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Angular Momentum and Rotational Energy

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Figure: http://i.imgur.com/E2D1hkW.png

    A massless rod of length 2[itex]R[/itex] is attached at the middle to a pivot point that allows it to rotate in the vertical plane. Masses [itex]m[/itex] and 2[itex]m[/itex] are attached to the rod at the locations depicted in the figure. Initially the rod makes an angle of [itex]\theta[/itex] with the vertical. A lump of clay, with a mass of 3[itex]m[/itex] and a velocity [itex]v[/itex], strikes the end of the rod opposite to where mass [itex]m[/itex] is attached with and sticks to the rod.


    a. What is the magnitude of the angular velocity ω for the rod just after it is struck by the clay?

    b. What is the magnitude and direction of the net torque [itex]\Gamma[/itex] on the rod just after the clay mass sticks to the rod

    c. Does the kinetic energy of the system change when the clay mass impacts the rod? If so, does it increase or decrease and by how much?


    2. Relevant equations

    [itex]\Gamma[/itex] = I [itex]\frac{dω}{dt}[/itex]

    [itex]\Gamma[/itex] = r x F

    Erotational = [itex]\frac{1}{2}[/itex]Iω2

    Angular Momentum= L = r x p , where p is the linear momentum.

    L = Iω

    3. The attempt at a solution

    [itex]For a:[/itex] There is no external torque on the system if we think of the system as the clay and rod so we can use conservation of momentum.

    Initially, before the clay hits the rod, the rod has no angular momentum but the clay does and is given by:

    Lbefore collision = r x p = [itex]rvmsin\alpha[/itex] where [itex]\alpha[/itex] is the angle between [itex]r[/itex] and [itex]v[/itex] (Figure for this: http://i.imgur.com/GY9E1hZ.png )

    But since sin[itex]\alpha[/itex] = [itex]\frac{Rsin\theta}{r}[/itex], replacing this in the previous equation we get

    Lbefore collision = [itex]vmRsin\theta[/itex]

    After the clay hits the rod, the rod and clay combo have an angular momentum given by:

    Lafter collision = Iω , where I is the moment of inertia.

    Finding the moment of inertia:

    I = mR2 + 2m([itex]\frac{1}{2}[/itex] R)2 + 3mR2

    = 4mR2 + ([itex]\frac{1}{2}[/itex]mR)2

    =[itex]\frac{9}{2}[/itex] mR2

    By the conservation of angular momentum we have:

    Lbefore collision = Lafter collision

    vmRsin[itex]\theta[/itex] = [itex]\frac{9}{2}[/itex] mR2ω

    Solving for ω we get

    ω = [itex]\frac{2}{9}[/itex][itex]\frac{vsin\theta}{R}[/itex]

    However, since the net force on the system (the clay and rod again) is 0 we should be able to get ω through the conservation of energy

    Ebefore collision = Eafter collision

    [itex]\frac{1}{2}[/itex]mv2 = [itex]\frac{1}{2}[/itex]Iω2

    But then we get a completely different ω:

    ω = [itex]\frac{v}{R}[/itex] [itex]\sqrt{\frac{2}{9}}[/itex]

    Clearly one of them is incorrect and it seems to be the second equation for ω since it does not depend on the angle[itex]\theta[/itex]. If [itex]\theta[/itex] was 0 I would imagine the torque to be larger and thus ω be larger

    [itex]For b[/itex]: I'm having an even harder time with this one. My train of thought was this: since we don't know the Force we can't use [itex]\Gamma[/itex] = r x F, we also can't use [itex]\Gamma[/itex] = I [itex]\frac{dω}{dt}[/itex] since we don't have an equation for how ω changes during the impact. After the collision, ω is constant because there is no net torque. So that leaves us with the energy equation Erotational = [itex]\frac{1}{2}[/itex]Iω2 but how would we relate this to torque?

    [itex]For c[/itex]: Since there is no net force on the system (clay+rod) energy is conserved and the only energy associated is kinetic energy and so kinetic energy is conserved.

    But from question a, we used this assumption and got a different (incorrect?) ω.

    Any help or suggestions would be greatly appreciated.
     
    Last edited: Nov 3, 2013
  2. jcsd
  3. Nov 3, 2013 #2

    TSny

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    Hi larkmen, welcome to PF!

    You never want to assume that mechanical energy is conserved in a collision. Note that part (c) of the question is to check whether or not the KE is conserved. Hopefully, you've covered elastic vs inelastic collisions and what happens to some of the KE when you have an inelastic collision. As a general rule, if you have a collision where objects stick together, then KE will not be conserved. But even if the objects don't stick together, KE is not necessarily conserved.

    For (b), draw a free body diagram for the rod just after the collision and identify all of the forces acting on the system.
     
  4. Nov 3, 2013 #3
    Thanks for the reply TSny!

    I see what you mean by it being an inelastic collision and so the KE isn't conserved. I hadn't thought about that before, that definitely helps clear a lot up.

    As for part b, after drawing the free body diagram, I get this http://i.imgur.com/sn06rfF.png (imagine the 3m has collided already) which means [itex]\Gamma[/itex] = [itex]FRsin\theta[/itex]. I'm having a hard time finding F, one way I was thinking was to calculate the change in momentum of the 3[itex]m[/itex] particle and using F = 3m[itex]\frac{dp}{dt}[/itex] but because everything is constant before the collision and constant after the collision I am unsure as to how to go about doing this. Am I going in the right direction?
     
  5. Nov 3, 2013 #4

    TSny

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    After the clay sticks to the rod, you can just consider the system to be the massless rod with three lumps of mass. The collision is over so you don't need to worry about the collision force of 3m with the rod.
     
  6. Nov 3, 2013 #5
    But if we ignore the force of the collision wouldn't the free-body diagram have no forces on it, since there is no gravity? Or is there gravity and I'm reading the problem wrong?
     
  7. Nov 3, 2013 #6

    TSny

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    The rod rotates in a vertical plane. So, I think you are meant to include gravity.
     
  8. Nov 3, 2013 #7
    Ah, so yea we do probably have to include gravity. If we do include gravity then the equation for

    [itex]\Gamma[/itex] = + mgRsin[itex]\theta[/itex] - 2mg[itex]\frac{1}{2}[/itex]Rsin[itex]\theta[/itex] - 3mgRsin[itex]\theta[/itex] = -3mgRsin[itex]\theta[/itex]. (measuring positive torque as counter-clock wise)

    Shouldn't this depend on v however? Since if the 3m mass is moving faster, it would mean ω is larger which implies a larger torque?

    Thank you so much for the help so far.
     
  9. Nov 3, 2013 #8

    TSny

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    Looks good.

    The torque just after the collision does not depend on v. But during the collision, the torque due to the collision force does depend on v.
     
  10. Nov 3, 2013 #9
    Ohhhh, I see. I was reading the question wrong. Thank you so much for your help!
     
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