# Difficult circular motion problem

1. Dec 7, 2011

### utm01

1. The problem statement, all variables and given/known data

http://imageshack.us/photo/my-images/266/problembd.jpg/

2. Relevant equations

F=mv^2/c

3. The attempt at a solution

I drew the free body diagram and found

TSin(theta) = mv^2/r

r = LSin(theta)

TSin(theta) = mv^2/LSin(theta)

In the y direction Fy = mg - TCos(theta) = 0
So mg = TCos(theta)

I tried to turn T = mg/cos(theta) into T=mg/(1-Sin(theta)) and sub it in but it doesn't seem to work.

Any help would be appreciated!

2. Dec 7, 2011

### Spinnor

Eliminate T, then you can get the trig functions on one side as a function of all the given information.

3. Dec 7, 2011

### sandy.bridge

Hello. You're on the right track. Rather than trying to get cosine in terms of sine, go the other way around. For example, you have:
$$T=\frac{mv^2}{Lsin^2\theta}$$
and
$$T=\frac{mg}{cos\theta}$$
Put sin^2 in terms of cosine, and equate both the formulas that you have for tension. You should be able to come up with a quadratic equation in terms of cosine.

4. Dec 8, 2011

### utm01

How did you get $$T=\frac{mv^2}{Lsin^2\theta}$$

I solved for T in the y direction and got

$$T=\frac{mg}{cos\theta}$$

$$\frac{mg}{cos\theta}*{sin\theta}$$=$$=\frac{mv^2}{Lsin\theta}$$

I'm stuck after this I can't seem to get the sin into a quadratic equation

5. Dec 8, 2011

### Spinnor

So you have,

mv^2/Ls^2 = mg/c (where s = sin and c = cosine)

so,

cv^2 = Lgs^2 = Lg(1 - c^2)

now you can use the quadratic equation

6. Dec 9, 2011

### utm01

oh ok! Thank you guys!!