Another bucket of water, centripedal force problem

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PhysicsDaoist
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Homework Statement


This is a classic problem with a slight twist on followon questions -
A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the top of the circular loop, the speed of the bucket is 4.00 m/s. a) Determine the acceleration, the net force and the individual force values when the bucket is at the top of the circular loop.
b) What if the velocity becomes 2 m/s? What is the tension (Ft) of the rope?
c) Can the tension be negative? Will cirular motion still hold when Ft is negative?

Homework Equations


Ft = F_net - mg
F_net = m v^2/r


The Attempt at a Solution


I can solve (a) easily -> a_c = v^2/r = 16 m/s^2; F_net = 24 N (down); Ft = 9.29 N.
and plugging in eqt. I also get (b)
a_c = 4 m/s^2; F_net = 6 N (down); Ft = -8.72 ! negative tension??
c) I have problem with this? In fact, I ran this in excel with velocity all the way down to 0.1 m/s and of course F_tens are negative, what does this mean? Does it mean that the bucket is not traveling in circular motion?
What if this is NOT a rope but roller coaster cart in frictionless track? does that mean the normal force acting on the cart change direction when velocity drops below certain threshold? Can someone help me with the free-body-diagram here?
 
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Tricky business. For (a) it takes an Fc downward of 24 N to hold the bucket in circular motion. Gravity provides 14.7 N of this; the tension must provide the remaining 9.3 N.
For (b) it only takes 6 N to maintain circular motion and gravity still pulls with 14.7 N - too much - so the bucket will fall out of circular motion and spill the water. The string can't push upward to reduce the downward force to the required 6 N.

A roller coaster rides on top of the track, doesn't it? If so, it is in no danger of falling because the track holds it up. But in the (a) case it would be launched into the air unless the track can also hold it down.