# Vertical Pendulum Questions

1. Jan 29, 2014

### GoodOldLimbo

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

A conical pendulum has length (L) 1.5m and rotates at 4ms-1. What is it's angle (θ) to the vertical?

2. Relevant equations

r = Lsinθ

tanθ = v2/Lsinθg

tanθsinθ = v2/Lg

sinθsinθ/cosθ = v2/Lg

sin2θ/cosθ = v2/Lg

sin2θ + cos2θ = 1

3. The attempt at a solution

I've followed the list of equations given:

L = 1.5m
v: 4ms
g: 10ms2 (In our workbook, we usually round it up to 10)

r = 1.5sinθ

tanθ = 42/1.5sinθg

tanθsinθ = 42/1.5 x 10

I get stuck here and I'm not sure how to continue. I get the impression that I've approached this the wrong way entirely :P

Any help? It would be most appreciated.

Thanks

2. Jan 29, 2014

### Sunil Simha

There's nothing wrong with your approach. Why don't you go ahead with it?

3. Jan 29, 2014

### GoodOldLimbo

Well, when I get to sin2θ + cos2 θ = 1, I assume you can cancel out the square because the square root of one is one. But if it's sinθ + cosθ = 1, I'm not really sure how to go about that. cos-1 x 1 = 0, and sin-1 x 1 = 90, so θ = 90? Even then, I'm not really sure how to write that.

4. Jan 29, 2014

### BvU

tanθ=sin/cos, so you get sin^2/cos. Write sin^2 = 1-cos^2 and you end up with a quadratic equation in cos.

5. Feb 8, 2014

### GoodOldLimbo

I'm still stuck on this question. Is anyone willing to help me? I'm still not sure how to go from sin^2 θ / cos θ = 106.7 to sin^2θ + cos^2 θ = 1.

6. Feb 8, 2014

### BvU

Screeech. That's me hitting the brakes. I suppose our posts 3 and 4 crossed, because otherwise I would have been sleepless since!

You learned about Pythagoras with his simplest triangle ? Like 3^2 + 4^2 = 5^2 ? And you still post something like that, amounting to "therefore 3+4=5" ?

Or, conversely: 3+4 = 7 so 3^2+4^2 = 49 ? I hope not!

Did you read my post #4? If you did, for a while already you'd have in you notebook the relationship

${1-\cos^2\theta\over\cos\theta} = {v^2\over Lg}$

So you would definitely not have 106.7 on the right hand side, because that is ${v^2 g\over L}$. Easily mistaken if typing things like tanθ = 42 /1.5sinθg without brackets!

Equation solving skills make you see a simple quadratic equation in $\cos\theta$ here. If you don't see it, rewrite using $x$ instead of $\cos\theta$.

If you don't see it yet, multiply with $x$ on the left and on the right. Later check that $x\ne 0$