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A few more variables need to be factored in: * the speed at which the ball is moving at the top of its path, v. * gravitational field strength, g. * the mass of the ball, m. (The mass of the string is neglected.) The ball's acceleration towards the centre of its circular trajectory is given by v2/r therefore the force required to keep it in this trajectory is mv2/r from Newton's second law of motion. This force is supplied jointly by the weight of the ball, mg, and the tension in the string, T. Therefore T = (mv2/r) - mg The minimum possible value of T is actually zero. This will occur when the ball is moving so slowly that its centripetal force can be supplied entirely by its weight, without pulling on the string. In other words: v2m/r = mg v2/r = g Therefore, to achieve minimum tension use: vmin = sqrt(gr) If the velocity falls below sqrt(gr) the ball will drop inwards from its circular path.

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Q: If a ball is tied to the end of a string and swung in a vertical circle of radius r under the action of gravity what is the tension in the string when the ball is at the top of its path?
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