**Problem Statement:**

A ball of negligible size is attached to a string of length R. A man rotates the string in a vertical plane so that the ball describes a circular motion. Demonstrate that for the string to be tensioned at the highest point of its trajectory, its minimum speed at the lowest point of the trajectory must be v_{A} = (5gR)^{1/2}.

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**Solution:**

We will first calculate the minimum speed of the ball so that the string is tense at the highest point (B) of the trajectory. To do this we write Newton’s second law for the forces acting on the ball at this point. In the following figure we have represented the forces and axes that we will use to make the projections.

Newton’s second law for the ball at point B is:

Then we project on the axis perpendicular to the trajectory (n in the figure):

The normal acceleration appears in the second member of equation (1) because it is by definition the projection of the acceleration vector on the perpendicular axis.

After substituting the magnitude of the normal acceleration in equation (1) we get:

When the string is no longer streched, its tension is canceled. By imposing this condition in equation (1) we obtain the minimum speed at this point for the string to have tension:

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With this value we can apply the principle of conservation of the energy for the ball between points A and B. The tension does not produce work because it is perpendicular to the displacement. On the other hand, weight is a conservative force. Therefore the mechanical energy of the ball is conserved between these two points:

The height of the ball at point B is twice the radius of the trajectory. After substituting this value and the speed of the ball at point B given by equation (1) we obtain:

And finally, we can isolate the minimum speed at point A from equation (2):

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