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

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The discussion revolves around calculating the angle of a conical pendulum with a length of 1.5m rotating at 4 m/s. Participants share equations relevant to the problem, including relationships involving sine and tangent functions. One user expresses confusion about progressing from the equations to find the angle, particularly regarding the use of trigonometric identities. Others clarify the importance of correctly applying the relationships and suggest rewriting the equations to identify a quadratic form. The conversation emphasizes the need for careful algebraic manipulation to solve for the angle accurately.
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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
 
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There's nothing wrong with your approach. Why don't you go ahead with it?
 
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.
 
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.
 
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.
 
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##
 
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