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Truncated cones

  1. May 16, 2006 #1
    I have been presented with this problem. I somewhat know what I need, I just don't know how to get it :blushing:
    The problem:
    A truncated cone, top diameter of 1m bottom diameter of 1.5m and a height of 10m. With a given density(I do not have it with me at this moment, I do not remember exactly what that number is). Find the weight of the object and the force it would hit the groud with, I assume when falling over.

    I would like to know where, or if anyone knows, I can find the appropriate formulas for solving this. I have found how to get the volume of the object, and I can get the weight of the object. What I do not have any idea about is calculating the force the object hits the ground with when falling. I assume torque would have something to do with it, but I do not know what type of formulas to use or where to find them. Thanks for any help with this.
  2. jcsd
  3. May 17, 2006 #2


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    I would advise a recheck of the stated problem. In general, it's not easy to predict how much force a falling object will exert as it hits the ground, because force is the derivative of momentum with respect to time. I know of no way to calculate the time taken to decelerate the object in a collision, since it depends on unknown factors (like elasticity of the materials involved).

    What we might be able to calculate is the impulse (nett momentum change) of an object as it hits the ground. Even for this, it must either be assumed that the object comes completely to rest with respect to the ground after the collision (completely inelastic collision) or more info must be given about the rebound velocity, etc.

    In considering the final velocity of the leading edge of the truncated cone just before it hits the ground, you can find that from conservation of energy (equate the final rotational kinetic energy = 1/2*moment of inertia times angular velocity^2 to the change in gravitational PE of the center of mass.).
    Last edited: May 17, 2006
  4. May 17, 2006 #3
    Sorry for the first post, I meant the energy that it will hit the ground with, not the force.
  5. May 17, 2006 #4


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    That's completely doable. Just locate the center of mass, use geometry to find the change in height of the COM from the upright position to when the leading edge has touched the floor and use [tex]PE = -mg\Delta h[/tex] to find the energy with which the object hits the floor.
    Last edited: May 17, 2006
  6. May 17, 2006 #5
    Thanks, wow I feel dumb now. I know I have done that before. The only thing I don't think I have done is finding the COG of an object... well if I have I don't remember. Thanks for helping me.
  7. May 17, 2006 #6


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    The COM of a cone may not be trivial to find. The basic method is simple enough : observe that the COM has to lie on the line of rotational symmetry (there's only one) along the y -axis, mark off a point [tex](0,y_0)[/tex] to denote the COM, set up two integrals of the first moment of mass of cyclindrical slices below and above the COM point and equate them. The uniform density will cancel out.

    I say "not trivial" because my scrap paper work seems to indicate that you'll need to solve a quartic to get [tex]y_0[/tex]. Well, try it and see what you get.

    I don't know if there's an easier way. I've seen some stuff on the net saying you can equate cross-sectional areas above the COM line to below it to find the COM but I think this is oversimplified since the moment of mass is not being considered (this is a somewhat irregular, sloping solid).
  8. May 17, 2006 #7
    Could you explain how I can do that?
    I was thinking making half the volume equal to the volume of the truncated cone with the unknown of the top radius, but then I realized the height is also an unknown. I am really kinda clueless on how to find the COG because I have never done it before.
    Also I have found some websites where you put in the 2 radii and the height and it gives you the volume(12.44m^3), but when actually calculating it out subtracting the cone from the top cone that doesn't exist in the truncated cone I am getting a different number(8.8m^3). I am not sure I am doing something right. I am using the formula v=(1/3 PI R1^2 H1)- (1/3 PI R2^2 H2) Where:
    R1= base radius
    R2= top truncated cone radius
    H1= total height
    H2= height of top cone

    I feel like I am just a little off and about to understand it all as soon as I realize one little aspect. hehe
    Again thanks for the help.
  9. May 17, 2006 #8


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    I preferred to find the COM of the entire cone, then consider how the COM got shifted after truncation.

    Actually, it's pretty easy to do the COM of the whole cone, it's only a linear equation after doing the algebra.

    Basically, the sum of the moments of mass of a rigid solid about the COM is zero. The COM of the cone is along the central long axis, let it be a distance [tex]y_0[/tex] from the base. Let the central triangular cross section of the cone be positioned along x, y axes with the center of the base at the origin.

    An element of mass is a cylinder of radius x, height dy and COM at (0,y) The distance of the element's COM from the cone's COM is [tex]y-y_0[/tex]. The cone is homogeneous with mass density [tex]\rho[/tex]. Therefore, we can set up the integral

    [tex]\int_0^H {\frac{1}{3}\rho \pi x^2(y - y_0)}dy = 0[/tex]

    Now observe from similar triangles that [tex]\frac{x}{H-y} = \frac{R}{H}[/tex]. Hence express x in terms of y. Put that expression into the integrand, integrate and solve for [tex]y_0[/tex] in terms of H.

    To find the COM of the truncated cone, observe that this solid can be thought to be composed of a whole solid cone minus the top. Set up an equation to relate the mass moments of the whole cone and the "negative mass" of the truncated part to find the COM of the truncated cone.

