What RPM Must a Shaft Reach for Arms to Extend Perpendicular?

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To achieve perpendicular arms on a rotating shaft, the necessary RPM can be calculated using the relationship between centrifugal force and gravitational force. The centrifugal force acting on the arms, which weigh 2.5 lbs each and are 10 inches long, must counterbalance the gravitational force for the arms to extend horizontally. The discussion reveals that while achieving perfect perpendicularity is theoretically impossible, one can get very close by rotating at a high frequency, approximately 100 Hz for massless arms. Additionally, when arms are held at a 45-degree angle with a rope, the tension in the rope varies depending on the rotation speed, with potential scenarios where the angle is less than, exactly, or greater than 45 degrees. Overall, the analysis involves balancing forces to determine the dynamics of the system as it rotates.
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Hey guys. Heard about this forum from a friend and decided to join up. Just so you guys know, some of my questions are pretty random and weird , hope i don't annoy you fellas...

Alright, first question i got.

http://img170.imageshack.us/img170/5953/00000000zd3.png


say you have 2 arms connected to a shaft, the arms are hinged which allows them to move up and down . The arms weigh 2.5lbs each and are 10 inches long. How do you figure out at what rpm the shaft has to spin at in order for the arms to be perpendicular to the shaft?
 
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welcome to physics forums =)
good question, I am no physicist or anything but maybe I can help you get started until one gets here

Fc=m(w^2)r

Fc= centrifugal or inertial force
w=angular velocity
r=radius

this force acts to throw the center of mass as far away from the axis of rotation as possible, this maximum distance occurs when the plane of the rotation is perpendicular to the axis of rotation

I guess there must be some torque involved to lift up the arms... this is where I reach a dead end
 
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Welcome to PF, Treborly.
I don't know the answer to your question, but it's the basis of the 'fly-ball' governers used on steam engines. When the arms got too high, they throttled down the steam.
 
Danger said:
Welcome to PF, Treborly.
I don't know the answer to your question, but it's the basis of the 'fly-ball' governers used on steam engines. When the arms got too high, they throttled down the steam.


ya, same thing is used on some elevators to engage the emergency brakes.
 
Quick answer: Never.
Longer answer: The direction of the arms will equal the direction of the sum of gravitational force and centripetal force (note that both are vectors).
Reason for the quick answer: You want to achieve a direction (1,0) which is not possible due to the sum of the forces always being (Fc,Fg) with Fc being centripetal and Fg being gravitational force. In other words: There's always a non-zero component downwards.
 
Longer answer:

The correct answer is "never." But you can get as close as you like--so close that, for all intents and purposes, the thing looks like it is horizontal.

For example, suppose that you wanted to be within 0.001 radians of horizontal.

In that case, you only need to rotate the shaft at about 100Hz, and that isn't too tough.
 
Oops, sorry. I did that calculation for massless arms of a length 1 meter long, not 10 inches. In the picture it looks like there are some weights at the end of the arms...

I figured that that meant the arms could be treated as is all their mass was at the end. In which case the mass doesn't matter. For the case of 10 inches instead of 1 meter one can use the formula

<br /> \nu=100\sqrt{\frac{1}{L}} \quad (\textrm{Hz})<br />
where 'L' is whatever 10 inches is in meters...
 
alright cool, thanks.
 
how do you get that? can you describe it in terms of angular velocity...or get that angle in the equation somehow?
 
  • #10
let's see... I just did a balance of forces:

Call the angle that the arm makes with the shaft \theta. Then there are two forces, the tension T and the gravitational force mg. The mass at the end of the shaft moves in a circle of radius r=L\sin(\theta) where L is the length of the arm. So we have
<br /> T\sin(\theta)=m\frac{v^2}{r}=m\omega^2 L\sin(\theta)<br />
thus
<br /> T=m\omega^2 L<br />

And the other component of the forces gives
<br /> T\cos(\theta)=mg<br />

Plugging one into the other to eliminate T gives
<br /> \cos(\theta)=\frac{g}{L\omega^2}<br />

but, if \omega is large we see that \cos(\theta) must be small. I.e. \theta must be near \pi/2 thus if I write \pi/2-\theta\equiv\epsilon ((where actually, \epsilon is either cos(theta) or (\pi/2 - theta) in radians--they are roughly equal near theta=\pi/2...just like sin is roughly equal to its argument near zero)) then I have
<br /> \omega^2\approx\frac{g}{L\epsilon}<br />

Then I chose \epsilon=10^-3 and used g=10m/s^2...
 
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  • #11
=) thank you for that
 
  • #12
Another question related to this topic...

Take the diagram in my first post. You raise each arm to 45°, then tie a rope to the end of each arm which then connects back to the main shaft also at 45°. As you spin the shaft the arms rise, until the rope stops it from going past 45°. How to you figure out how much tension is put on the rope at a given rpm?

http://img178.imageshack.us/img178/1305/00000000sy4.png
 
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  • #13
Same method. Balance the forces.
 
  • #14
I can't figure that out. Can you do this one for me? each arm is 10 inches long , arm weighs 3 lbs each, shaft spins at 2600 rpm. How much tension in lbs is on the rope.
 
  • #15
well, if its 45 degrees... all the tension goes to the arm and there should technically be no tension in the rope
 
  • #16
I think he wants to consider the case when it is spinning so fast that there is tension in the rope.

In general, either: 1. There will be no tension in the rope and the angle will be less than 45 degrees; or 2. The angle will be just exactly 45 degrees and there will be just exactly no tension; or 3. The angle will be 45 degrees and there will be tension (but the tension depends on the frequency of rotation).

Of course, cases 1 and 3 form the great majority. In case 3 one will see three forces: the tension in the arm, the tension in the rope, and the gravitational force. These three forces may be subjected to the usual Newtonian analysis to obtain expressions for the tensions in terms of the frequency of rotation.
 
  • #17
in that diagram... isn't it the same situation as just having one rope perpendicular to the axis of rot. in between the arm and rope... instead of having an arm and rope converging at 45 degree angles... -i- instead of <|>... since its 90 degrees, you could pretty much ignore gravity and say that the centripetal force is the only force on it, then divide that force by two, and that's the tension in the bottom rope
is that what you meant by case 3?
where does frequency of rotation come into the equations?
 
  • #18
olgranpappy said:
I think he wants to consider the case when it is spinning so fast that there is tension in the rope.

In general, either: 1. There will be no tension in the rope and the angle will be less than 45 degrees; or 2. The angle will be just exactly 45 degrees and there will be just exactly no tension; or 3. The angle will be 45 degrees and there will be tension (but the tension depends on the frequency of rotation).

Of course, cases 1 and 3 form the great majority. In case 3 one will see three forces: the tension in the arm, the tension in the rope, and the gravitational force. These three forces may be subjected to the usual Newtonian analysis to obtain expressions for the tensions in terms of the frequency of rotation.

ya, that's what i meant.
 

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