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Question

Physics Question on work, energy and power

A body initially at rest and sliding along a frictionless track from a height hh (as shown in the figure) just completes a vertical circle of diameter AB=DAB \,=\, D. The height h is equal to

A

54D\frac{5}{4} D

B

32D\frac{3}{2} D

C

75D\frac{7}{5} D

D

DD

Answer

54D\frac{5}{4} D

Explanation

Solution

As track is frictionless, so total mechanical energy will remain constant
T.M.EI=T.M.EFT.M.E_I = T.M.E_F
0+mgh=12mvL2+00 + mgh = \frac{1}{2} mv_L^2 + 0
h=vL22gh = \frac{v_L^2}{2g}
For completing the vertical circle, vL5gRv_L \geq \sqrt{5 g\, R}
h=5gR2g=52R=54Dh = \frac{5g \,R}{2g} = \frac{5}{2} R = \frac{5}{4} D