Register to reply 
Torque on a hub as a result from a flexible arm 
Share this thread: 
#1
Jan2013, 07:05 PM

P: 19

I'm working on a research project as a graduate student in engineering. I'm modeling a set of equations in Simulink and MATLAB derived from a journal entry.
I'm a little rusty on classical mechanics as my specialization is electrical engineering. I want to model the angular torque on a hub as a result from a flexible arm that has mass (and therefore inertia). The hub is a flat (or nearly flat) solid cylinder that also has mass. The arm is a type of flexible material that can move on its own (piezoelectric to be specific). The flexible arm will have a sinusoidal motion of some kind, this motion will translate into torque on the hub. I would like to go from the motion of the arm to an output of the torque on the hub. It doesn't have to be specific, just a general way of solving or going about such a problem. I have included a diagram to help demonstrate what I mean. The axis of rotation (though not attached to anything, no friction) is labeled "O". Any help with this would be much appreciated! 


#2
Jan2013, 11:56 PM

Homework
Sci Advisor
HW Helper
Thanks
P: 9,815

Do you mean a torque arising from the reaction against the arms as they move through some fluid, or from the inertia of the arms themselves (or both)?
If the first, you need some expression for the force per unit length that arises for a given rate of movement of a section of the arm. 


#3
Jan2113, 12:03 AM

P: 19

Sorry, I should have specified that. This would be in an ideal vacuum, so only from the inertia of the arms themselves.



#4
Jan2113, 12:27 AM

Homework
Sci Advisor
HW Helper
Thanks
P: 9,815

Torque on a hub as a result from a flexible arm
That's relatively easy, then.
Consider an element ds of an arm. At time t it will be (in polar relative to hub centre) at some position (r=r(t), θ=θ(t)). The tangential acceleration is ##r\ddot{\theta}+2\dot{r}\dot{\theta}##. See http://en.wikipedia.org/wiki/Polar_c...ector_calculus If the element has mass dm (=ρds if uniform density) then the reactive torque is ##(r\ddot{\theta}+2\dot{r}\dot{\theta})r.dm## 


#5
Jan2113, 12:34 AM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 6,477

The arms flap, and the hub spins.



#6
Jan2113, 12:43 AM

P: 19

I see, this makes sense. Will this account for the flexibility of the arm as well? I suppose I would specify that motion somehow. Also, this gives torque from the arm, so I can translate this directly to where the arm is mounted on the central hub correct? The hub will have a moment of inertia as well, and the end result I'm trying to reach is its angular motion around the axis "O" (in the center). In the end I would have a torque from the arm on one side of the equation and the torque applied to the hub on the other. Solving for the motion on the hub would it come out to angular acceleration? I really appreciate the help, this seems to be a problem that's not easily found by searching or flipping through text books. 


#7
Jan2113, 12:50 AM

P: 19

While I'm here, let me pose a conceptual question:
If these arms were to have a motion profile in which they move much faster in one direction than the other, translating to a higher acceleration in one direction (but for a shorter time) in one direction, can this result in an overall angular velocity of the hub? It wouldn't be smooth I imagine, but can a rotation be created in this way? It doesn't seem possible to me, but I believe one of the research sources is indicating this to be the case. 


#8
Jan2113, 01:01 AM

Homework
Sci Advisor
HW Helper
Thanks
P: 9,815




#9
Jan2113, 01:23 AM

P: 19

Neglecting the modes of the flexible arm, the equation looks like this: ##J\ddot{\theta}+ {\sum}({\rho}\ddot{\theta}qin^2 + 2{\rho}\dot{\theta}qin\dot qin + Jn\ddot qin)## Where J is the inertia of the hub, Jn is the inertia of the arm, and qin is the motion of the arm. The summation (I'm not versed in Latex) is i=1 to k, I think this is for point mass along the arm. 


#10
Jan2113, 01:24 AM

P: 19

Sorry, all that is equal to the external torque. I was trying to take this equation equal to zero and produce a rotation. But it didn't make sense.



#11
Jan2113, 06:44 PM

P: 102

The system can't gain angular momentum without an external couple force being applied.



