# Refresh my memory on centripital force/acceleration

1. Apr 27, 2005

### benji

Alright so we did this stuff at the beginning of the year and haven't touched it until now and I can't remember exactly how it works...

Say a ball is being spun in a vertical circular path and you need to draw a FBD for it at the highest point and lowest point on the circle.

At the top you would just have the force of gravity pulling down, correct? Or would you have the centripital force pulling the ball downward towards the center of the circle too?

At the bottom you would have the force of gravity pulling down and the centripital force pulling the ball towards the center of the circle, correct?

Hrm... But this is screwy in my head.

Because now I need to figure the minimum velocity the ball could have without leaving the circular path at the top of the circle... So my net force equation would look like: (damn I wish I knew how to do all the fancy markup on the forum)

Fnet=Fg=ma

So my minimum velocity would end up being v=sqrt(gR)?

Then at the bottom what would my net force equation look like if the maximum tension the string could have without breaking was Tmax and I needed to derive an equation for Vmax (the maximum speed the ball can have without breaking the string).

I know I'm wrong with something here...

2. Apr 27, 2005

### James R

The centripetal force is the NET force on the object (i.e. the vector sum of all the real forces acting).

At the top you will have gravity pulling down and some other force also pulling down (e.g. the tension in the string the ball is attached to). The NET force is downwards (centripetal).

The tension would pull up, and gravity would pull down. The sum of the two forces gives a NET centripetal force upwards.

Your net force equation at the top, for example, would look like this:

$$F_{net} = \frac{mv^2}{r} = mg + T$$

(taking the downward direction as positive).

You can do the net force at the bottom.

3. Apr 27, 2005

### benji

Thanks James, really appreciate it.