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Homework Help: Find mass and center of mass of ice cream cone

  1. May 17, 2010 #1
    1. The problem statement, all variables and given/known data
    A toy manufacturer wants to create a toy ice cream cone by fitting a sphere of radius 4 cm inside a cone with a height of 8 cm and radius of the base of 3 cm. The base of the cone is concave, but the rest of the cone is solid plastic so that with the sphere attached there is no hollow space inside. The sphere and cone are constructed from special materials so that the density of the sphere is proportional to the distance from the tip of the cone, with constant of proportionality 1.4, and the density of the cone is proportional to the distance from its vertical axis, with constant of proportionality 1.8. Find the mass and center of mass of the ice cream cone. (Hint: decide on a position for the whole ice cream cone. If the center of the sphere is the origin, then the tip of the cone cannot also be at the origin. It is a good idea to have the cone oriented vertically the way you would normally hold one.) Also, provide a plot showing the entire ice cream cone.

    Note from professor: Instead of dealing with these
    integrals, I will allow you to change the density functions given so
    that instead the densities for both the cone and sphere are
    proportional to the square of the distance described. With these
    simpler density functions (without the square roots), the integrals
    should be much simpler to solve.

    2. Relevant equations

    Sphere: z = x^2 + y^2 + z^2
    Cone: z = sqrt(x^2 + y^2)
    Vsphere = 4/3*pi*r^3
    Vcone = (pi*h*r^2)/3
    density = mass/volume

    center of mass:
    x(bar) = Myz/m
    y(bar) = Mxz/m
    z(bar) = Mxy/m

    where M is the moment about a coordinate plane.

    3. The attempt at a solution

    Not exactly sure where to start... If I have no mass how can I begin to solve for this? Should I start by solving for the volume that the sphere+cone take up?

    Any help is greatly appreciated!
  2. jcsd
  3. May 17, 2010 #2


    Staff: Mentor

    Here is the information you need to use to get the mass.
    The mass of the sphere = density * volume, with the density as specified above. The mass of the cone can be computed similarly. Also, keep in mind the hint given by your professor.

    In both, the density is not constant (i.e., is a function).
    Your equation above is incorrect. You have z as a function of x, y, and z. What you should have is x^2 + y^2 + z^2 = r^2.
  4. May 17, 2010 #3
    Thanks for your reply Mark44!

    Dont I still have two unknowns? density and mass? I guess my problem now is finding the equation for density. Do I need to set the equations equal to and find out where they intersect?

    I've solved for both volumes:

    Vcone = 24*pi
    Vsphere = (32/3)*pi

    Last edited: May 18, 2010
  5. May 18, 2010 #4


    Staff: Mentor

    Your density functions will be in units of mass/volume. Due to the way the density functions are defined, the center of mass will be along the central axis of the cone+sphere, which I'm assuming you will orient so that the z-axis passes vertically through the center of the cone and sphere.

    It's not enough to get the volume of each of the two solids. The formulas I gave before should be interpreted as incremental mass, [itex]\Delta m[/itex], which you will need to use in each integral.

    Of course you need the equations of the cone and sphere. Hopefully you have already graphed them.
  6. May 18, 2010 #5
    how do these equations look:

    cone- z2 = (64/9)(x2+y2)

    sphere- 4 = x2+y2+z2

    r=0 to 3
    theta = 0 to 2pi
    z = 0 to 8

    p = 0..2
    theta = 0..2pi
    phi = 0..pi
    Last edited: May 18, 2010
  7. May 18, 2010 #6
    i've attached a picture of the cone/sphere graphed.

    Attached Files:

  8. May 18, 2010 #7


    Staff: Mentor

    The equations in post #5 don't go with the graphs in post #6. Your equation for the sphere has its center at the origin. The graph of the sphere has its center at somewhere around (0, 0, 11).

    Also, your equation for the sphere has a radius of 2. In the problem description, the radius is supposed to be 4.
  9. May 18, 2010 #8
    I updated one of my last posts with some integrals... But back to the graph..

    The cone has a height of 8. The total height from the bottom of the cone to the center of the sphere is 8+sqrt(7). And the radius is still 4 for the sphere, and still 3 for the cone
  10. May 18, 2010 #9


    Staff: Mentor

    Read what I said in post #7.

    As for your integrals, I think you are jumping the gun. It doesn't look to me like you have the density functions figured out yet.
  11. May 18, 2010 #10
    Okay, then how about I change them...

    sphere- 16 = x2+y2+(z - (8+sqrt(7)))2

    cone- x2 + (y-3)2 = (z-8)2
  12. May 18, 2010 #11


    Staff: Mentor

    Where did the 8 + sqrt(7) come from?

    Why is the y - 3 term in the equation for the cone?
  13. May 18, 2010 #12
    I think i've about given up on this one. I clearly have no idea what I'm doing. I was trying to rearrange the equations to fit my graph. I saw my professor, he gave me an equation of

    z^2 = 64/9 * x^2 + y^2 for the cone

    and said that the height from the center of the sphere to the bottom of the cone was 8+ sqrt(7). the radius of the cone is 3, the radius of the ball is 4... To find the distance from the center of the sphere to the cone-- that leaves us with a triangle that has 4^2 = x^2 + 3^2.

    x= +- sqrt(7). Add this to the height of the cone and we get 8+sqrt(7)

    I've got an example in my book that calculates the density+mass+center of mass, so I tried adapting that to this problem and that is why I 'jumped the gun' with the integrals.

