STEP III 1990 question 13 solution

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Consider a general point P which makes an angle with the upwards vertical.

The forces acting on the particle are its weight, acting downwards with magnitude mg, and the reaction force R acting towards O. Let the radius of sphere be r.

The particle leaves the surface when R = 0.

Since the velocity is tangential to the circle, the resolved velocity vector makes an angle of to the horizontal.

Therefore, the velocity and position vectors are:

Let the collision occur at some point C.

We're trying to find the angle which has to exceed, so we can consider C at the same height as O, and find the resulting value of ; then has to be larger than this for the collision to be below O.

The horizontal distance to C is ; the horizontal velocity is a constant . Therefore the collision occurs at .

Inserting this value of t into the equation for the height:

The change in height is . Equating the two you get:

When you multiply this all out and simplify, you end up with ; the only real solution to this is , so when the collision is at the same height as O, .

For C to be below O, .

Alternate Method

Energy is conserved, so

Where v = velocity at P, V = velocity at C.

This simplifies to .

Using the velocity vector, we find

When this is simplified you end up with the same cubic equation in terms of .

Solution by Dystopia.