Homework Help: An issue with Conical Pendulums

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1. Nov 6, 2014

ohmanitsDAAAAAN

1. The problem statement, all variables and given/known data
A conical pendulum with an unelastic tether has a mass of 4.25 kg attached to it. The tether is 2.78 m. The mass travels around the center every 3.22 seconds.

What angle does the rope make in relation to its original position?
m=4.25 kg
T=3.22 s
L=2.78 m

2. Relevant equations

FTx=4(pi^2)R/T^2m
FTy=mg
R=Lcos(phi)

3. The attempt at a solution
To find the angle, I decided to use equations for force tension, then set the equations equal to each other using trigonometric functions, cosine on the equation for FTy and sine on the equation for FTx. The cosine on FTy canceled out with the cosine on the inserted equation for R, as well as the mass on both sides, leaving me with 4(pi^2)Lcos(phi)/T^2 on one side of the equation, and g on the other. However, I seem to have hit a snag. I cannot use inverse trig functions, as I do not have phi yet. I either messed up the symbolics, or something else is amiss.

2. Nov 6, 2014

ohmanitsDAAAAAN

It should be mentioned that the height, radius, velocity, centripetal acceleration, and angle are unknown.

Also unrelated yet mentionable, the mass in question is a cat a disgruntled Ukrainian tied up.

3. Nov 6, 2014

Simon Bridge

i.e. I don't know what you mean by "I use the force-tension equations" ... shouldn't you be using physics?

Recap: You have the mass m, the period T of the rotation, the length L of the tether, and you know it is a cone - so you just need a free-body diagram. What sort of motion does the mass execute? What should the forces add up to.

If the half-angle at the apex of the cone is $\alpha$, then $L\sin\alpha = R$, the radius of the base...
Then $F_T\sin\alpha=F_c$ is the centripetal force, and $mg=F_T\cos\alpha$
... is that where you are up to? I think you have $\phi = \frac{\pi}{2}-\alpha$ giving you cosines where I have sines.

You should be able to get $F_T$ and $\phi$ by simultaneous equations.

A disgruntled cat tied up as a conical pendulum seems pretty unmentionable to me...

4. Nov 7, 2014

ehild

It is confusing that you use T both for tension and time period. If FT is the tension, you can eliminate it by dividing the x and y components, giving $\frac{4\pi^2R}{gT^2}=tan(\phi)$. Substitute the third equation for R. And you certainly know how the tangent and cosine of an angle are related.