Centripetal Force Not Include Force Needed To Oppose Weight? In a question where a mass swings on a string, it asks at the the lowest point in the swing, what the centripetal force acting on the mass is?
Now the answer it wants is gained just by filling in the formula F = (mv^2)/r
And after that it asks for the tension in the string, which it wants you to add the weight of the mass to the previous answer.
What confuses me is, shouldn't the centripetal force at the lowest point be equal to the (mv^2)/r + mg? As the centripetal force is the tension?
 A: I think that you should be careful using the term “centripetal force” although in this case it is used and is to be interpreted as “the force which produces the centripetal acceleration” and this is mass $\times$ centripetal acceleration $= m \times \frac{v^2}{r}$.
If $T$ is the tension in the string then using Newton’s second law with up as positive one can write $T-mg= m \frac{v^2}{r}$ from which the tension can be found.
A: Let's start with Newton's Second Law.  It says that net force is the product of mass and acceleration.  An acceleration can, and most often is, the result of a number of forces added together.
Centripetal force is the force that results in centripetal acceleration i.e. acceleration towards the center of rotation.  Centripetal force doesn't have to be one force, rather it is the net force towards the center.  In many cases, like the mass on a string swung in a vertical circle, there are several forces acting radially.  It's not accurate to say that any one of them is the centripetal force.
If the mass were at the top of the circle would you say that gravity was the centripetal force?  Only if the object were travelling at the minimum possible speed where the tension had dropped to zero.  In general the centripetal force would be the sum of tension and gravity.  The same is true at the bottom of the circle.
A: There are two forces acting on the mass: $mg$ pointing toward the ground, and the tension $T$ pointing in the opposite direction. The (modulus of the) centripetal force is the difference of (the moduli of) these two forces $$R = T-mg.$$ 
Therefore, since $R = \displaystyle\frac{mv^2}{r}$, you can find $T$:
$$T = mg + \frac{mv^2}{r}.$$
Notice that I wrote $T-mg$ and not $mg-T$. Indeed, the centripetal force must point toward the sky, and hence $$T > mg.$$
For the static case, you have that $mg = T$. In fact, in this particular case, you don't have any centripetal force!
