Centre of Mass of a Uniform Cuboid -- Show that it is at the Center

  • Thread starter patrykh18
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  • #1
patrykh18
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Homework Statement:
Starting from a definition of Centre of Mass, show explicitly that for a uniform cuboid the centre of mass is at the centre.
Relevant Equations:
Provided below.
So, I volunteered to run a seminar to first year students in my college. They got a question like this for homework recently and a lot of them made a mistake in the calculation. I am not asking for help with the question itself because I know how to do it. However, a lot of students made a mistake that I shown in the image below. I'm curious about what would be the best way to explain to them why that is a wrong approach (without going into too much details about mathematics).
1605044028976.png
 

Answers and Replies

  • #2
Woahhhhhh, why you putting volumes in the limits?
 
  • #3
patrykh18
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Woahhhhhh, why you putting volumes in the limits?
Well I am going from x1 to x2 but it's all multiplied by yz
 
  • #4
PeroK
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Well I am going from x1 to x2 but it's all multiplied by yz
It should be multiplied by the constant cross sectional area, surely?
 
  • #5
patrykh18
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It should be multiplied by the constant cross sectional area, surely?

Yeah yz is the constant cross area
 
  • #6
I was trying to figure out how to answer this but there's too much wrong at the moment to make a start. You have ##x##'s in the limits for your integration with respect to ##m##, you use ##V## both as the volume of the cuboid and as an integration variable, you set ##V=xyz## at one point, when this is clearly incorrect [##x##, ##y## and ##z## are coordinates..., and I don't even know which ##V## you're trying to refer to], you have some weird limits. Too much to untangle for me, sorry.
 
  • #7
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Yeah yz is the constant cross area
The standard approach should integrate with respect to ##x##, and not ##V##. In any case, you cannot have ##V## as the fixed volume of the cube and an integration variable. That's a problem that encourages the invalid cancellation.
 
  • #8
patrykh18
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I was trying to figure out how to answer this but there's too much wrong at the moment to make a start. You have ##x##'s in the limits for your integration with respect to ##m##, you use ##V## both as the volume of the cuboid and as an integration variable, you set ##V=xyz## at one point, when this is clearly incorrect [##x##, ##y## and ##z## are coordinates..., and I don't even know which ##V## you're trying to refer to], you have some weird limits. Too much to untangle for me, sorry.

Yeah. I solved a lot of integrals. I know I need to distinguish between ##V## the volume and ##V## the integration parameter. I just never asked myself why you fundamentally have to do it.
 
  • #9
patrykh18
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The standard approach should integrate with respect to ##x##, and not ##V##. In any case, you cannot have ##V## as the fixed volume of the cube and an integration variable. That's a problem that encourages the invalid cancellation.

Yeah, if I did this question I would naturally distinguish between those two but I never asked myself why that is fundamentally the case.
 

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