# Motion in a vertical circle and conservation of energy

A stone  tied to a string of length l is whirled around a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at the lowest position and has a speed u. What is the magnitude of change in its velocity as it reaches a position where the string is horizontal?

Let’s assume that the potential energy at the lowest position be zero. So, when the string is horizontal, the stone has risen by a vertical height l, the length of the string which is also the radius of the vertical circle.

If v is the magnitude of velocity at the horizontal position, then according to the law of conservation of energy,

KE+PE at the lowest position = KE+PE at the horizontal position $\frac{1}{2}mu^{2}=\frac{1}{2}mv^{2}+mgl$

From the equation above, v-u can be calculated. $v=\sqrt{u^{2}-2gl}$

The following links will help you for deeper understanding and you can browse through some solved problems from the topic too.