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Work and Kinetic Energy of a Trebuchet

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data
    This time you construct a trebuchet. A counterweight (M = 550kg) is at one end a distance ℓ2 = 1.5m away from the pivot. The child (m = 10kg) is a distance ℓ1 = 3m away from the pivot. The mass of the uniform rod connecting the two is mrod = 35kg. When released from the horizontal position, what will be the child’s velocity when she gets to the top?

    There are two extra things to take into account: the rotational kinetic energy of the rod (breaking it up into two rods ℓ1 and ℓ2 might help here), and the gravitational work done on the rod. This will be Wg = -mgΔy as usual, but where Δy is the vertical displacement of the center of the rod.

    Here's a picture of the problem:
    http://img46.imageshack.us/img46/2929/v8xv.jpg [Broken]

    2. Relevant equations
    Sum of Work = Change in Kinetic Energy
    KE = 1/2 * m * V2
    Wgravity = -m * g * Δy
    ω = v / r


    3. The attempt at a solution
    This has to be solved without using potential energy.

    The example the teacher gave in class was without the rod having a mass, but this time it has a mass so I'm not sure how to factor that in.

    Here's the example from class with the massless rod:
    http://img607.imageshack.us/img607/3025/vw4t.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 9, 2013 #2

    CAF123

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    You can model the system as a uniform rod pivoting about it's centre of mass with two point masses at either end. You will need the moment of inertia of a point mass and that of the rod about an axis through it's centre of mass.
     
  4. Dec 9, 2013 #3
    I'm not sure how to do that, but I tried to follow the example the teacher gave in class and I got it filled out up to the point where I have to fill in the KE for the rod that I split in to two.
    Here's my work: http://img543.imageshack.us/img543/3576/4zm6.jpg [Broken]

    Could you please help me fill in the KE of the rod part?
     
    Last edited by a moderator: May 6, 2017
  5. Dec 9, 2013 #4

    haruspex

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    The worked example uses potential energy, but in the OP you say it is not to be used. Wg = -mgΔy is an example of using PE, not an alternative. Indeed, there's no point in calculating KE if you're not allowed to use PE.
    Without PE, the only way I can think of is to develop the torque and angular acceleration equations. That will require some calculus.
     
  6. Dec 9, 2013 #5
    Thanks for clearing that up. I see now that PE is used. I just can't figure out how to do the KE side of the equation for the rod.
     
  7. Dec 9, 2013 #6

    CAF123

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    You forgot about the rotational kinetic energy contribution of the two masses. The moment of inertia of a point mass about an axis and a rod rotating about its C.O.M are results you can just use from a textbook or course notes.

    I am not sure if the analysis of the gain in potential energy of each rod segment is correct since each mass element of the rod will not be at the same height.
     
    Last edited by a moderator: May 6, 2017
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