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Profile image of Eshan
8 Years ago
Dear student,

Apologies for the delayed reply.

To be able to just complete the circle, the body’s centripetal acceleration must be provided by its weight only.

Hence\dfrac{mv_T^2}{r}=mg\implies v_T=\sqrt{gr}

Now, since the work done by the constant horizontal force from the bottommost point to the topmost point is zero(net displacement is vertical and force is horizontal), the loss in kinetic energy in going up will only go into the gain in potential energy.

Hence\dfrac{1}{2}mv_B^2-\dfrac{1}{2}mv_T^2=mg(2r)

\implies v_B^2=v_T^2+4gr=5gr

Hence minimum velocity required by the body at the bottommost point isv_B=\sqrt{5gr}

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