#12
Jan2113, 10:31 PM

Homework
Sci Advisor
HW Helper
Thanks
P: 9,815

1. Arms flattened against hub, anticlockwise, say. System stationary. 2. Swing arms outstretched clockwise until flat against hub again. System stationary once more, but hub has rotated anticlockwise. 3. Curl tip of arm back, anticlockwise, but keeping it flat against itself. Slide tip anticlockwise around hub until arm back in original position. System stationary once more. Hub rotated clockwise during step 3, but not as much as it rotated anticlockwise at step 2. 


#13
Jan2113, 11:45 PM

P: 102

My point was that if the arms are moving in opposite directions, there is a couple due to the equal but opposite forces across the body as a result of the symmetry. If the arms are waving clockwise, the central body must turn counterclockwise to keep the total angular momentum at zero. Everything stops rotating when the arms reach the end of their travel.



#14
Jan2213, 07:34 PM

Homework
Sci Advisor
HW Helper
Thanks
P: 9,815




#15
Jan2213, 10:04 PM

P: 19

Sorry, I've been away. Yes, what I am working asks the question on whether or not a long term change in angular position of the hub can be created by some motion of the arms. They would move in the same direction (both CCW or CW when they move).
The arms would consist of a piezoelectric material, so they can only deflect in the same fashion either direction. They can't replicate the way a cat would do this to change position. The arms can't be shorter in one direction than the other. If you could imagine the inverse of the diagram shown above, that's exactly the way it would look if the arms were to move the other direction. My question was whether or not how these arms move could create a long term change in the position of the hub around the "O" axis. I had the impression that a very fast motion, say in the CCW direction, and a very slow motion (estimate a 4th of the time it takes for the CCW motion) in the CW direction would produce a higher angular acceleration in the CCW direction than the CW direction and therefore higher torque. However, it seems that the higher torque would occur for a smaller time than the lower torque which should equal out to a zero net change in position of the hub over time. This is a flapping motion if you will. The paper I'm referencing for this is attached, but perhaps I'm reading it wrong. It isn't my background so I'm a bit unsure of how I'm reading it. Equation (20) is the Lagrange equation of motion that I was using to attempt to prove this theory. 


#16
Jan2213, 10:07 PM

P: 19

I should note that (zeta) on the right hand side is equal to zero for my purpose, as there are no outside torques present. I'm trying to create one.



#17
Jan2213, 11:43 PM

P: 102

The math in that paper is a bit over my head. But I can offer this that you need to consider the nature of angular momentum and that it can be defined as the product of torque x time. So you can move an object slowly with over a long time with a small torque or move it fast with a large torque for a short time and still get the same angular momentum. Since the arms have nothing to push against in space, the system can't gain any angular momentum by flapping the arms. The speed doesn't really matter. The center body will shift in the opposite direction of the arms to conserve the total angular momentum. You won't be able to cause the body to spin, but you'd be able to change it's rotation angle.



#18
Jan2313, 01:16 PM

P: 19

I see that we can't create an angular velocity without morphing the shape of the arms. But I need to be able to bring this before my advisor and explain it well. I might have simply misunderstood his reasoning behind using this paper to begin with. With the equation of motion as you posted (the same form of what I posted just after now that I look at it), can the differential equation be solved in the usual way? Is there a general solution for that second order differential since the position "r" is also a function of time as is theta? It's been a while since calc IV, but everything I can find in books and reference only point to examples where the function (say "r" in this case) is being multiplied by constants, not timevarying functions such as theta. Wolfram doesn't like this guy either. These seem to be a bit ugly, but I do like the challenge. Thanks for all the help from all of you up until now by the way, it's been invaluable. 


Register to reply 
Related Discussions  
Motor Torque, Starting Torque, Stall Torque, Load Torque  Mechanical Engineering  3  
Dyna result doesn't match Test result  Mechanical Engineering  1  
Dyna result doesn't match Test result  Engineering Systems & Design  2  
Dyna result doesn't match Test result  Materials & Chemical Engineering  0 