Center of Mass constant with near-correct attempt

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Homework Help Overview

The problem involves calculating the constant \( c = \frac{I}{MR^2} \) for a wagon wheel composed of a rim, spokes, and a hub, with specific masses and dimensions provided. The discussion centers around the moment of inertia calculations for each component and the overall setup of the problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations for the moment of inertia for the hub, rim, and spokes, questioning the assumptions made about the spokes' rotation point and the application of the parallel axis theorem.

Discussion Status

Some participants have provided guidance on re-evaluating the distance used in the parallel axis theorem and have expressed uncertainty about the calculations. Multiple interpretations of the setup are being explored, particularly regarding the spokes' rotation and the parameters used in the calculations.

Contextual Notes

There is mention of potential discrepancies in the answers obtained and the possibility of a glitch in the homework program. Participants are also considering the implications of significant figures in their calculations.

jamesbiomed
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Homework Statement



A wagon wheel is made entirely of wood. Its components consist of a rim, 16 spokes, and a hub. The rim has mass 5.1 kg, outer radius 0.90 m, and inner radius 0.86 m. The hub is a solid cylinder with mass 3.1 kg and radius 0.12 m. The spokes are thin rods of mass 1.1 kg that extend from the hub to the inner side of the rim. Determine the constant
c = I/MR2 for this wagon wheel.

Hi folks, this will be my last problem, promise.

Homework Equations



I for hub=> I for solid cylinder=> 1/2 M R^2
I for wheel=> I for thick hollow cylinder or wheel (I THINK...) =1/2 M (R1^2+R2^2)
Otherwise, I for hollow cylinder or hoop is MR^2
I for thin rod is 1/12 M h^2. These are rotating about their end though, so by parallel axis theorem, their moment is : 1/12Mh^2+M(h/2)^2

I(net)=cMR^2
c=I/MR^2

So there you have it.

The Attempt at a Solution



The correct answer is .356
My answer is .344

Here's what I did:

For Hub:1/2(3.1)*.12^2=.002232
For Wheel: 1/2(5.1)(.90^2+.86^2)=3.95148
For thin rods: 16 *{(1/12)(1.1)(.74)^2 + (.74/2)^2(1.1)}=3.212586

Then added them all to get 7.1832
Then divided them by their combined mass (16*1.1+5.1+3.1)=25.8 and the entire radius^2
so 7.1832/(25.8*.9^2)=.344

I've tried with less sig figs, and it doesn't work. There's something off, but its small. I have a feeling a quick once-over might spot it. Many thanks!
 
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I should note, I've given it many, many once-overs myself and can't find it. I don't think its alebraic, I think its conceptual, somewhere
 
Hi jamesbiomed.

Are you sure the spokes are rotating about an end? Aren't they rotating about the center of the wheel? But the spokes do not extend all the way to the center.
 
Last edited:
Tsny, I think you're right. That makes a lot of sense, and changes h. Thanks for the help!
 
Not h, but d distance during parallel axis theorem.
 
Darn it, still off.. this time a little high.

I added radius 12 to the distance in the P.a.th. Was that the correct adjustment?
 
I think that's right. I get that d = .49 m for the parallel axis theorem.
 
right, that's what I'm getting. Then I'm still using .74 for the h in 1/12mh^2.

So even though that needed fixing, there was something else. I just reworked the whole thing start to finish and still got the same answer.

For the R in 1/MR^2, would you say that's .9?

It could be a glitch in the HW program, so I might give up soon
 
Not to be pessimistic or anything, but that's all I can find
 
  • #10
I don't see anything wrong. I get an answer of .43 for c.
 
  • #11
That's what I get too. Not sure how they're getting their's. I feel good about the process, so I'll just run it by my prof in the morning. Thanks for the help.
 
  • #12
OK. Good luck with it.
 

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