Centrifigal force? Can I get some feedback on this?

Question

Say you have a stick/rod/bar. It is rotating like a two-blade propellar. One blade of equal size opposite eachother spinning around an axis. If at the end of each blade I have a 50lb weight. Now lets say I swap one blade with a 25lb weight. How much further out would I have to stretch one blade to compensate for the reduce weight so that each blade is balanced? So that I could still spin the propeller, even with two different weights, and not have it loose it's equalibrium? Any answers or suggestions appreciated. Thank you in advance.

Answer 1

You don't need a bunch of complex math.  Just think of a see-saw. One kid weighs 50 lbs, and the other weighs 25. Were would you put the fulcrum to balance the forces?

50........................25

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|<--x --> | <---2x --->|

Sorry, using Arial font throws off the diagram.

Answer 2

simply draw a diagram

lets say the length for the weight to the center is l for each side

no initially you have 50*l = 50*l

if you reduce 25 pounds from side

then

you have 25*x= 50*l

if you know l then plug in and solve

else leave it in terms of l

and x= 2*l

and ya dont take this Centrifigal force thing to seriously  because there is no such this its just a simplification of some thing complex you will learn that in college level dynamics class which i took.

Answer 3

This is one situation where you cannot ignore the cross-section and density of the rod, nor the density and geometry of the weights on the end.  Assuming a uniform cross-section of rod and cubical weights of equal density,

EDIT:

(ρ(w))((s(w1))^3)(s(w1)/2 + L1) + (ρ(r))(A(r))L1 = (ρ(w))((s(w2))^3)(s(w2)/2 + L2) + (ρ(r))(A(r))L2

Answer 4

Over the edge is right.

You can do this one simply by using Archimedes principle

F d = F d

force x distance to fulcrum on one side = force time distance to fulcrum on other side.

this simplifies to w d = w d where w is weight.

Say the length of the rod is 1 meter, then this equation becomes:

25(x) = 50(1-x)

solve for x but you should get that the balance point is 1/3 of the way along the rod.

Use the equation above for finding any balance point, changing the 25 and 50 to your weights, and changing the 1 in the equation to the actual length of the rod you are using.

Hope this helps.

ADDED

I did assume that your situation was idealized: first that the rod was uniform, second that the weights have a center of gravity at the distances you specified and that the don't move as the rod spins.

Answer 5

The most important thing is to get your static balance right.  You want the center of mass to be over the center.  So if the mass is half as much, it needs to go out twice as far just to balance the static torques.

N = mgr

Centrifugal force is:

F = m omega^2 r

So that same ratio that balances the static torques will balance the centrifugal forces.

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Now what that will may do, however, is throw off your dynamic balance.

The moments of inertia won't be equal on either side.  I think that might cause your blade to want to nutate, which is going to be tough on your bearings.  Maybe not. Your principle axis is still the same.  You're centrifugal forces balance, so you're probably ok.

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I'll star this so DrD looks at it.  He's an engineer and probably can tell you for sure what that's going to mean without having to think about it.  But I think you're ok.