# 1 Question with regarded to Centripical Forces

1. Apr 20, 2006

### JayDub

Hey there, so we started learning centripical forces today. We learned the basic formulas:

$$Fc = \frac{v^2}{r}$$

and

$$Fc = m \frac{4 pi^2 r}{T^2}$$

http://elarune.net/admins/josh/q-angle.jpg

That is the question, I had to get rid of stuff that i had written in, probably not appropriate for the forums. I am not sure how I would go about solving this question. I am supposed to derive the relationship for it. Any help would be appreciated. Thank you.

2. Apr 20, 2006

### moose

You can solve this using the rotational acceleration formula and gravity. Read the question again and try to think about what will have to be equal.

3. Apr 21, 2006

### JayDub

Actually, I think I may have it, would this be it?

[bare with me while I try to do that tex graphic stuff]

$$Fx = Fn sin(\theta) + mew Fn cos(\theta)$$

$$Fy = Fn cos(\theta) - mew Fn sin(\theta) - mg$$

which to find the Fc we would solve it to

$$Fc = \sqrt{\frac{r g (sin(\theta) + mew cos(\theta))}{cos(\theta) - mew sin(\theta)}}$$

Last edited: Apr 21, 2006
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