Physics of a TreadWheel [Abstract Concept?]

[012anonymousx]

Homework Statement

[EDIT] Cleaned up the problem.

A tread wheel crane is an ancient device used to lift heavy objects. It consists of a large circle (like a hamster wheel) for a person to walk inside. The large circle connects to the middle where there is a small circle that is attached a rope that goes over a pulley and hooks onto some mass.
http://www.lowtechmagazine.com/2010/03/history-of-human-powered-cranes.html

Givens:
Outer circle radius = 6m
Inner circle radius = 0.5m
My weight: 500N
My speed: unknown.
I stand 1.5m away from the center of both circles

What is the maximum weight I can lift?

Homework Equations

F = (m)(v^2)/r
a(radial) = (v^2)/r
a(radial) = (r)(w^2)
v(tang) = (r)(w)

The Attempt at a Solution

I draw the two circles and placd a dot 1.2m from the center on the edge of the outer circle. The distance from the edge to the center is the radius of the big circle.

You can work out that the tangential force is mgsin(theta) given the geometry. Similarily, you can get a force directed outward mgcos(theta)

I'm thinking that F(Radial) is equal to F(outward). But I also thought perhaps you can solve v(tangential) given the tangent force...

But going ahead with the F(Radial) thing,

F(radial) = F(outward)
500cos(theta) = m(v^2)/r

That is where I was able to get to, because I do not know what mass is in this context. Nothing is really spinning around. Maybe my whole approach is wrong. I would greatly appreciate insight and help!

 

[rcgldr]

v^2/r is normally associated with centripetal acceleration. What is wanted in this case is the tangental force at the radius of the small circle.

 

[012anonymousx]

Yeah, I figured that after I thought about it a lot.

So F = ma, and we have the tangental force at the radius of the bigger circle.

But f = ma, and once again, we don't have m and we shouldn't even have a. The circle should not be accelerating. The angular velocity anyway.

 

[rcgldr]

But f = ma, and once again, we don't have m and we shouldn't even have a.

You have the weight that is supposed to be lifted by a rope attached at the radius of the inner wheel.

 

[rezeew]

The tread wheel crane is a prime example of how simple machines can be used to accomplish tasks that would otherwise require a great deal of effort. In terms of physics, the tread wheel operates on the principles of rotational motion and torque.

The maximum weight that can be lifted by the tread wheel crane depends on a few factors, including the radius of the outer and inner circles, your weight, and your speed. Based on the given information, we can use the equation F = (m)(v^2)/r to determine the maximum weight that can be lifted. However, it is important to note that this equation assumes a constant speed and circular motion, which may not be the case in this scenario.

To accurately determine the maximum weight that can be lifted, we would need to consider the dynamics of the tread wheel, including the forces acting on it and the motion of the person inside. This would require a more detailed analysis and may involve considering other factors such as friction and the strength of the materials used.

Overall, the tread wheel crane is a fascinating application of physics and serves as a reminder of how simple machines can be used to accomplish tasks that would otherwise be impossible.