)=(1)/(2)` |
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