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Center of mass for a cylinder with water

  1. Oct 31, 2009 #1

    LMH

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    Hi! :)

    This is my first post here, hoping this will help me understand a few things. I'm really stuck at this problem and it is annoying me because I think I understand the theory, but just can't seem to be able to do it. So any help, hints and tips to solve this kind of problem is highly appreciated! :)

    Also, don't really know if I'm supposed to post this in introductory or advanced physics, but i figured this was most probable.

    1. The problem statement, all variables and given/known data

    A cylinder shaped tank is filled with water, the water then goes out a hole in the bottom of the cylinder. The height and mass of the water is given by x and m and the height and mass of the cylinder is given by H and M. What x corresponds to the lowest value of the centre of mass for the cylinder with water?


    2. Relevant equations

    I'm guessing this equation is relevant:

    R = [tex]\frac{1}{M}[/tex] [tex]\sum[/tex] miri


    3. The attempt at a solution

    I have thought a lot about this and figured that the centre of mass will start at the middle, when the tank is full. Then as the water pours of of the tank, it will lower. But at some point it will start to increase again, because the mass of the cylinder will become dominant over the mass of the water. I figured the x I'm looking for might be the one where the two masses are equal, does this make sense?

    Either way, I can't seem to be able to express myself mathematically, and then I'm just as far. I'm supposed to express the answer x in terms of M, H and m.

    Please help :)
     
  2. jcsd
  3. Oct 31, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi LMH ! Welcome to PF! :smile:

    (are you at LMH? :wink:)
    The mathematical way is to proceed step by step …

    start by saying let the height be x, then find the position of the c.o.m. of the water, then use your formula to find the overall c.o.m.

    What do you get? :smile:
     
  4. Nov 1, 2009 #3

    LMH

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    No, I'm not at LMH :) Didn't even know what it was, so I had to google it :p

    Anyway, thanks for replying! I have thought a bit more about this and what you said. The c.o.m of the water must always lie at the center of the water volume i suppose, that is at x/2. Following that logic, the c.o.m of the cylinder lies at H/2. I then put these two heigths in the formula and get:

    R = [tex]\frac{1}{M+m}[/tex] ( m[tex]\ast[/tex][tex]\frac{x}{2}[/tex] + M [tex]\ast[/tex][tex]\frac{H}{2}[/tex] )

    I then ask myself: "How do I find the smallest possible R? Aha! The derivative!"

    But a problem the occurs when I try to differentiate R because, at least in my head, both m and x are variables. How do I handle that?

    I then thought about the formula for the speed of the center of mass which is analogous to the one with R, but again, how do I find the speed of the c.o.m of water?

    Stop me if I'm thinking to complicated or in the wrong direction please, because I really feel there is something I'm missing here :)
     
  5. Nov 1, 2009 #4

    tiny-tim

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    Hi LMH! :wink:

    Yes, but variable m is a function of variable x …

    m = … ? :smile:
     
  6. Nov 1, 2009 #5

    LMH

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    m = d * V ?

    where d is the density of water and V is the volume given by x2pir2 ?
    (Could not seem to get the pi to work, it appeared as a superscript for some reason)

    But then I'm thinking: "d and r are not given in the exercise, and I'm not supposed to include them in my answer."

    or.. ?
     
  7. Nov 1, 2009 #6

    tiny-tim

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    (have a pi: π :wink:)

    Try keeping d and r as letters, and writing it all out …

    maybe the answer won't depend on them after all. :wink:

    What do you get? :smile:
     
  8. Nov 1, 2009 #7

    LMH

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    I get something complicated :tongue2:

    I tried to put m = dx2πr2 in:

    R = [tex]\frac{1}{M+m}[/tex] ( m[tex]\ast[/tex][tex]\frac{x}{2}[/tex] + M [tex]\ast[/tex][tex]\frac{H}{2}[/tex] )

    Then I tried to differentiate that, but it all became rather ugly pretty quick.

    I got something like:

    R' = [tex]\frac{ xM + dx^2 pi r^2 - M Hx/2 }{( M + dx 2 pi r^2)^2}[/tex]

    and then I was stuck..
     
  9. Nov 1, 2009 #8

    tiny-tim

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    I'm confused … what is your m as a function of x? :confused:

    Don't you get something like R = (Ax + B)/(Cx + D)?
     
  10. Nov 1, 2009 #9
    You put m = dx2πr2, is that [itex] m = d x^2 \pi r^2 [/itex] ? The volume of a cylinder is height * area of crosssection, so it should be [itex] m = d x \pi r^2 [/itex]

    You better calculate R and R' again. The R' you should get is indeed something like what you had, so it won't look any better.

    note that if you have R' = f(x)/g(x) = 0 than you must have f(x) = 0
     
  11. Nov 1, 2009 #10

    LMH

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    Hmm... I don't know why I had that 2 in my formula for m, but my algebra was way off anyway.

