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B Using center of mass for loop de loop

  1. May 12, 2016 #1
    A quick question, I remember my professor said to use the center of mass for solving the classic loop de loop problem. For example, he wanted us to use the distance from the surface of the loop to the center of the object (instead of just from the loop to the surface of the block) for the height in PE = mgh. Thinking back on this, it seems pretty weird that this was done. What do you guys think?
     
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  3. May 12, 2016 #2

    Doc Al

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    Often the size of the object is assumed negligible compared to other distances in the problem. But when you calculate changes in the gravitational PE of an object, it makes sense to use changes in the center of mass position.
     
  4. May 12, 2016 #3
    So lets say the mass is at rest at the highest position, and is just about to fall. So its potential energy would be say mg(51) instead of mg(50)? And at the zero reference point (ground), its potential energy is still mg(1)?
     
  5. May 12, 2016 #4

    Doc Al

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    If you measure PE from ground level, when the object is resting on the ground its PE = mgd (say). Then when it's at the highest point, resting on top of the track at height h, its PE would be mg(h+d). For some things it doesn't matter, since the change in PE will still be mgh.

    It depends on exactly what you are trying to calculate.
     
  6. May 12, 2016 #5
    I see, I just find it weird that its potential energy would be non-zero at ground level. When would one want to use the center of mass to calculate PE?
     
  7. May 12, 2016 #6

    Doc Al

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    Realize that the zero level is arbitrary. Only changes in gravitational PE have meaning.

    Depends on how accurate you want to be and what you are asked to calculate.
     
  8. May 12, 2016 #7
    In the classic problem, where it asks to calculate the energy needed to get pass the loop, is this required? I'm having trouble thinking up with situations where it would want to be done. What is the size difference to make this matter?
     
  9. May 12, 2016 #8

    Doc Al

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    Think about it. What's the PE initially? What's the PE of the object when it passes the top of the loop? (Note that the object is now below the track.) Since centripetal acceleration is involved, what radius will you use?

    It depends how accurate you want to be. Many versions of the problem don't even give the dimensions of the object, so you are expected to ignore it.
     
  10. May 12, 2016 #9
    For without accounting for the center of mass:
    y = 2r + 0.5r

    For accounting for the center of mass,
    y = 2r + 0.5r - 2x

    (y is the minimum height needed to pass the loop, r is radius of the loop, and x is the distance to the center of mass)
    So it would matter right?
     
  11. May 12, 2016 #10

    Doc Al

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    OK.

    How did you arrive at this?

    Sure. As to how much it matters, that depends on how x compares to r.
     
  12. May 12, 2016 #11
    mg(y+x) = mg(2r-x) + 0.5mrg
    y + x = 2r - x + 0.5r
     
  13. May 12, 2016 #12

    Doc Al

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    Check that 0.5mrg term. Recall my comment about what radius to use.
     
  14. May 12, 2016 #13
    0.5mrg should be 0.5m(r-x)g?
     
  15. May 12, 2016 #14

    Doc Al

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    Right.
     
  16. May 13, 2016 #15

    CWatters

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    Perhaps it helps to think about the case where the object is an irregular shape or even flexible with no obvious "surface" to measure from. Something like a rain cloud....

    Cloud.png
     
  17. May 13, 2016 #16

    Doc Al

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    That will certainly make things more interesting!

    I, of course, was speaking of the simpler case of a rigid object (a cart, for example) with a well-defined center of mass.
     
  18. May 13, 2016 #17

    CWatters

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    With a cloud the centre of mass might be the only think you can estimate the position of.
     
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