A) 2
B) \[\frac{1}{2}\]
C) 1.0
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
Let F is spring force and \[F=kx\] In position D, spring in equilibrium \[N+mg\,-F=0\] ? In bottom most position. \[N+F-mg=\frac{m{{v}^{2}}}{r}\](since spring is elongated) \[x-\frac{3R}{4}-\frac{R}{2}\,=\frac{R}{4}\] in position D and \[x=R-\frac{3R}{4}\,=\frac{R}{4}\] in bottom position B hence F is same in D and B position. Now from and \[N+(N+mg)-mg=\frac{m{{v}^{2}}}{r}\]\[2N=\frac{m{{v}^{2}}}{r}\]\[\frac{N}{\frac{m{{v}^{2}}}{r}}=\frac{1}{2}\]You need to login to perform this action.
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