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You have a light string. At one end of the string there is a ball modelled as a particle. The string has negligible mass. The other end of the string is fixed at a point and the ball is undergoing circular motion.

At the bottom, the centripetal force = mg - T. At the top, the centripetal force = T + mg.

m = mass of ball

g = gravitational field strength

T = tension

If an exam asks you: What is the reaction force at the bottom? What is the reaction force at the top?

Ball goes in a vertical circular path. Resistive forces = 0.

Is the reaction force always 0? The ball isn't in contact with anything but the string. Isn't the centrifugal force fictitious? Thanks.

Edit: Velocity is not constant due to conservation of energy, so the centripetal force is not constant. I was wondering if there is a reaction force acting on the ball, and if so, what is the value of that reaction force.

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The tension on the string will have an equal and opposite "reaction force" from the mass.

Note that the tension $T_{top}$ at the top is in general not the same as the tension $T_{bottom}$ at the bottom of the loop. Using the same symbol $T$ either implies that the tension is the same (which means something must be done to change the speed of the particle as it goes around), or it is just messy. Since the point in the middle is "fixed" per your statement of the problem, it must be that the particle is undergoing free motion, and we need to account for the change in velocity.

You can start with the velocity at the top of the orbit - this must be sufficient to offset the force of gravity, but tension in the string could be zero. This puts a lower bound on the velocity, and

$$\frac{m\ v_{top}^2}{r} > m\ g\\ v > \sqrt{r\ g}$$

Now the velocity at the bottom will be greater because the potential energy from the top is converted to kinetic energy at the bottom:

$$\frac12mv_{top}^2 + mgh = \frac12mv_{bottom}^2$$

And the tension at the bottom must be sufficient to account for the additional velocity as well as the force of gravity - which is now pointing outwards.

The total tension at the bottom has to be

$$T = mg + \frac{mv_{bottom}^2}{r}\\ =mg + \frac{mv_{top}^2 + 2mgh}{r}$$

I am still not completely clear what is intended with the "reaction force" on the mass - but it is reasonable to say that it is the force of the string on the mass, which is equal to the tension and is given by the above expression.

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    $\begingroup$ I agree the question is not very well phrases. Your answer is better than mine anyhow, so deleted mine. $\endgroup$ – Bernhard Aug 22 '14 at 17:38
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A ball in a circular path is constantly accelerating(just a change in motion) so using $$F=ma$$ We can tell that if the ball has mass, and is moving in a circle, it will have some force exerted on it.

Since the ball is tied to the string, the string pulls on the ball with the same amount of pull, as the ball is pulling on the string. The force at the top of the string should be the same as the force at which the ball is pulling on the string.

I hope that helps.

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