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Volumes can be amusing

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data
    http://www.math.rutgers.edu/~greenfie/mill_courses/math152/pdfstuff/w2.pdf

    problem 2


    2. Relevant equations

    Work is the integral of force..


    3. The attempt at a solution

    Problem 2:

    Basically I know how to calculate the work for the cylinder. Since they have the same volume, we know the volume of the cones, but since they are the same depth underground, then how is the answer going to be different for any of these three. I know the work is the integral of the force, and the force is basically found by taking the volume * density * g...so why are they going to be different and how can I know?

    I also know it relates to distances from the center of mass to surface (i.e. one is 5m one is 2.5m and one is 7.5 m), but I'm not sure how I can apply this....I think I should use similar triangles, but I am a little unsure how to approach this.
     
  2. jcsd
  3. Feb 1, 2009 #2

    Dick

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    They want you to compute the work for each volume anyway, so you may as well get started. You have to integrate density*g*h*A(h) dh for each of the three shapes, where A(h) is the cross-sectional area of the shape at a depth h and h varies from 0 to 10. It's not really all that hard.
     
  4. Feb 1, 2009 #3
    I will take g=10 for simplification purposes.

    For the cylinder, we have it as

    integral (0 to 10) 270000* pi* h dh=
    13500000* pi J

    Now we look at the inverted cone and we have integral(0 to 10) 30000 * pi * r^2 * h dh

    But my problem is, how do I find what r is for the cone?

    Do I just do (1/3)pi*r^2*h=pi*r^2*h

    But actually this will become (1/3)*r^2=3^2
    So r=sqrt (27) for the cone...is this correct and if so how will it differ depending on whether or not the cone is inverted?
     
  5. Feb 1, 2009 #4

    Dick

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    r is going to depend on h in two different ways for the two cones. They are both linear in h. Since they tell you the volumes are equal you'll want to find the base radius of the cones that makes them equal to the cylinder first.
     
  6. Feb 1, 2009 #5
    Ok, that's what I said, that r=sqrt(27) makes them equal to the cylinder. Now comes the part where I'm confused.
     
  7. Feb 1, 2009 #6

    Dick

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    Ah, right. Ok for the first cone r=sqrt(27) at h=0 and r=0 at h=10. And it's a linear function of h in between. Can you write down r(h)?
     
  8. Feb 1, 2009 #7
    Ok so I see it as a line...We have pts. (0, sqrt 27) and (10,0) so the slope is:
    -sqrt(27)/10

    And so

    r(h)= -sqrt(27)/10*h + sqrt(27)

    And I suppose the other cone would be found in a similar manner...just the points would be (10,sqrt 27) and (0,0)

    Is that correct? And if so, I'm still a bit confused :
    Do I just integrate that from 0 to 10 of density*g*1/3*pi*h*r(h)^2 dh with r(h) being that funky function above?
     
  9. Feb 1, 2009 #8

    Dick

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    Exactly. You integrate density*g*h*pi*r(h)^2 dh (notice I left out the extra 1/3). You can do the cylinder using the same formula if you set r(h)=3.
     
  10. Feb 1, 2009 #9
    Ok great...thank you very much for your help.


    The one last thing is....so we don't include the 1/3? That's because we already used it when finding r(h), correct? (just want to be completely sure).
     
  11. Feb 1, 2009 #10

    Dick

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    I don't know where the (1/3) came from. I just know it shouldn't be there.
     
  12. Feb 1, 2009 #11
    I got it since the volume of a cone is (1/3)* pi* r^2* h and since we are looking at a cone in this problem and we need to integrate the volum*density*g.
     
  13. Feb 1, 2009 #12

    Dick

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    The only formula you need is integral density*g*h*A(h) dh. You can apply that to all three shapes.
     
  14. Feb 1, 2009 #13
    Got it now, thanks.
     
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