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!!