Gyroscope Analysis for Non-Symmetrical Body?

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SUMMARY

This discussion focuses on analyzing the dynamics of a non-symmetrical body using gyroscopic principles, specifically in the context of Problem 8 from an assignment related to rigid body dynamics. Key equations discussed include the inertia of a slender rod, represented as (1/12)ml², and the general moment equation Mx = H(dot) + HyΩz + HzΩy. Participants clarify the nature of the xyz reference frame, noting that the x-axis is non-rotating while the y and z axes rotate with the yoke. The conversation emphasizes the importance of correctly applying Euler's equations of motion to derive accurate results.

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  • Understanding of rigid body dynamics and Euler's equations
  • Familiarity with angular momentum calculations
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This discussion is beneficial for students and professionals in mechanical engineering, particularly those focusing on dynamics, robotics, and aerospace engineering, as well as anyone involved in the analysis of gyroscopic systems and rigid body motion.

murstr
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Hello, first of all I know this post for help is not the clearest. I am looking for quick help but not necessarily the answer. I know I can solve the math of the problem but I am having trouble analyzing and setting up what I have. So I am asking for help with the approach.

Homework Statement



Problem number 8 on this link: http://home.iitk.ac.in/~mohite/Assignment_05_AE688.pdf

Homework Equations



Inertia of slender rod is (1/12)ml2

The Attempt at a Solution



So here are the problems I have run into and would like to discuss:

1) do I consider xyz a non rotating reference frame?
I found angular momentum with respect to the xyz

Hx = ωsin2(phi)
Hy = -ωsin(phi)cos(phi)
Hz = phi

2) is Ω = (ωi + pk)?

3) The hint to the problem is to apply the general moment equation:

Mx = H(dot) + HyΩz + HzΩy

I get as far as assuming all terms are zero but HyΩz.

Which then gives me an answer: ωp(1/12)ml2sin(phi)cos(phi)

Thank you for the help.
 
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Hello, murstr. Welcome to PF!

murstr said:
1) do I consider xyz a non rotating reference frame?

I think the xyz axes shown in the figure rotate about the axis of the yoke (the x-axis). So, the rod always lies in the xy plane of these axes. The x-axis is therefore not rotating relative to the inertial frame of the laboratory, while the y and z axes rotate with the yoke.

I found angular momentum with respect to the xyz

Hx = ωsin2(phi)
Hy = -ωsin(phi)cos(phi)
Hz = phi

Note that the expressions on the right do not have the correct dimensions for angular momentum. (Also, I don't see how you got these expressions.)

2) is Ω = (ωi + pk)?

This is the correct expression for the angular velocity of the rod. However, the angular velocity of the yoke frame (i.e., the xyz frame in the figure) relative to the lab frame is just ##\vec{Ω} = ω \hat{i }## (see below for why this might be important.)

3) The hint to the problem is to apply the general moment equation:

Mx = H(dot) + HyΩz + HzΩy

Should there be a subscript on the first term on the right? Are the signs of all the terms correct?

This appears to be one of http://en.wikipedia.org/wiki/Euler's_equations_(rigid_body_dynamics)]Euler's[/PLAIN] equations of motion . If so, note that the Euler equations refer to principal axes that rotate with the rod. So, the x, y, z subscripts in this equation are not referring to the xyz axes shown in the figure.

[EDIT] After thinking some more, I think you can use your (corrected) equation for Mx in the xyz coordinate system shown in the figure (without going to the principal axis frame of the rod). But you will need to work with components of the moment of inertia tensor for the rod in this frame. And Ωz and Ωy would then be components of the angular velocity of the xyz coordinate system relative to the lab frame. As noted above, these two components are zero. If you are careful, you can get the correct answer for the problem fairly quickly this way.

The advantage of going to the principal axis frame of the rod (Euler's equations) is that the moment of inertia tensor is simplified in this frame. But, overall, I found the calculation to take about the same amount of work using either the xyz frame or using the principal axis frame.
 
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