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Trebuchet Problem

  1. Apr 2, 2008 #1
    (I ignored the template because it didn't really seem to fit my problem, sorry about that)

    I'm currently doing a coursework assignment investigating the how the range of a model trebuchet varies according to the mass of the counterweight. Its pretty basic stuff (especially compared to a lot of whats on this forum), but i've spent the best part of today doing it and i'm still stuck.

    Anyway, the problem:

    Essentially i don't really know how to tackle the trebuchet in a mathematical model/equation. The trebuchet i'm using is pretty basic, it doesn't use a sling, the projectile (in this case, a marble) just sits in a slot on one arm and a counterweight is hung from the other. The beam is horizontal when the system is released. The arm with the projectile on it is considerably longer than the counterweight arm. The marble is released when the beam becomes vertical and is stopped by a metal bar. The problem I'm having is my limited physics knowledge can't get me from the force applied by the counterweight to the velocity the marble is released. I tried two approaches:

    I looked at it in terms of moments/torque but in this case i don't know how to translate the torque in the system to the marble itself.

    I also tried looking at in terms of energy. Ie. the counterweights GPE transferred to kinetic energy in the other arm. But I can't see how this accounts for the fact that the other arm is longer and that the projectile is positioned at the very end of this arm.

    Sorry if this is very wordy and not very mathematical, I'm just trying to convey as much information about it as possible. Also, I'm not looking for someone to give me the answer, just a clue, or a point in the right direction will do.

  2. jcsd
  3. Apr 2, 2008 #2


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    I would have thought energy was the obvious approach but remember that all the PE of the weight doesn't go into the ke of the projectile.
    The PE of the weight goes into ke at the bottom of the swing, it doesn't matter if it stopped by a bar or not.
    From this you have the rotation rate of the bar ( since the weight and bar are moving horizontally at the bottomof the swing and you have the totalV form ke=pe)
    If you know the relative lenghts of the upper and lower arms from thr pivot you have the linear speed of the top of the arm and hence the projectile.
    Then it's just a matter of the point in the arc that the projectile is released - and the normal firing a projectile at a certain angle and speed / cannon type problems.
  4. Apr 2, 2008 #3
    Thankyou very much xD

    I follow all of that fine except i don't understand how I get the rotation rate.
    Last edited: Apr 2, 2008
  5. Apr 2, 2008 #4


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    Assume that the weight is attached to the bottom of the pivoting rod.
    You know the total speed of the weight/arm at the bottom of it's swing because 1/2mv^2=mgh. At the bottom of the swing it is going exactly horizontal so the bottom of the arm is going horizontal at this speed. If you know the length of the lower part of the arm to the pivot then you have the angular velocity (ie rotation rate) of the arm.

    It's even simpler since the linear speed of the top of other part of the arm is just the linear speed of the weight multiplied by the ratio of the lengths of the two sides of the arm. (If the axle was in the middle it's pretty obvious that they are the same ! )
  6. Apr 4, 2008 #5
    you honestly have no idea how much that helps.

    thankyou xD

    EDIT: how do i change the title to [solved] ?
    Last edited: Apr 4, 2008
  7. Apr 6, 2008 #6
    sorry for the double post but i thought i had this figured out when i still haven't (sorry).

    When I do this, in each equation (1/2mv^2 and mgh) I use the mass of the counterweight as 'm'. I couldn't see a way to factor in the mass of the rod without it getting very complicated, due to the fact that different bits of the rod fall different distances.

    However, this means that when it comes to working out what 'v^2' is, I have to do


    this causes the masses to cancel out. Which is a problem, as what I am investigating is how the mass effecs the range.

    Obviously the mass DOES effect the range, so what am I doing wrong here? I assume its the fact that I'm effectively treating the rod and the projectile as being 'light', but a way to factor in their masses is beyond my limited knowledge.

  8. Apr 7, 2008 #7


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    The easy way to factor in the mass of the rod is to treat the upper and lower halves as two point masses at half the distance between the pivot and the end ( effectively average along the arm)
    But the mass of he counterweight or arm doesn't matter - all object fall at the same speed! (assuming it is heavy enough to overcome friction)
  9. Apr 7, 2008 #8
    But surely while - yes - it would fall at the same speed if it was on its own, the effect of the projectile + projectile arm, acting in the OTHER direction around the pivot/fulcrum, will mean it moves slower?

    (also note that the projectile arm is much longer than the counterweight arm)

    not sure if i'm getting across what i mean so i included an equally unhelpful diagram:


    I'm sorry if i'm just being really stupid here.....
  10. Apr 8, 2008 #9
    sorry to keep bumping this, but I know i'm never going to get an answer sat half way down page 2.
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