**Problem Statement:**

One of the attractions of a water park consists of a circular track without friction of radius R with a swimming pool below (see the figure). A bather slips from the top of the track without initial speed, determine the value of the angle θ when he leaves the track.

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**Solution:**

When the bather stops being in contact with the track, the normal force is canceled. Therefore, If we write Newton’s second law for the forces acting on the bather when he leaves the track, we can calculate his speed in that moment. And by using the principle of the conservation of energy we will then be able to calculate the angle θ.

First we draw the forces that act on the bather when he slides on the circular track as well as the axes that we will use to make projections for Newton’s second law:

The forces acting on the bather are the weight (because he is close to the surface of the Earth) and the normal (for being on the track). Therefore Newton’s second law for this situation is written as follow:

After doing the projection on the axes tangent and perpendicular to the trajectory we obtain:

We have represented the projections of the weight on these axes in the following figure:

From equation (1) we would obtain the tangential acceleration of the bather.

We will use equation (2) to determine the speed of the bather when he leaves the circular track. At this point the normal becomes zero because there is no more contact between the bather and the track. We impose this condition in equation (2) and substitute the magnitude of normal acceleration to obtain:

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We will now apply the principle of conservation of energy between points A (at the highest point of the circumference) and B (point at which the bather takes off from the track).

The bather’s mechanical energy between points A and B is conserved, since the normal force produces no work because it is perpendicular to the motion and the weight is a conservative force. Therefore between points A and B we can write:

After substituting the value of the heights and the speed of the bather at point B given by equation (2) we obtain:

And we can deduce the angle θ after simplifying equation (3):

Observe that this result does not depend on the mass of the bather nor the radius of the circular track.

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