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Homework Help: Center of mass varying volume

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data
    If x is the height of the remaining soda at any time in a can of soda, find x (in terems of M, H, and m) when the center of mass is the lowest. {{The soda is draining out of the bottom of the can}} Also, the Can and soda have uniform density.

    So the can is set up so that H=height of the can, M=Mass of the soda can (just the can not the soda) and m= the mass of the soda in the can.

    I know that I need to take the integral of the equation dm/dx, well i dont know, but i think this is true. I did a problem similar to this with a pyramid but had the volume and density so it was quite a bit different.

    I know that often it isn't good to just give the answer, and i agree with that fully. But i would love if someone could give me a decent sized nudge to get me there. I have the answser, thanks to the back of the book, but it looks so insane that I don't think any of what I have been doing has been in the right direction.

  2. jcsd
  3. Jan 16, 2009 #2


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    Yes, quite an interesting question! As the pop drains out, the center of mass will fall and it may well go below the center of the can for a while.

    The first step is to write a formula for the center of mass. Then you can analyze it and find its minimum.

    There appears to be something wrong with the answer you wrote - it's dimensions aren't quite right - better check it.

    Oh, quite messy! I got M and m mixed up badly, so started over using a and b instead of M and m. Also replaced b/h with k to reduce fractions during the calc. I ended up with an answer similar to the one you gave MINUS another term. I suppose that means I have an error. Hope you will post the correct answer for me.
    Last edited: Jan 16, 2009
  4. Jan 16, 2009 #3
    I am sorry I think that when i was writing the square root something got messed up. The end part is -1, not to the negative 1 power. Is that what you meant?

    I believe I am going to have ot find the center of mass and for this I am going to have to integrate for Dm then solve for x and take the derivative of this, so that I can find the critical number, but what equation to use for the center of mass?
    Last edited: Jan 16, 2009
  5. Jan 16, 2009 #4


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    I don't see any need for an integral. The center of mass is

    sum of (massxheight) for all masses
    sum of masses

    You can treat the can as one mass located at half its height.
    Same for the pop - a fraction of m. So no need for density and integration.

    The nasty part is that the total mass on the bottom depends on x.

    Yes, I get that answer with the minus 1 outside the square root.
  6. Jan 16, 2009 #5
    no integral huh?

    Well, that is something else, because my teacher actuallythought we neededd to integrate something also. I might also add, the teacher couldn't get this problem at the moment haha. Delphi, you think you could show me what your starting equation looked like for this? If it isn't toomuch trouble..
  7. Jan 16, 2009 #6


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    Well, I have to be very careful not to spoil the problem for you. Especially good problems like this one which are really important to your learning. How about just a hint . . .

    In the numerator, mass of can x average height + mass of pop x half height
    In the demominator, mass of can + mass of pop

    The trick is to express the mass of the pop when it is filled to height x as a fraction of m (mass of pop when can is filled).
  8. Jan 16, 2009 #7
    haha you sound like my physics teacher with the spoiling good problems. I will consider this hint and hopefully get it. Just a quick question, why would I do it at half height? Would it be because the center of mass for the can without the soda is at H/2 so that is where the mass would be of the can?
  9. Jan 16, 2009 #8


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    Yes, that is very close. Certainly the center of mass of the can is H/2, but the center of mass of the pop actually varies with x. It is at the center of the pop, not of the can.
  10. Jan 16, 2009 #9
    ... This is the worst feeling ever. Getting hints thrown at me and I still just cant seem to get it.

    The trick is to express the mass of the pop when it is filled to height x as a fraction of m (mass of pop when can is filled).

    I have been staring at this portion of your reply for like 20 minutes now and I just honestly can't grasp it. I dont know. Maybe it will come to me later. Thank you very much for the help though Delphi, At least I know ballpark of what I have to do.
  11. Jan 16, 2009 #10


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    Okay, the mass of the pop is x/H times m. A fraction of the full pop mass.
    Think about it when full: x =H and the x/H*m becomes the full m.
    When empty, x = 0 and the mass becomes 0/H*m = 0.
    When half full, you should get half of m. It works!
    Yes, I am a retired high school physics teacher. Kind of miss it.
  12. Jan 16, 2009 #11
    Thanks Delphi, I can't believe that i wasn't gettig that after the hints you were telling me, really kicking myself after that one. I ended up writing down the starting equation but have yet to take the derivative, looks like its gonna be pretty messy. Thank you very much Delphi, woulda prolly been doing it for the next few hours, can't stand when i dont get a problem.
  13. Jan 16, 2009 #12


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    Yes, the x in the denominator means you'll have to use the quotient rule.
    Save yourself some mess by multiplying it all by 2 - the minimum of 2C is the same as the minimum of C. And you can let k = m/H to clear some fractions and put it back as m/H in the final answer. It isn't too bad that way.

    There is a book all you students should read - "Outliers" by Malcolm Gladwell. He very convincingly writes that anyone of reasonable intelligence is not very good at anything until he or she spends thousands of hours practising it. After 10 000 hours, one becomes a world class expert. So keep working and don't worry - you're getting better at it all the time. It may efficient to study solutions to problems if you can find some in the library - that way you gain experience without so much frustration.
  14. Jan 16, 2009 #13
    I definetly agree with what you said about practicing something. The only issue is that with math and physics, I always get everything so quick, so spending alot of extra time happens to get very frustrating to me. I am learning as I get older to try to understand that learning takes time, but it is sort of hard to adjust to, in grade school and early high school plenty of people breeze through it until classes catch up.

    With that being said, I have been solving the derivative for this puppy and so far with 4 sheeets of computer paper I haven't gotten it. I keep working it through and end up getting an X only on the bottom, which since I am finding when the derivative is zero, the bottom wouldn't matter, I would be setting the numerator to 0. I am not sure what I am doing wrong, but I will prolly be able to get it better after I sleep seeing as how it is 12 at night. Thanks again Delphi, I hope this brought back some old times teaching at High school for you.
  15. Jan 16, 2009 #14
    So i got ((MH+Kx)/2)/(M+Kx) with K being m/H like you had said.

    When I work through the derivative, I get a KX on the top, but I assume I'd need an X^2 somewhere seeing as how the answer is a square root. I dunno if its my derivative that is wrong or my starting equation.

    I would also like to point out that this is way beyond our learning at this point. One of those stretch questions that nobody is really supposed to get ya know? They are the most fun.
  16. Jan 18, 2009 #15
    Hey guys I was wondering if anyone could run through this real quick and find out if the solution to it requires the use of quadratic equation. I am getting a derivative with an X and an X^2 that doesn't seem to be factorable so i dont know if I am doing something in correctly or if I would have to use quadratic to get this (which seems also impossible because my c value seems negative, whereas A is positive, so negative in the square root???) Is my original derivative equation the incorrect part?
  17. Nov 9, 2010 #16
    yes you need to solve a quadratic equation at some point
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