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Trebuchet physics problem

  1. May 13, 2017 #1
    1. The problem statement, all variables and given/known data
    The trebuchet is a siege engine that was employed in the Middle Ages to smash castle walls or to lob projectiles over them. A simplified version of a trebuchet is shown in the following figure. A heavy weight of mass M falls under gravity, and thereby lifts a lighter weight of mass m. The motion of the mass M is blocked as shown in the figure, which launches the lighter mass m; the blockade forms an angle θ with the vertical. The mass of the blockade is much larger than all other masses. The shorter arm of the trebuchet is of length H , whereas the longer arm is of length l; the whole beam (both arms) are of mass µ.
    2. Relevant equations
    conservation of energy

    3. The attempt at a solution
    ok i did the solution but i got a super complicated expression for omega i will is it supposed to be that way? and is it possible to get a numerical value for omega from the given information? i have just described my solution in words if you want my actual equations i would gladly post thanks

    x y a b are heights above ground
    I moment of inertia
    v translational velocity of masses

    Ei = Ef
    Mgx + μ(L+H)gy = Mga + μ(L+H)gb + 1/2 I ω2 + 1/2 Mv2 + 1/2 m v2

    y = (L+H)/2 sin tan-1(H/L)
    x = (L+H) sin tan-1(H/L)

    a = H(1 - cosθ)
    b = H + (L-H)/2 cosθ

    I = 1/12 μ(L+H)3 + μ(L+H)(L-H)2/4

    v of M : ω* H
    v of m : ω * L

    is this correct?
  2. jcsd
  3. May 13, 2017 #2


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    μ is given as mass of the beam, you shouldn't multiply it by (L+H) to get the mass.

    The expressions can get a bit lengthy, yes.
  4. May 13, 2017 #3


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    I haven't worked out the problem (yet), but this item in your attempt attracted my attention. Expressions such as ##\sin \left [ \tan^{-1}(H/L) \right ]## are ugly and can be simplified. You are looking for ##\sin \theta## where ##\tan \theta = H/L = opp/adj.## Well,
    $$\sin \theta = \frac{opp}{hyp}=\frac{H}{\sqrt{H^2+L^2}}.$$
    $$\cos \theta = \frac{adj}{hyp}=\frac{L}{\sqrt{H^2+L^2}}.$$

    On edit: Your expression for Ef does not include the potential energy of mass m.
    On second edit: Angle ##\theta## mentioned in this post is not the same as angle ##\theta## given by the problem. It is the angle formed w.r.t. the horizontal by the arm in its initial position.
    Last edited: May 13, 2017
  5. May 13, 2017 #4
    oh yah got confused with mass and linear mass density sorry about that

    oh ya i forgot about potential energy of small mass
    and simplification too

    i dont understand what you mean by they are different θ they are the same in my working

    because for a it is the height above the ground which H(1-cosθ) like a pendulum)

    and for b it can be calculated similarly
    and finally for small mass m the final height above ground is H + L cosθ
    i dont get what you mean by they are different angles
  6. May 14, 2017 #5


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    The heights of what and when are x, y, a, b?
    I is moment of inertia about what axis?
    v is linear velocity, but it is different for m and M. Use different symbol for them.
  7. May 14, 2017 #6
    sorry for the ambiguity was in a rush that day let me explain myself

    x is the initial height of mass M above the ground
    y is the initial height of centre of mass of rod above the ground
    a is the final height of mass M above the ground
    b is the final height of centre of mass of rod above the ground
    new variable i forgot to include
    c is the final height of mass m above the ground

    I is the moment of inertia about where the rod is pivotted i had used the parallel axis theorem

    vm = ωL
    vM = ωH

    am i right now?
  8. May 14, 2017 #7


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    H/L=sin(φ), x=(L+H)sin(φ) and y=sin(φ)(L+H)/2. What are the initial and final potential energies?
    Initial state:
    Final state

    Last edited: May 14, 2017
  9. May 15, 2017 #8
    initial potential energy :
    μ sinφ (L+H)/2 g + M g (L+H) sinφ

    final potential energy
    μ (H + (L-H)/2 )cosθ g + M g ( H - H cosθ) + mg(H + Lcosθ)

    final kinetic energy
    I ω2 + 1/2 M (ω H)2 + 1/2 m (ωL)2

    is this correct ?

    edit :
    what did you use to draw that thanks next time it would be helpful for me to post diagrams thanks.
  10. May 15, 2017 #9


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    What is your zero of potential energy? If it is the ground, then the initial PE is correct. The final PE is not. You need to add distance a (see the diagram posted by @ehild) to all the final vertical distances.
  11. May 15, 2017 #10


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    It is "Paint" drawing program of Windows.
  12. May 16, 2017 #11
    yes my zero potential energy is ground
    but i don't see what height i need to add as far as i know i have accounted for all the heights
    please point out to me


    Attached Files:

  13. May 16, 2017 #12


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    See figure, and check your heights.

  14. May 16, 2017 #13


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    I did in post #9.
  15. May 17, 2017 #14
    a = H - Hcosθ

    M is a above ground

    centre of rod is
    H + (L-H)/2 above the ground

    the small m is Lcosθ + H above the ground
    i still don't see why you need to add another a to the heights
  16. May 17, 2017 #15


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    a and c are correct, but b is not. The angle is missing.

  17. May 17, 2017 #16
    oh ya forgot the cosθ

    but isn't this the heights i have i been saying since the start so is it correct? thanks!
  18. May 17, 2017 #17
    just for clarification i drew this please don't mind my ugly drawing skills thanks!
    is this corrects?
  19. May 17, 2017 #18


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    The drawing is all right. So what is b? Where is cos(theta)?
  20. May 17, 2017 #19
    cos theta is the vertical height from edge of blockade to b
    that is b is
    H + (L-H)/2 cosθ
  21. May 17, 2017 #20


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    correct at last.
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