Why am I subtracting/adding normal force?

[gibson101]

For the following question, I am given the weight of a ball being twirled on a string, the radius of the string, and the velocity. And I am asked to calculate the tension of the string when the ball is at the top of it's path, and the bottom of it's path. I know that tension is a centripetal force so I used the formula Fr=mv^2/r and then added or subtracted mg (normal force = mg). I don't understand why I am subtracting mg when the ball is at the top, nor adding at the bottom, since the centripetal force and normal force are in opposite directions at the bottom then I would think I would subtract them then. And now that I think about it, there isn't a normal force, cause the normal force is always perpendicular to the surface that the object is on, but there is no surface! Can someone please enlighten me on this! Thanks.

((.3)(3.9)^2/(.8)) + or - (.3)(9.8) = 8.6 or 2.76 Newtons

 

[Pengwuino]

Yes, don't consider normal forces since in your case, there is no normal force; you just have your centripetal force and gravity.

Remember, what you're looking at is $m\vec a = \Sigma \vec F$.

You have circular motion so you know the left hand side is $m{{v^2}\over{r}}$ in magnitude. Look at both cases and remember the directionality of your vectors.

Top of the circle:

At the top of the circle, your total acceleration must be downward correct? So that's a negative value. The gravitational force is downward so that's negative (I assume 'g' is +9.80m/s^2). The tension must be down as well. So Newton's 2nd Law reads:

$$-m{{v^2}\over{r}} = -T - mg$$

Isolate T and you have your answer.

Bottom of the circle:

Now in this case, the acceleration must be pointing up right? So the left hand side is positive this time. Gravity is still going down so it has a negative and the tension must be going up so it's positive. So the 2nd law reads:

$$m{{v^2}\over{r}} = T - mg$$

Now do you see why you add and why you subtract?