# Sin^2 tetherball

1. Mar 19, 2013

### tony873004

I'm working on the exact same problem described in this 3 year old thread:
except the length of my rope is 3.8 m and my velocity is 5.6 m/s.
A tetherball is on a rope 3.8 m long and moves in a circle with velocity 5.6 m/s. Solve for the angle the rope makes with the pole.

I follow the explanation in the old thread up to this point.

$$\begin{array}{l} \left( {\frac{{mv^2 }}{{r\sin \theta }}} \right)\cos \theta = mg \\ \\ \frac{{\cos \theta }}{{\sin ^2 \theta }} = \frac{{rg}}{{v^2 }} \\ \end{array}$$

When I rearrange the top expression into the bottom expression, I don't get sin^2. I just get sin. It becomes very simple to solve for θ at that point. How did sin end up getting squared?

My full attempt is this:
$$\begin{array}{l} T\cos \theta = mg \\ T\sin \theta = \frac{{mv^2 }}{r}\,\,\,\, \Rightarrow \,\,\,\,T = \frac{{mv^2 }}{{r\sin \theta }} \\ \frac{{mv^2 }}{{r\sin \theta }}\cos \theta = mg \\ \frac{{\sin \theta }}{{\cos \theta }} = \frac{{v^2 }}{{gr}} \\ \tan \theta = \frac{{v^2 }}{{gr}}\,\,\,\, \Rightarrow \,\,\,\,\theta = \tan ^{ - 1} \frac{{\left( {5.6\,{\rm{m/s}}} \right)^2 }}{{\left( {9.8\,{\rm{m/s}}^{\rm{2}} } \right)\left( {3.8\,{\rm{m}}} \right)}} = 40^\circ \\ \end{array}$$

The back of the book says 48.4 degrees.
When I continue the problem with the sin^2, I get the correct answer. The logic makes sense. But where did that extra sin come from?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 19, 2013

### TSny

You used the length of the rope for r. Can you find the relationship between r and the length of the rope?

Last edited: Mar 19, 2013
3. Mar 19, 2013

### tony873004

Thanks. I missed that! The earlier explanation used r for both rope length and radius. r = r sin theta -->> r(circle) = r(rope)sin theta.