Q. Cart on Track which consists of a vertical circular loop?

Here we have a cart on a track which consists of a vertical circular loop. Of course we don't want the cart to fall o? the track at the top of the loop so it must have sufficient forward velocity that its centrifugal force keeps it in contact. We aspiration to find the height from which the cart must be dropped on the leading ramp to satisfy this requirement.

We know from the preceding problem that the velocity of the cart at the top of the loop must be

v_t = √gR

Where g is the acceleration due to gravity and R is the radius of the loop. As we are assuming that friction is negligible the kinetic energy at point P must the same as the kinetic energy at the top of the loop. This energy have to be provided by the conversion of potential energy to kinetic energy from the point of release of the cart to point P.

Thus mgh = mv_t²/2 or

h =v _t ²/2g.

Substituting

h =R/2.

This is the height above point P from which the cart have to be dropped. The total height above ground is R/2 + 2R = 5R/2.