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Find the center of mass of a cone with variable density...

  1. Nov 1, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the center of mass of an inverted cone of height 1.5 m, if the cone's density at the point (x, y) is ρ(y)=y2 kg/m.

    2. Relevant equations

    The formula given for this problem is rcm=1/M * ∫rdm, where M is total mass, r is position, and m is mass.

    3. The attempt at a solution

    I've read you can do this with a triple integral instead, but I don't fully understand that. At my university, triple integral is in Calc3, and we only need Calc1 for this physics course.
    Something also seems wrong about the question. It wants a specific numeric answer, but don't you have to know the radius to determine the total mass?
     
  2. jcsd
  3. Nov 1, 2016 #2
    Hint: Your intuition is incorrect. You have all the information you need. And, you don't need to do this with a triple integral.
     
  4. Nov 1, 2016 #3
    You can assume the radius of the base of the inverted cone as R then M = Integral (y = 0 to 1.5)[ypi*{(y*tan alpha)^2}*dy], where alpha is semi vertical angle of cone which is given by tan alpha = (R/h). You can find ycm in terms of M and R and then find the ratio of y/h to get the numerical value.
     
  5. Nov 2, 2016 #4

    haruspex

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    You would, but luckily you do not need to find the total mass.
     
  6. Nov 2, 2016 #5
    You can find total mass in terms of R or alpha and also center of mass, say Y in terms of M and alpha You can also express 1.5 m maximum y in terms of M and alpha and then find the ratio Y/ 1.5 to get Y. I think M and alpha may get eliminated.
     
  7. Nov 2, 2016 #6
    Or as others are suggesting balance the clockwise and anticlockwise moments of disks above Y and below Y. You need to equate two integrals tan alpha will cancel out on both sides
     
  8. Nov 2, 2016 #7
    Balancing two integrals is far above my pay grade, so I'm not sure why we're even given this question. We've never seen anything similar in class.

    I've been examining http://physics.stackexchange.com/questions/38624/determining-the-center-of-mass-of-a-cone, which comes up with two balanced formulas. The problem I see in their final formula is that ρ ended up being on both sides and cancelled out. I think their formula also assumed the balance point is somewhere on the z-axis, which wouldn't be accurate in my case, because the mass is not centered on any given x-y disc.

    I'm just not informed enough in calculus to understand this one.
     
  9. Nov 2, 2016 #8
    Does their formula simply calculate only the z portion, meaning that no matter what you plug in for ρ, the z-axis offset is always going to be 3/4h? And if so, don't I just calculate the x,y parts without considering z at all? It seems like the axis of the center of mass is going to be curve from the top of the cone to a point on the bottom which is not center, given that ρ(x,y)=y2. I'm hearing that r cancels out, but if that's the case, how do I know the bounds of y?

    Given the rest of the homework, and because this is not the last problem we were given, it seems to me that this should be something we can calculate with the formula given for the homework. One interval using just the ρ given. The answers offered so far are way outside the bounds of the class, so I don't understand what I am missing.
     
  10. Nov 2, 2016 #9

    PeroK

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    Are you sure that the cone isn't lying along the y-axis?
     
  11. Nov 2, 2016 #10

    kuruman

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    Try this. Think of the inverted cone as a stack of disks of thickness dy. The mass of one such disk is ##dm = \rho \pi r^2 dy##. Express ρ and r in terms of y. Then add up all the disks (it's a one-dimensional integral in y) to get an expression for the total mass. This gives you the denominator in the expression for the CM. To get the numerator, simply multiply the previous integrand by y and integrate once more. Take the ratio and you are done. The answer is extremely simple.

    My assumption is that the cone axis is along y.
     
  12. Nov 2, 2016 #11
    Nevermind. Yes, the teacher just confirmed that there is no z axis, and that y is the "cone" axis, and it isn't a cone, it's just a triangle. Makes sense now.
     
  13. Nov 2, 2016 #12

    PeroK

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    It still works as essentially a 1-D problem even if it is a cone. The mass is distributed along the y-axis in the same way. So, it could be done with simple mathematics if you can see this relationship and use the symmetry of the cone

    It may be a mistake, therefore, to think that you need more sophisticated mathematics to solve this problem for a cone.
     
  14. Nov 2, 2016 #13
    This teacher has a problem with the difference between 2D and 3D. He never defines which axis we're talking about. He seems to think of everything in 2D when he teaches the class, but then says words like "cone". Obviously, if you don't define which is x, y, and z, all you have is guessing left. That just simply isn't science. Like Sudoku puzzles that come down to a flip of a coin between two answers. That's not what I paid for at the news stand :) Cheapens the experience...

    Thanks to everybody for trying to help out. I'm told that in geophysics higher classes, I actually will make use of cone density formulae with x or y having non-linear density, but by then, I will have finished higher calculus, and my brain won't hurt as much. So they say...
     
  15. Nov 3, 2016 #14
    OK, so if anybody is still interested in this, I'm told that the key point of data in the question is to notice a mistake. That mistake is that density ρ is actually linear, not volumetric. Note that density is in kg/m not kg/m3. So part of this experiment was to test if we remembered chapter 1 (about Dimensional Analysis), and could spot the error. Density was incorrectly stated as ρ (for volumetric density), and should have been spotted as needing to be in μ (mu). Had we spotted that, we would know instantly that "y" represents z axis of the cone. Vector r is then replaced with just y in the equation. So then the formula given is then the exact formula we needed, nothing more. The answer is then:

    ycm = interval 0->1.5 y(y2)dy / interval 0->1.5 (y2)dy

    ...which an online calculator tells me is

    ycm=(81/64)/(9/8)
    ycm=1.125

    So the answer for rcm=(0m,1.125m)
     
  16. Nov 3, 2016 #15

    kuruman

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    Forgive me, but how can ρ(y)=y2 be in kg/m? What are the units of y then? They can't be [kg/m]1/2 because then what is the meaning of "point (x,y)"? A constant is needed to multiply y2 so that the units come out right. Yes, the given expression is dimensionally incorrect, but without that constant and its interpretation, I think the given dimensions of kg/m are meaningless. How is one to know that kg/m is correct and ρ is not as opposed to ρ is correct and kg/m is not?
     
  17. Nov 4, 2016 #16
    It is in km/m because it is defined that way. There are three types of density. Most people know about volumetric mass density, because that is what we use in basic chemistry. But there is also surface mass density σ, which is for use with sheets of known thickness with a density already calculated for a unit of volume of that sheet. So σ=density * thickness = mass / area. And there is also linear mass density, which is mass per unit length, often written with μ or λ = density * area = mass / length.

    In this case μ(y)=y2 kg/m is not a calculation, but a sliding constant. It's dimensionally correct automatically by definition. You don't plug in y as a unit variable, you plug in y as a non-unit number. The function adds the units for you. In this case y is not talking about the y axis, but a variable input which may have previously been h for height, or L for length. It doesn't matter. All that matters is that the output is branded as kg/m. The solving function given in the original post shows us where we use it. The symbols dm are equal to derivative of mass, which is equal to linear mass density. That was another clue.

    So in this equation, the density formula for point (x,y) is μ(x,y)=y2, meaning that the density function does not care about x, so we can drop it from the function, leaving just μ(y)=y2. Which in turn helps define what the meaning of y is (which axis direction), because the density is talking about a length, and changes with position, so we know at this point we're talking about a dimension which runs along the cone axis from the pointy part. And yes, as you said, who is to know what part is incorrect, except that when we examine the formula we're supposed to use, we know we are working only in one dimension, so we know only the symbol for density was incorrect.
     
    Last edited: Nov 4, 2016
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