Carnival ride : chair swings from cable attatched to overhang, spinning on axis

[conorwood]

Homework Statement

The chair is 10 m from the rotating vertical axis. The solid overhang stretches 6 meters from the axis. (making 4 m from end of overhang to chair on the radial axis). The chair spins at a constant speed at 1 revolution per 10 seconds. Find the angle θ the cable makes with the vertical axis.

Homework Equations

ƩFr = mar = m(ω^2)r
ƩFz = maz
ƩFt = mat = 0 (constant speed)

The Attempt at a Solution

I have not had troubles with the equation, but I have had issues setting up the problem.

I originally thought that I might be able to find θ if I analyzed the chair as if the cable was attached directly to the z axis. This would give me an angle let's call β. From here I could use trig to find θ. This is only true if the relationship between the situation where the cable is attached directly to the z axis and the situation where the cable is attached to the over hang looks like this:

|-\
|-β--\
|-------\
|----------\
|-------------\
|----------------\
|________________\
(sorry for the bad diagram. Its a triangle ignore the --)

where the bottom side is length 10 and

|-\
|-θ-\
|----\
|-----\
|------\
|-------\
|_______\

Where this bottom line is 4 and the heights are equal.

I doubt this is true. I would guess that the heights wouldn't be equal, and thus I would not be able to find θ this way.

My question is how I would find this true angle, or more generally, how would I deal with any radial problem where a mass is hung by a rope from a spot a certain distance x away from the center of the circle.

Thank you

 

[tiny-tim]

hi conorwood!

I originally thought that I might be able to find θ if I analyzed the chair as if the cable was attached directly to the z axis. This would give me an angle let's call β. From here I could use trig to find θ.

i think β and θ are the same

try solving the equations …

you'll probably find that the radius of revolution is the only length that matters