Using inertia tensor in conservation of angular momentum

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Homework Help Overview

The discussion revolves around the application of the inertia tensor in the context of conservation of angular momentum, specifically regarding a problem that involves calculating moments of inertia for a thin plate and their use in angular momentum equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the inertia tensor and its components as presented in a solution manual, expressing confusion over the labeling of moments of inertia and their application in calculations.
  • Some participants clarify the notation used for moments of inertia and discuss the implications of the parallel axis theorem.
  • Questions arise regarding the correct application of the final angular momentum equation and the role of the unit vector in the context of the problem.

Discussion Status

Participants are actively engaging in clarifying the concepts related to moments of inertia and their calculations. Some guidance has been provided regarding notation and the interpretation of the equations, but the original poster continues to seek clarity on specific calculations and their implications.

Contextual Notes

The original poster mentions a lack of instruction on this topic in class and expresses frustration over the discrepancy between the solution manual and their understanding. There is an indication of uncertainty regarding the values and labels used in the problem setup.

Mark24
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Hi all, this is my first post. I'm under the gun at the moment because we were just told yesterday to expect a similar problem to the one below on our final, yet we were not taught how to work such a problem in class. (Our teacher does not write the final hence the discrepancy.)

I have a solution manual I am trying to learn how to do this problem with the help of, but it is not going very well. My book only vaguely touches on the inertia tensor so I'm very confused about what's going on in this problem. I've posted the two pages from the solution manual. I keep trying but I can't seem to duplicate the numbers the solution manual gets.

1)The problem statement, all variables and equations:
Page 1: http://i43.tinypic.com/2j4d761.jpg
Page 2: http://i41.tinypic.com/w1mhg.jpg

2) Attempt at solution
The 'I1' calculation seems to be Izz for a thin plate from the back of my book, where Izz = 1/12*m(a2 + b2). Yet the manual labels that value I1 and seems to use it in place of where Ixx should be in the inertia tensor (Imat.) I don't know if this is an error or not.

And also for Ioa, I'm using the formula Ixx*ux2 + Iyy*uy2 + Izz*uz2 where the u's are the unit vectors for ro/a. Since ux is zero, I should get: Ioa = Iyy*uy2 + Izz*uz2. Again, I'm unsure of which is Iyy and Izz from the solution manual due to the obscure labeling of I1/I2/I3, and I'm unsure of how to identify which is which from the picture. (that's something I never figured out since we are usually given the values)

If anyone can offer some tips or suggestions on this problem, I would be very grateful. After spending about 4 hours staring at it, googling, etc. I'm really at my wits end.
 
Last edited:
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Welcome to PF!

Mark24 said:
The 'I1' calculation seems to be Izz for a thin plate from the back of my book, where Izz = 1/12*m(a2 + b2). Yet the manual labels that value I1 and seems to use it in place of where Ixx should be in the inertia tensor (Imat.) I don't know if this is an error or not.

Hi Mark24! Welcome to PF! :smile:

The 1 2 and 3 refer to x y and z, so the I1 is Ixx, which they get by adding Iyy and Izz (the parallel axis theorem).

I don't understand why you're confused … 1,2,3 ~ x,y,z is fairly simple notation :confused:

Anyway, look at the diagram :smile: …​

Izz clearly depends on a, and it's through the c.o.m., so it's small, so it's 1/12

while Iyy clearly depends on b, and it's not through the c.o.m., so it's larger, so it's 1/3 :wink:
 
Thanks tiny-tim, how they derived the moments of inertia makes sense now after you explained it. Since the inertias seem to be correct but I'm still getting the wrong answer, I must be messing up the final equation - the one listed in the second picture as:
Imat(0,0,w1)oa = Ioa*w2

For Ioa, I get Ioa = Iyy*uy2 + Izz*uz2 = .192

For Imat, I get a diagonal matrix with values: .4167, .2667, .15. w1 is 2 rad/s.

If I do Imat*(0,0,w1) = Ioa*w2, I get w2 = 1.56 in the z direction and 0 in the y, which is wrong. Any idea what I am missing? The solution has an 'oa' at the end on the left hand side of the equation. oa is the unit vector, so I don't understand why it would be needed in the cons. of angular momentum equation.
 
Last edited:
Hi Mark24! :smile:

(have an omega: ω :wink:)
Mark24 said:
If I do Imat*(0,0,w1) = Ioa*w2, I get w2 = 1.56 in the z direction and 0 in the y, which is wrong. Any idea what I am missing? The solution has an 'oa' at the end on the left hand side of the equation. oa is the unit vector, so I don't understand why it would be needed in the cons. of angular momentum equation.

Imatrix converts the angular velocity vector (0,0,ω1) to the angular momentum vector, Imat*(0,0,ω1).

The external forces are at O and A, so angular momentum is only conserved about the OA axis, so we need the OA-component of the angular momentum vector:

Imat*(0,0,ω1) "dot" oa :wink:
 

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