physics problems~~~~urgent

First, let the mass of the whole cart (together with passengers) be m.

Then let its velocities at the bottom and top of the circular track be (V1) and (V2) respectively.

Comparing them using the relation of loss in k.e. = gain in g.p.e., we have:
[m(V1)^2]/2 - mgh = [m(V2)^2]/2
with h = 40 since it is the twice of the radius of the circular track.
Hence, [m(V2)^2]/2 = [m(V1)^2]/2 - 400m by taking g = 10 m s^-2

Now, consider the moment when the cart is at the top of the track. Over there, 2 forces are acting on it which include the normal reaction of the track and its weight, both in downward direction.

So to speak, if the normal reaction force is not zero, then it means that the cart is still in contact with the track at the top and hence it will not fall down.

Thus we have an inequality [m(V2)^2]/r > mg (derived from the fact that centripetal force required is greater than the weight of the cart)

So from the previous equation: [m(V2)^2] = [m(V1)^2] - 800m and sub this relation into the inequality:
[m(V1)^2]/r - 800m/r > mg
(V1)^2/20 - 40 > 10 by substituting r = 20 and g = 10
(V1)^2/20 > 50
(V1)^2 > 1000
V1 > 31.6 m s^-1

Note: In fact for any vertical circular motion, the minimum of the bottom velocity for completing the circular motion is sqrt(5gr).

(mv^2)/r = mgh --------------- eqn 1
substitute m, r, g and h then cal v

In order to hold the passengers at the top, the centripetal force at least should cencel the downward acceleration i.e. eqn 1
or kinetic energy change to gravitational energy