    After doing all that, just tip the cone over to the side and use geometry/trig to see how high the COM is now above the ground. The change of height will immediately give you the change in gravitational PE.
  10. May 17, 2006 #9
    I don't know if I understood you completely, but I solved for x in the second equation and substituted in the first and got.
    Y0= y - (3/p PI (R^2+Y^2-2RY). Im thinking that is not right. What I had begun to do before was take 1/2V= 1/3 PI H (R1^2 + R1*R2 + R2^2) in a system of equations, one for the top and one for the bottom. Solve for R in terms of H then solve for H which I thought would have given me to height of the COG. I am guessing I am wrong there.
    Like I said, I have never attempted to find the COG of an object.
  11. May 17, 2006 #10


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    There's one more important consideration I may have overlooked (what I posted previously about COM is still correct). Give me some time to figure this out.
  12. May 17, 2006 #11


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    You should get [tex]y_0 = \frac{1}{4}H[/tex] for a complete (untruncated cone).

    Have you learnt basic integral calculus yet? You need it to do this. I don't think you integrated the expression, did you?

    What I just realised and worked out was that from the point of stability with the base square on the ground to the point of complete stable rest in a tipped state, the PE of the COM can either increase or decrease depending on the dimensions of the object.

    The PE decreases (tipped is a more stable state) for :

    [tex]R < \frac{H}{2\sqrt{2}}[/tex]

    and the PE increases (tipped is a less stable state) for :

    [tex]R > \frac{H}{2\sqrt{2}}[/tex]

    (When equality occurs the PE remains equal between the initial and final states).

    In any case, during the initial part of the tipping, the COM is moving higher from the ground, and therefore the gravitational PE is increasing. The energy for this is coming from the conversion of biochemical energy in your muscles (manual effort).

    The COM reaches a maximal height above the ground (maximised PE) at a point I've named "unstable tipping point". Before the cone is tipped to this angle, it can right itself if released. Beyond this point, if released, it will complete the tip to lie flat.

    This point occurs when the base of the cone subtends an angle [tex]\theta[/tex] with the ground where [tex]\tan \theta = \frac{4R}{H}[/tex]. At this point of unstable tip, the COM lies at a height of [tex]h_{max} = \frac{1}{4}(\sqrt{16R^2 + H^2})[/tex] above the ground.

    At this point, the cone is at maximal gravitational PE, and it will always lose energy as it tips over to the ground. This is a spontaneous, self sustaining movement. It is best to consider what happens from this point to the point of complete tipping, so there is where you should start your analysis of energy change from.

    At the point of complete tip, where the sloping edge lies flush with the ground, the COM lies at a height of [tex]h_{min} = \frac{3RH}{4\sqrt{R^2 + H^2}}[/tex]

    It is possible to prove that for all admissible values of R and H, [tex]h_{max} > h_{min}[/tex]

    Hence the required energy change (and the kinetic energy with which the cone hits the ground) is [tex]-\Delta PE = -mg(h_{min}-h_{max}) = \frac{1}{4}mg((\sqrt{16R^2 + H^2})- \frac{3RH}{\sqrt{R^2 + H^2}})[/tex].
    Last edited: May 18, 2006
  13. May 17, 2006 #12


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    BTW, that entire analysis was for a whole (untruncated cone). You can see that dealing with a truncated cone is even more involved. You have to start with a different expression for the COM, then work through the algebra. It's tedious, no doubt.

    In a truncated cone of top radius r and bottom radius R and truncated height of [tex]H'[/tex], the COM is at [tex]y_0'[/tex] from the base where

    [tex]y_0' = \frac{1}{4}H'(3(\frac{r}{R})^2 + 2(\frac{r}{R}) + 1)[/tex]

    In addition for a truncated cone, you must be aware that the analysis may be *much* more complicated than simply working out the new COM and proceeding from there. The first unstable tipping point is the same and the situation is the same till the sloping edge hits the ground. However, from this point on, it is possible that it can overbalance and *roll* truncated end over truncated end, if the momentum carries it far enough. I'm not going to analyse this possibility even though I know how to do it in a qualitative way, it just seems way too tedious.

    Where did you get this question from, BTW? Even a whole cone seems bad enough, a truncated cone seems very sadistic?!
    Last edited: May 18, 2006
  14. May 18, 2006 #13
    Thanks for explaining it further. It is basically my whole exam for High school Physics class. I have not gone over intergal calculus yet. I am pretty sure he just wants the difference in COM height when the truncated cone is standing upright and when it is laying on the ground so that the PE can be found.
    That is interesting to know that PE can be calculated as an object rocks onto a corner before falling over, I can only hope I learn that later.
    I asked him about it and he said he just wants the PE of the object sitting there and on the ground because he was trying to model a tree and having the tree cut in one side so that it basically just falls over with a little push, without rocking over. Thanks so much for helping, I have learned alot, maybe one day I will have an understanding as well as you do.
  15. May 18, 2006 #14


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    You're welcome. :smile:

    If your teacher just wants to model a process he's interested in, this should give him a very good starting point. He can use the truncated COM formula and work through the rest of the stuff to get the final formual (unwieldy but conceptually very easy). Maybe print this out or direct him to this forum.
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