    If the tripple integral of the constant of proportionality multiplied by the formula of the equation doesn't get me the mass, then im not sure what does.

    my book reads:

    p(x,y,z) = K*(x^2+y^2+z^2) for density.

    where K is a constant


    int(int(int( p(x,y,z) dx )dy)dz)

    the tripple integral of the density equation for the mass.
  14. May 18, 2010 #13


    Staff: Mentor

    This looks fine. The assumption here is that the tip of the cone is at the origin.
    That's fine. I just didn't see where you got it, and now I do. This makes the equation of the sphere x2 + y2 + (z - (8 + sqrt(7))2 = 16.
    But not for this problem. In this problem, the density varies in different ways for the cone and the sphere.

    Before setting up the triple (one p in triple) integrals, see if you can find formulas for the density functions as described in your first post, and using your instructor's suggestion. You'll need one function for the cone density and another for the sphere density.
  15. May 19, 2010 #14
    Okay, I'm assuming it has something to do with the slope of the cone, and the increasing radius/size as you move up from the bottom of the cone... Since the radius is 3 and the height is 8, this gives us a slope of 8/3.

    I'm assuming that I need to use that 8/3 in my equation, I'm not sure where though...


    \int^{2pi}_{0}\int^{8}_{0}\int^{(8/3)x}_{0} (1.4*r^2) r dr dz dtheta

    Not sure about the sphere, though...
  16. May 19, 2010 #15


    Staff: Mentor

    Double integrals should suffice for both the cone and sphere.

    For the cone, the region over which integration is performed is {{r, theta) | 0 <= r <= 3, 0 <= theta <= 2pi}.

    The density function is as you have it, 1.4*r^2 (using the note from your professor). The integral for the mass of the cone should look like this:
    [tex]m = \int_{\theta = 0}^{2\pi} \int_{r = 0}^3 (z_{top} - z_{cone}) 1.4 r^2 r dr d\theta[/tex]

    What I'm showing as ztop is the z value at the top of the cone, which is the z value on the lower half of the sphere. zcone is the z value at a point on the surface of the cone.
  17. May 19, 2010 #16
    Okay very cool. I suppose now I have to find out how deep the cone is in the center (along the z axis) due to concavity.

    Thanks you for your help thus far. I am going to keep going at it! I'll post back soon with an update.
  18. May 19, 2010 #17


    Staff: Mentor

    Yes, that's basically it. The top surface of the cone is the bottom surface of the sphere. You don't have to find how deep the top surface of the cone is - you just need the z value, which will be the z value on the lower half of the sphere.

    Still ahead, you need to find Mz, the moment about the xy plane, and the mass and moment for the sphere.
  19. May 19, 2010 #18
    The center of mass of the cone is somewhere along the Z axis. This yields a problem when dealing with double integrals doesn't it? I cant substitute a Z into the equation when dealing with polar coordinates.

    I believe the equation is: 1/mass Int(Int((z*density)rdr)dtheta)

    edit: i've been reading that the center of mass of a cone is simply h/4... But since my cone is concave, I'm not sure how to go about it.
    Last edited: May 19, 2010
  20. May 19, 2010 #19


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    Yes, it does. You need a triple integral.

    Just add the dz integral between the lower and upper limits (cone and lower sphere) as the inside integral (r dz dr dtheta).
  21. May 20, 2010 #20
    Okay, got that, now what should I do for the sphere? Finding the density of the sphere should be relatively easy... Its not just the triple integral of volume * density constant *radius?
  22. May 20, 2010 #21


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    No, it isn't. The density isn't constant. You have a constant of proportionality given. Before you worry about that, your first step should be to set up a triple integral in either cylindrical or appropriate spherical coordinates with an integrand of 1 to calculate just the volume of the sphere which should come out

    [tex] V = \frac {4\pi r^3} 3 =\frac {256\pi} 3[/tex]

    Once you have that working you can add density or moment arms etc.

    And, out of curiosity, what did you get for the center of mass of the cone? Are you using Maple to do the heavy lifting in this problem?
    Last edited: May 20, 2010
  23. May 20, 2010 #22
    I am indeed using maple. I've already found the volume of the sphere,using triple integrals-

    int(int(int(vsp(rho, theta, Phi), rho = 0 .. 4), theta = 0 .. 2*Pi), Phi = 0 .. Pi) = 256Pi/3

    for the mass of the cone and the center of mass i got:

    f := ((4-sqrt(7))*1.4)*r^3
    mass:=Int(Int(f, r = 0 .. 3), theta = 0 .. 2*Pi) = 241.2300214

    cmass := 1.4*z*(4-sqrt(7))*r^3
    centerOfMass:=(Int(Int(Int(cmass, z = 0 .. 8-sqrt(7)), r = 0 .. 3), theta = 0 .. 2*Pi))/(241.23) = 14.33399078
  24. May 20, 2010 #23


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    That is the volume of a sphere and it is equal to the volume of your sphere because they are the same size. But you didn't calculate it using your sphere which is not at the origin. You are going to have to set it up using the sphere above the origin because where it is matters when you consider the density and hence the mass.

    Your cone extends from the origin up to z = 8. Does it bother you that your calculated center of mass is outside the cone?
  25. May 20, 2010 #24


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    I have to leave for the rest of the afternoon now, so here are some additional comments about your cone.

    What happened to the triple integral we were talking about earlier?

    I don't understand your density. Furthermore, the limits you have given in this last triple integral don't describe your cone. They would describe a cylinder. You haven't used either of the equations for the top and bottom surfaces of the cone. They matter, don't they? Back to the drawing board.
  26. May 20, 2010 #25
    Yes it bothers me. This assignment was due two days ago and clearly I do not understand it. My book examples don't help me at all and I've got nothing to go by. Intuition?

    As for the "back to the drawing board" -- I don't thinks so. These repeated attempts post due-date are just brewing frustration. There is no way for me to understand how to do a problem like this because I've never seen one done before.

    Thanks for your help,
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