What is the Angle of a Conical Pendulum Rotating at 4 m/s with a 1.5m String?

[GoodOldLimbo]

Homework Statement

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

Homework Equations

r = Lsinθ

tanθ = v²/Lsinθg

tanθsinθ = v²/Lg

sinθsinθ/cosθ = v²/Lg

sin²θ/cosθ = v²/Lg

sin²θ + cos²θ = 1

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

 

[Sunil Simha]

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

 

[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.

 

[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.

 

[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.

 

[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##.

Spoiler

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