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

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Homework Help Overview

The problem involves a conical pendulum with a length of 1.5 meters that rotates at a speed of 4 m/s. The original poster seeks to determine the angle (θ) of the pendulum with respect to the vertical.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster outlines several equations related to the conical pendulum and attempts to apply them but expresses uncertainty about progressing from a certain point. Some participants affirm the validity of the original approach while others suggest alternative methods for manipulating the equations.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's attempts. Some guidance has been offered regarding the manipulation of trigonometric identities and the formation of a quadratic equation, but there is no explicit consensus on the next steps.

Contextual Notes

Participants are working with rounded values for gravitational acceleration and are discussing the implications of their assumptions and calculations. There is a noted confusion regarding the application of trigonometric identities and equation solving techniques.

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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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