How to Solve for CosA and T in a Conical Pendulum?

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SUMMARY

The discussion focuses on solving for CosA and T in a conical pendulum scenario, where V represents the tangential speed of the particle, m is the mass, g is the acceleration due to gravity, and L is the length of the string. The equations provided include tcosA = mg, tsinA = mv²/R, and R = LsinA. To isolate CosA, participants suggest eliminating R from the second equation using the third equation and applying the identity sin²A = 1 - cos²A, leading to a second-order equation for further analysis.

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sAXIn
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Hello all , I encouter a problem solving this one :

We are given a conical pendulum with : V - tan. speed of particle
m - mass of rotating particle
g - gravity acceleration
L - leght of the string

We need to find CosA , T by what is given above !
A- is the angle between the string and vertical line !

So : I wrote : tcosA=mg
tsinA=mv^2/R
R=LsinA

but I can't present cosA without sinA or something
Please Help !
 
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sAXIn said:
So : I wrote : tcosA=mg
tsinA=mv^2/R
R=LsinA
That's fine. You can simplify and solve for T. Use the 3rd equation to eliminate R from the 2nd equation. Then realize that [itex]\sin^2\theta = 1 - \cos^2\theta[/itex].
 
okay got it I get 2nd order eq.
thanks a lot
 

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