Q.A particle of mass m performs vertical motion in a circle of radius r. Its potential energy at the highest point is ____. (g is acceleration due to gravity)

Using energy conservation between the two points and Newton's second law radially at each, T_horizontal − T_top = 3mg.

Let the lowest point of the vertical circle (radius $r$) be the reference for height. At the horizontal position (level with the centre, height $r$ above the bottom), the string tension alone supplies the centripetal force (gravity is tangential there): $T_h = \dfrac{mv_h^2}{r}$.

At the top (height $2r$ above the bottom), both tension and gravity point toward the centre: $T_{top} = \dfrac{mv_{top}^2}{r} - mg$.

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Taking the lowest point of the vertical circle as the reference (zero PE), the highest point is a height $2r$ above it, so PE $= mg(2r)$.

In vertical circular motion of radius $r$, the particle traces a full circle in a vertical plane. If the lowest point of the circle is taken as the reference level for gravitational potential energy, the topmost point is diametrically opposite, at a vertical hei …