Tetherball - find theta and tension

[JJBladester]

Homework Statement

A 450-g tetherball A is moving along a horizontal circular path at a constant speed of 4m/s. Determine (a) the angle θ that the cord forms with pole BC, (b) the tension in the cord.

Homework Equations

\sum F_{y}=ma_{y}
\sum F_{n}=ma_{n}=m\frac{v^{2}}{r}

The Attempt at a Solution

Find θ and T.

Given:
m=0.45kg
v=4m/s

FBD:

Sum of forces equations:
(1)\sum F_{y}=-mg+Tsin\theta =0 \to Tsin\theta =mg
(2)\sum F_{n}=Tcos\theta =m\frac{v^{2}}{r} \to T=m\frac{v^{2}}{rcos\theta}

Substitute expression for T found in (2) into (1):
\left (m\frac{v^{2}}{r} \right )tan\theta =mg
\theta =tan^{-1}\left (\frac{gr}{v^{2}} \right )

This is where I get stuck. I know that if I can find r, I can find θ and subsequently T. How do I find r?

 

[rl.bhat]

You have found that

tanθ = gr/v^2.

Check this expression. In FBD angles indicated are different.

In the problem, the length of cord L is given. So sinθ = r/L.

Put it in the expression and find θ.

 

[JJBladester]

Check this expression. In FBD angles indicated are different.

I fixed my FBD so that θ is in the correct place:

sin\theta=\frac{r}{1.8}\rightarrow r=1.8sin\theta

\sum F_{y}=-mg+Tcos\theta=0

(1) Tcos\theta=mg


\sum F_{n}=Tsin\theta=m\frac{v^{2}}{r}

(2) T=\frac{mv^{2}}{rsin\theta}

Substituting (2) into (1) yields:

\left (\frac{mv^2}{rsin\theta} \right )cos\theta=mg

\frac{cos\theta}{sin^2\theta}=\frac{1.8g}{v^2}

How do I simplify the \frac{cos\theta}{sin^2\theta} to solve for θ?

 

[Doc Al]

How do I simplify the \frac{cos\theta}{sin^2\theta} to solve for θ?

Hint: Look up the Pythagorean trig identities. (I'm sure it's one you already know.)

 

[JJBladester]

Hint: Look up the Pythagorean trig identities...

I did this:
\frac{cos\theta}{sin^2\theta}=\frac{cos\theta}{1-cos^2\theta}

Making the equation:
\frac{cos\theta}{1-cos^2\theta}=\frac{1.8g}{v^2}

However, this doesn't get me any closer to getting theta out on its own.

 

[Doc Al]

However, this doesn't get me any closer to getting theta out on its own.

Sure it does. Hint: Rearrange that equation and solve for cosθ. (You'll get a quadratic.)

 

[JJBladester]

You'll get a quadratic...

Thanks to both of you for the persistence. With your help, the solution presented itself:

\left (\frac{1.8g}{v^2} \right )cos\theta+cos\theta-\left (\frac{1.8g}{v^2} \right )=0

cos\theta=\frac{-1+\sqrt{1^2-4\left (\frac{1.8g}{v^2} \right )\left (\frac{-1.8g}{v^2} \right )}}{2\left (\frac{1.8g}{v^2} \right )}

\theta\approx49.9 degrees

T=\frac{mg}{cos\theta}=\frac{\left (0.45kg \right )\left (9.81m/s^2 \right )}{cos(49.9)}\approx6.85N

These answers line up with the book's answers. Again, thanks to both of you for the help.