a block of mass m slides without friction along a looped track. if the block is to remain on the track, even?

Question

at the top of the circle (whose radius is r), from what minimum height h must it be released?

Answer 1

The instant before the block is dropped it has some gravitational potential energy,
PE = mgh
as the block falls the PE in converted into Kinetic energy,
KE = 1/2 mv^2
as the block rises up the looped track its KE is converted back into PE.
If the block is to remain on the track even at the top it must have a speed fast enough so that the centripetal force is greater than or equal to the force of gravity at the highest point.
Centripetal force = m * v^2 / r

Set the centripetal acceleration to be the gravitational acceleration at the top of the track to give calculate the minimum speed necessary,
v^2 / r = g
Solving for v gives us,
v = sqrt (g*r)

Now plug this back into the KE equation,
KE = 1/2 m (sqrt(g*r))^2 = 1/2 * m * g * r
Now set KE = PE
mgh = 1/2 * m * g * r
m and g cancel,
h = 1/2 * r

This means that at the very top of the loop the block must have a velocity equal to that gained by a drop of 1/2 the radial distance.
This means that the block needs to be dropped 1/2*r HIGHER than the top of the loop with radius r. Since the loop's higher/diameter is 2r, the total drop height would need to be 2*2 + .5*r = 2.5*r.

Think of it this way, if the block is dropped from 2.5*r, it gains all that KE from the loss of its PE, but then as it ascends the loop it looses 2*r worth of the KE, leaving only .5*r worth of the KE left...the bare minimum needed to stay on the loop without falling off.

Answer 2

If the track had some type of railing on it and since it has no friction it can go one forever and still remain on the track forever.