    [itex]
    m = d x \pi r^2
    [/itex]

    That's what I'm working with :smile:

    So, I put that in to:

    R = [tex]\frac{1}{M+m}[/tex] ( m[tex]\ast[/tex][tex]\frac{x}{2}[/tex] + M [tex]\ast[/tex][tex]\frac{H}{2}[/tex] )

    I then tried to differentiate that and used:

    Which in the end gave me:

    2xM + d π r2x2 - MH = 0

    Then I thought that since:

    [itex]
    m = d x \pi r^2
    [/itex]

    I could just put that back in and get:

    2xM + mx = MH

    which would then give me:

    x = [tex]\frac{MH}{2M + m}[/tex]

    but the answer I'm looking for is:

    x = [tex]\frac{M}{m}[/tex] * H * ( [tex]\sqrt{1 + \frac{m}{M}}[/tex] - 1 )

    It's probably my algebra, I'm a bit rusty, but am I on the right track?

    I really appreciate your help with this! :)
     
  12. Nov 1, 2009 #11
    You can just solve 2xM + d π (r^2)(x^2) - MH = 0 for x with the quadratic formula.

    Since your m depends on x, it can't appear in the answer. I think m was meant to be a constant, and equal to mass of the water if the cylinder was full. The mass of the water, if the height of the
    water is x would then be [itex] (x/H)m [/itex]

    you'll need to replate [itex] d \pi r^2 H [/itex] with m in your answer.

    BTW, do not use x2, but use x^2 or [itex] x^2 [/itex]
     
  13. Nov 2, 2009 #12

    LMH

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    Hmm.. no m is definetly a part of the answer, because:

    x = [tex]\frac{M}{m}[/tex] * H * ( [tex]\sqrt{1 + \frac{m}{M}}[/tex] - 1 )

    according to my book :smile:

    tiny-yim:

    I get something like:

    R = (A(x^2) + B)/(Cx + D) = f(x)/g(x) :smile:

    I have tried to differentiate this using:

    f'(x) = [tex]\frac{f'(x)*g(x)-f(x)*g'(x)}{g(x)^2}[/tex]

    And then setting the nominator = 0

    which gives me:

    2Mx + d π r2x2 - MH = 0

    I tried using the quadratic formula, but ended up having to take the square root of:

    4(M^2) + 4dπ(r^2)MH

    and I don't really know how to do that. :smile:
     
  14. Nov 2, 2009 #13

    tiny-tim

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    oh i see … m is the total mass of the water when the tank is full (depth H) …

    so the mass at depth x is xm/H.

    Does that make any difference?
     
  15. Nov 2, 2009 #14

    LMH

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    Why do you say that m is the total mass when the tank is full?

    I don't understand, my original thought was that the m is the mass of the water at any given time. Making it a variable and not a constant? Or does that not make sense?

    But if what you say is the case:

    I don't really understand what do to with this :confused:

    Do I put in my formula for R?
     
  16. Nov 2, 2009 #15

    tiny-tim

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    Yes, I know, but looking at the solution makes it clear that that is wrong …

    m is a constant.
    Yes, instead of m, use xm/H …

    I did that, and I got the answer given. :wink:
     
  17. Nov 2, 2009 #16

    LMH

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    Allright, I now tried to put xm/H in for m in:

    R = [tex]\frac{1}{M+m}[/tex] ( m[tex]\ast[/tex][tex]\frac{x}{2}[/tex] + M [tex]\ast[/tex][tex]\frac{H}{2}[/tex] )

    and eventually got:

    (x^2)m/(H^2) + 2xM/H - M = 0

    did you get that?

    I tried to solve that using the quadratic formula, but I can't seem to manipulate my answer in to what I'm looking for, so I'm wondering if I might have the wrong starting point for finding my x :smile:
     
  18. Nov 2, 2009 #17

    tiny-tim

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    (try using the X2 tag just above the Reply box :wink:)

    Yes, that's what I got :smile:

    how can you not get the book answer from that? :confused:

    What did you get?​
     
  19. Nov 2, 2009 #18

    LMH

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    I got the right answer! :biggrin:

    I guess I just haven't been doing enough exercises because I do all these small mistakes that mess everything up, but this sure made me feel better! :smile:

    But! I don't understand why m is supposed to be the mass of the water when the tank is full, I'm not able to read that from the text which says, translated from norwegian:

    Express the answer in terms of the mass of the tank M, its height H and the the mass of the water m which is in the tank.

    But even if that had been stated more clearly, how did you come up with xm/H? :smile:
     
  20. Nov 2, 2009 #19

    tiny-tim

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    I'm glad it came out ok in the end! :smile:

    The question started …
    … and I suppose "the mass of the water m which is in the tank" refers to this original mass.

    The tank is the same shape all the way up, so the volume is proportional to the height, x.

    Since it must equal m when x = H, the factor of proportionality is m/H (because then Hm/H = m, which is correct), and the mass is mx/H. :wink:
     
  21. Nov 9, 2009 #20

    LMH

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    It's a bit late, but I get it now :)

    Thank you very much for your help with this!

    I had problems understanding how we could express m as a function of itself, but I realised that since x = H it actually says m = m.

    Things got busy last week, so I almost forgot this, but I had to come back and thank you! :)
     
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