An issue with Conical Pendulums

[ohmanitsDAAAAAN]

Homework Statement

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

Homework Equations

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

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.

 

[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.

 

[Simon Bridge]

I'm not following your working.
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...

 

[ehild]

Homework Equations

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

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.

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.