1 Question with regarded to Centripical Forces

[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.

 

[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.

 

[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

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