# Rotating sphere and circular trajectory: minimum speed

I have a sphere (mass = 3 kg), constrained to a fixed length rope, rotating (radius = 5 m) on a vertical plane. My textbook ask me about the minimum speed in the highest point in order to keep the circular trajectory.

Now, I know that in the highest point (v=speed):

$$F_c=mg+T$$ $$\frac{mv^2}{r}=mg+T$$ $$\frac{3}{5}v^2=29,43+T$$ I know that with a low speed I have a low Tension, so let's put T=0 and go on: $$v^2=49,05 \rightarrow v=7$$

The result is correct, but I have a little doubt: if the Tension is 0, why the sphere doesn't go along the tangent or start falling down? Exactly what is forcing the sphere to preserve the circular trajectory?

In my ignorant opinion, the minimum speed is a little more than 7.00 m/s, because the tension mustn't be 0, even a very small number, but not null..

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The tension is only zero for an infinitely short time. Immediately before and immediately after the apex of the swing the tension is non-zero. –  John Rennie Nov 5 '12 at 14:09
The tension may be nearly zero at the highest point, but its momentum keeps it going towards its tangent. –  prash Nov 5 '12 at 14:32
@JohnRennie but why it follows the tracjectory and doesn't go along the tangent? –  Surfer on the fall Nov 5 '12 at 14:36
@prash Exactly, towards its tangent! So it doesn't move along the circular trajectory: it's starting to go away! Am I right? –  Surfer on the fall Nov 5 '12 at 14:37
How can it fly away, if it is attached by a rope? It always "wants" to fly tangentially, but the rope stops it from doing so; that's what causes the tension. At the top, if T=0, the rope is not needed for an infinitesimally short time; after that T increases again, and the rope is needed again. If T<0 then yes, the ball will fall down. –  hdhondt Nov 6 '12 at 3:12