Though this question is old and has an accepted answer, I really don't feel the apparent confusion of terms and concepts in the question has been covered. Therefor let me add a try to clear this out.
In this case, the tension of the string acts as a centrifugal force that keeps the object moving in a circle.
Not centrifugal but centripetal.
- Centrifugal force is the feeling of being pushed outwards while in a circular motion (being pushed to the side of a car that turns).
- Centripetal force is the force (the resulting force) acting to keep an object in a circular motion. To do this, it must be pointing directly towards the center, perpendicular to the motion.
if centrifugal force is the only force acting on the object, why doesn't it collapse into the center? I am pretty sure this has to do with the object's inertia
Again, assuming you mean centripetal.
As mentioned above, the centripetal force is perpendicular to the motion. What happens when you move in one direction and is pulled to the side? This pull will pull you towards the center (towards the origin of the pull), only if you are standing still. If you are moving, then the pull will make you move towards the center, but at the same time you still move straight ahead. Your net movement is therefor not towards the center, but rather around it.
When you ask the question, you are making the mistake of connecting speed $\vec v$ directly with force $\vec F$. But remember that it should rather be $\vec a$ that is connected to $\vec F$. That means, force and velocity are not going in the same direction necessarily, only force and the change of velocity. And in this case, you change the radial velocity component, but keep the tangential velocity component constant - there is only acceleration (change of velocity) in the radial direction, but the net velocity is point in another way.
Also, what keeps the string taut during the swinging?
The explanation that the string is always taught is simple: The object doesn't move towards the center. When the centripetal force (the tension in the string) tries to pull the object closer, the object actually doesn't get closer. It only changes direction.
If you pull harder, then it will gain a larger radial acceleration, but since the speed must be kept constant (as there is no tangential acceleration to change the speed), the radius will change according to
$$a_{rad}=\frac{v^2}{r}$$
and the object comes closer. But if the force is just kept constant, then there is no change of radius during the motion.
There has to be some force countering the centripetal force that keeps the string tight.
No, why should there? I think you mix up the Newton's 1st and 2nd laws. The idea that counteracting forces must be present is when we have a situation of no acceleration; Newton's 1st law $$\sum \vec F=0$$
Then to reach a sum of $0$, all forces must be balanced by other forces. But since there is acceleration, namely radial acceleration in the direction of the centripetal force, then we must use Newton's 2nd law:
$$\sum \vec F=m\vec a$$
Here there is no requirement of a counteracting force to balance any other force. We just need force to cause this acceleration. And the centripetal force is doing exactly that: causing the radial acceleration.
but can a pseudo-force balance a real force?
No. Centrifugal force (not to confuse with centripetal force) is, as you point at, not a force. Just a "feeling" that you are being pushed out of the circle. Like when you are pushed to the side of car while turning. But this feeling is not a force, it is simply the acceleration that you feel. During a turn in a car, your body wants to just continue with the present speed and direction. The car then forces your body to follow along, and you are therefore pushed inwards by the car, which gives you the feeling of being squeezed into the side of the car.
Also, if centripetal were to be balanced by the pseudo-force, wouldn't the object not move at all (since the net force would be 0)?
Remember that it is not no motion but rather no acceleration (no change of speed) that is implied when the net force is zero. The speed should not be considered at all when looking at forces, and that is a key point to realize. Newton's 2nd law that combines forces with motion, is only connecting force with acceleration not with speed.
Best Answer
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.