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I have the situation described in this picture

enter image description here

I know the speed of the ball at the top of the loop ($v_{top} = 2.38 m/s$), and I have to demonstrate that the ball does not fall from the track at the top of the loop.

From what I understand, the two forces acting on the ball are the gravitational force ($F_g = m g$) and the normal force of the track, that provides the centripetal acceleration of the ball ($F_n = m \cdot a_c$, where $a_c = \frac{v^2}{r}$). Since the centripetal acceleration balances the tangential velocity $v_{top}$, that is, forces the ball to follow the circular path, why does not the ball fall under the effect of the gravitational force $F_g$?

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    $\begingroup$ The centripetal acceleration results from the net force on the ball which is the sum of the normal force and the force of gravity. If the ball is going just slow enough the normal force can be reduced to zero and centripetal acceleration is a result of the force of gravity alone. $\endgroup$
    – M. Enns
    Commented Jun 10, 2016 at 15:47

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At the top of the loop if the normal reaction on the ball due to the track is $F_n$ down and the weight of the ball is $F_g$ down then using Newton's second law

$$F_n + F_g = m\frac {v^2}{R}$$

where $v$ is the speed of the ball, $m$ is the mass of the ball and $R$ is the radius of the loop.

This equation tells you that the faster the ball is moving the larger is the value of the normal reaction.

However as the speed of the ball at the top gets less the normal reaction $N$ gets smaller until there comes a time when the normal reaction is zero and $$F_g = m \dfrac {v_{\text{minimum}}}{R}$$

where $v_{\text{minimum}}$ is the minimum speed that the ball can have and still keep in contact with the track.

If the speed of the ball is less than this minimum speed it will lose contact with the track before reaching the top of the loop.

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