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B Angular Velocity of a trebuchet

  1. Sep 13, 2016 #1
    Hello, I need some help regarding angular physics. I am working on a project and I want to be able to predict (to some degree) the velocity of the payload leaving the trebuchet. (Excuse my ignorance I am just a high school student)
    Lets say a trebuchet see diagram has a counter weight m1 distance r1 from the pivot point. I'll just assume that the mass and payload are attached to the arm for simplicity's sake. Also assume the arm makes a 30deg angle with the vertical (theta).

    I fairly certain that I cannot just use the E=mgh formula to equate velocity because I know some rotational physics are involved here.

    My attempt was to use Ek=1/2(Moment of inertia)(angular velocity) to find the final velocity.
    But I did not have angular velocity. Which is (change in angle/time) or (angular acceleration x time)
    I know that the counterweight is accelerating and therefor likely has rotational acceleration as well.
    I also believe that only the tangential acceleration/force acts on the arm and that the angle between the vertical (gravity) and the arm would constantly change (and also the tangential component) creating a non-constant acceleration.
    I thought that since we know acceleration due to gravity, we could figure out how much the counterweight had accelerated the payload to find its velocity. (sort of like finding velocity from an acceleration-time graph). (integration I think)

    Here's some math I tried
    Ek=1/2(I)(omega)
    =1/2[(m)(r^2)](omega)
    =1/2[(m)(r^2)][(alpha)(time)]
    =1/2[(m)(r^2)][(tangential acceleration x r)(time)]
    =1/2[(m)(r^2)][((g / cos theta) x r)(time)]

    (here is where the non-constant acceleration comes in because of the changing theta value)

    I've sort of hit a wall here. Because I have no idea how to figure that out.
    I thought about using torque (t=alpha x I) or work too.
    I hoping someone can help me out or point me in the right direction.
     
  2. jcsd
  3. Sep 14, 2016 #2

    Simon Bridge

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    Have you seen:
    http://www.real-world-physics-problems.com/trebuchet-physics.html

    Note: the trebuchet must obey conservation of energy because everything does ... if all the energy comes from the falling weight, then the gravitational energy lost in the fall goes to lifting the payload, and giving it initial kinetic energy at the top of the swing.
     
  4. Sep 14, 2016 #3
    Yes I have seen this page but It wasn't helpful because It didn't clarify what formulas to use.

    I am aware of the conservation of energy. Are you suggesting that I am actually able to use the E=mgh formula or Is there a different one for rotations?
     
  5. Sep 15, 2016 #4

    Simon Bridge

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    Gravity is a conservative force - this means that the energy change does not depend on the path.
    The only thing affecting the payload speed v upon release that depends on the mechanism is non-recoverable energy losses - friction, air resistance etc.

    So for counterweight mass M dropping through a distance d to raise a payload mass m through a height h, neglecting losses, neglecting the mass of the rebuchet components, the conservation of energy equation goes: ##\frac{1}{2}mv^2 + mgh = Mgd##.
    That equation just says that all the grav PE lost by the counterweight turns into grav PE in the payload and some kinetic energy too.
    In practise there are more places for the energy to go, so this equation needs to be modified. ie. the big lever may gain grav PE also.
    The effect of the sling is to increase the height possible at release without adding a lot of weight to the lever.

    That tells you about the speed at release ... but what about the direction at release?
     
  6. Sep 15, 2016 #5
    The direction (angle) at release is determined through statics, so I'm told. Something to do with the coefficient of friction I think. I don't know much about that either I'm afraid so i''l have to look into it.
     
  7. Sep 15, 2016 #6

    Simon Bridge

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    If you do it for a catapult (all rigid components) it's much easier - the direction depends on the angle the bucket got stopped.
    The trebuchet gets more complicated because it has non-rigid components. So it is all about the details of how the release mechanism works.
    But if you know how to control the release angle, the rest is just the physics discussed.

    Real life gets messy very fast ... generally we try to design things to minimize this. At some point, though, somebody usually has to do an experiment to measure something.

    I'm guessing you are building a trebuchet of some kind?
    For a HS project I'd suggest starting with something very rule-of-thumb ish ... and improve. Looks good if you identify parameters that need to be measured or tested to discover a better design. Definitely state all assumptions, and also state which direction measurements will go if your assumptions are bad. ie. the range will be smaller than predicted if you neglect energy losses in the launch and air resistance.
     
  8. Sep 17, 2016 #7
    I would normally ignore many of the "real life aspects" when approaching a problem like this, but I am in the International Baccalaureate program.
    http://www.ibo.org/
    And this project is my extended essay, and this project is supposed to "extend" beyond the curriculum and involve search and more complex aspects than discussed in class, so to leave out too many aspects of the model like the mass of the air, rotational friction in the axle, and a calculated release angle would not serve me well.

    I do think I have an answer for the release angle problem.
    I started with a Free-body diagram. Imagine the arm having a pin sticking out of one end with the loop hooked around the pin near the time of release.
    For simplicity's sake I made the pin vertical.
    The forces acting on the loop around the pin, are friction, normal and tension.
    We can ignore gravity and assume all Fnets= 0 because the mass of the loop is extremely small. See diagram below.
    In the x-dir we have ∑Fx = Fn-Ftx=0 and Ftx=Ftsinθ, Fty=Fcosθ and Ffmax=μFn
    So, Fn=Ftsinθ
    And the y-dir, where the y component of tension must overcome the force of friction:
    ΣFy=Fty-Ffmax
    =Ftcosθ-μFtsinθ
    μFtsinθ=Ftcosθ
    μ=1/tanθ
    tanθ=1/μ
    θ=atan(1/μ)

    (Ignore the dots)
    ...............Pin........Ft
    ...Fty..↑.....!........../
    ..........│...!......../
    ..........│...!.θ.../
    ..Fn....│...!..../
    ←--------!-/--------------→Ftx
    ...........│.!
    ...........│
    ...........↓
    ..........Ff

    So knowing that the separation between the pin and the loop+string=θ, we can adjust the pin angle relative to the length of the arm.
    This formula should give us the angle of release relative to the arm.
    The angle of the arm at this instance is when both the sling of the counterweight and the payload "stall" (this is where most of the energy is converted to kinetic).
    These Stalls can be synchronized by shortening or lengthening the strings holding the payload or counterweight.

    This is likely the only point that I will include the slings in my calculations.

    I now have to evaluate friction (I'll use bearings), the potential energy in the counterweight arm and the energy used to lift the payload arm, air resistance
     
    Last edited: Sep 17, 2016
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