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Ascending double cone

  1. Sep 18, 2014 #1
    1. The problem statement, all variables and given/known data
    We consinder a doble cone with a radius R and an angle α (pike) and the mass m. It is located on two rails with an opening angle β. The rails enclose the angle γ with the ground. A is the lowest point of the rails.
    First, the center of mass of the double cone is locatd vertically above the point A regarding the plane described by the two rails. The double cone ascends the rails. You can assume that base is in the center between the rails and that the line through the center of mass and the contact points with the rails is orthogonal to the rails.

    a) Why does the double cone ascends the rails? State the conditions for the angles α,β,γ so that this happen.
    b) Proof that the moment of inertia regarding the connecting line through through the pikes is I=3/10*m*R2
    c) Determine the function v(d). v is the velocity of the center of mass and d is the rolled distance on the rails.
    d) Calculate for the values α=50°; β=40°; γ=5°; R=10cm and m=100g the distance L which the double cone rolls on the rails and the maximal reached velocity.


    2. Relevant equations
    3. The attempt at a solution
    a) The reason is that the height of the center of mass decreases while the double cone is ascending the rails. Condition: tan(γ)<tan(α/2)*tan(β/2)
    b) Easy proof with the integral-formula of the moment of inertia.

    However, I do not have ideas how to solve c) and d). Can you help me?
     
  2. jcsd
  3. Sep 18, 2014 #2

    gneill

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    Consider conservation of energy. If you can determine the loss of gravitational PE with distance then it must go somewhere...
     
  4. Sep 18, 2014 #3

    BiGyElLoWhAt

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    Well d is related to c, you can't do d without c. This actually looks like a good time for the application of some conservation laws. Or you could do some calc, but I think that can be avoided. I'm not sure, I still need to work it out.

    I'm picturing something like so:
     

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  5. Sep 18, 2014 #4
    Hi gneill,
    Thank you for your help again.
    Of course we have the equation: m*g*Δhcm=1/2*m*v2+1/2*I*ω2 with ω=v/r.
    But i do not have a relation between h and d :(
     
  6. Sep 18, 2014 #5

    BiGyElLoWhAt

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    Is there any relationship between delta h and delta L? A trigonometric relationship perhaps?
     
  7. Sep 18, 2014 #6
    Yeah there must be one relation but i do not how to formulate with the angles...
     
  8. Sep 18, 2014 #7

    BiGyElLoWhAt

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    Look at the side view of my picture above. Draw yourself a right triangle, label the angles appropriately and it should be a simple relationship. SOHCAHTOA
     
  9. Sep 18, 2014 #8
    i am thinking of:
    Δh=x*(tan(γ)-tan(α/2)*tan(β/2))
    With x=cos(γ)*d
     
  10. Sep 18, 2014 #9

    TSny

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    That looks good to me.
     
    Last edited: Sep 18, 2014
  11. Sep 19, 2014 #10
    Okay, now I have the function v(d). What about d), how can I determine L, the maximal reached distance? The maximal reached velocity is v(L), isn't it?
     
  12. Sep 19, 2014 #11
  13. Sep 19, 2014 #12
    Nasu, I think you are right. However, does that change any results?
     
  14. Sep 19, 2014 #13
    No, my post was for bigyellowhat. I thought he may be confused about the geometry.
    I don't think you used that picture for your calculation, did you?
     
  15. Sep 19, 2014 #14
    No I thought of the double cone you have presented :)
     
  16. Sep 19, 2014 #15

    TSny

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    The maximum distance traveled is determined by where the double cone will no longer extend across the rails.

    Note that as the cone approaches this point the radius of rolling, r, approaches zero. So, what do you expect the speed V to do as the cone approaches this point?
     
  17. Sep 19, 2014 #16
    If the radius of rolling approaches zero, the velocity must be zero!
    That means I have to solve the equation v(L)=0 and than integrate the function v(d) from zero until L. Is that right? But the function v(d) does not become zero except for d=0
     
  18. Sep 19, 2014 #17

    TSny

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    You should be able to determine L from the geometry.

    You then find that v(d) goes to zero at d = L.
     
    Last edited: Sep 19, 2014
  19. Sep 19, 2014 #18
    My idea is to look when Δh=R.
    Another idea is to look when the double cone would drop between the rails: R/tan(α/2)<tan(β/2)*d
    Is one of the ideas right? Why is the v(d)=0 at this point?
     
  20. Sep 19, 2014 #19

    TSny

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    You are right that v(d) is not zero at the point where the cone drops between the rails.

    [EDIT: I now think v(d) = 0 at that point. Ugh!]
     
    Last edited: Sep 19, 2014
  21. Sep 19, 2014 #20

    BiGyElLoWhAt

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    Hmmm... Is double cone a standard geometric shape? I guess I'm just not familiar with it, I actually googled it before I posted that picture (doh!). Stupid google, leading me astray.

    And as far as the recent posts go, I think it makes physical sense that v is non zero at d=L. Why would the mass stop moving if it still has GPE w.r.t. the zero of your system? (if the center of mass is free to move down, neglecting friction, it will)
     
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