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Angular Momentum Question

  1. Jul 15, 2015 #1
    1. The problem statement, all variables and given/known data

    A rod of length l and mass M is suspended from a pivot, as shown The rod is struck midway alongs its length by a wad of putty of mass m moving horizontally at speed v. The putty sticks to the rod. Find an expression for the minimum speed v, that will result in the rod’s making a complete circle rather than swinging like a pendulum.

    iympw4.png

    2. Relevant equations


    3. The attempt at a solution

    Attempt at a solution:


    Use Conservation of Energy


    Kinetic Energy of Putty = Energy of Putty and Rod at the top of swing


    multiply everything by 2


    ##mv_i^2=[(\frac{l}{2})^2m+\frac{1}{3}Ml^2]w_t^2+2(m+M)gl\\\\##


    where w_t is the angular velocity at the top and v_i is the velocity of the putty originally


    at the top of the swing the minimum velocity is related to the centripetal acceleration being equal to g


    therefore,


    ##g=w_t^2l\\\\w_t^2=\frac{g}{l}\\\\##


    substitute into conservation of energy stuff and solve for v_i


    ##v_i=\sqrt {[(\frac{l}{2})^2+\frac{Ml^2}{3m}]\frac{g}{l}+\frac{2(m+M)gl}{m} } ##


    Answer in the back of the book…


    ##v_i=\sqrt {\frac{8(m+M)gl}{m^2}(\frac{1}{4}m+\frac{1}{3}M) }##
     
  2. jcsd
  3. Jul 15, 2015 #2

    gneill

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    Really? Is kinetic energy conserved over the collision? What type of collision is happening?
     
  4. Jul 15, 2015 #3

    andrewkirk

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    We can't use conservation of energy because energy in the system of rod and putty is not conserved. Why? Because the pivot does work on the rod and vice versa.

    Also, the centripetal acceleration at the top is not ##g## because there is also the stiffness of the rod pushing in the opposite direction.

    To solve this, we need to bite the bullet and do it using angular momentum, not conservation of energy (which is usually easier when it's valid). THe initial angular momentum of the rod and putty about the pivot will equal the ang mom of the putty about the pivot immediately prior to the impact. Based on the answer in the back of the book, it will get a bit messy.

    In fact I'm surprised that they have a closed form solution at all. It is a pendulum problem, and those are generally only analytically solvable for small angles where the approximation of ##\theta## to ##\sin\theta## can be used. That can't be used for this since the angle does not remain small.

    I can write the answer as an integral of ##\frac{\sin\theta}{\sqrt{1-\theta^2}}## but I'm pretty sure there's no closed form solution for that. Perhaps they have some clever trick for avoiding having to do the integral.
     
  5. Jul 15, 2015 #4

    gneill

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    Conservation of energy can be used provided that the inelastic collision is dealt with first (conservation of angular momentum in this case).
     
  6. Jul 15, 2015 #5

    haruspex

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    The pivot does no work on the rod. The force it exerts does not 'go' anywhere.
    Work is lost in the impact because internal forces within the putty do advance, compressing the putty in that direction.
     
  7. Jul 16, 2015 #6

    andrewkirk

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    I've been thinking about this aspect since gneill's post.

    My tentative conclusion is that there is work done, but the convention is to ignore it because unless the pivot is attached to something fairly small, the work is negligible.

    When the putty strikes the rod it imparts a force that would, in the absence of any other force, translate the rod sideways to the right. The reason it doesn't is because the pivot exerts a sideways force to the left on the central end of the rod, and the two forces between them start the rod rotating.

    Say the pivot is attached to a mass M2, which could be a trolley, a car, or even a building that is rigidly attached to an entire planet (in which case M2 is the mass of the planet). Then the leftwards force of the pivot on the rod accelerates M2 to the right, relative ot its initial rest frame. So M2 acquires a small rightward velocity and work is done. The reason this is trivial if M2 is large is that the combination of conservation of momentum and conservation of energy (less the energy of deformation from the inelastic collision) makes the post-collision velocity of M2 ##\frac{2v}{1+\frac{m}{M2}}##, which will be tiny. The work done will be the gain in kinetic energy of M2, which is proportional to the square of that, which is even tinier. So we ignore it, which is the same as assuming M2 is infinite..

    When I thought about it some more I realised that in many (most?) problems involving balls interacting with rigid surfaces there will be a similar case where, strictly speaking, energy of the 'moving' objects are not conserved. But by convention we implicitly assume that the rigid surfaces are attached to an infinite mass.

    Long story short: yes, for practical purposes we calculate as though the pivot is atached to an infinite mass and hence the pivot and the rod do no work on one another.
     
  8. Jul 16, 2015 #7

    haruspex

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    Yes, that is the approximation that is made whenever we deal with something regarded as an immoveable mass.
     
  9. Jul 16, 2015 #8
    The way to solve this - first find how much gravitational potential energy the system (rod/putty) must have to swing around 180 deg. Find the center of mass of the system to do this. This gain in potential energy is how much kinetic energy must come out of the collision (K = 1/2 Iω2) - or it will tell you how fast (angular velocity ω) the system must recoil. This, in turn, will give you ω immediately after the collision and thus the amount of angular momentum the system must after after the collision (L = Iω). Since there are no torques being applied from before to just after the collision, angular momentum will be conserved in the collision. You now have the angular momentum the putty must have relative to the pivot at the top of the rod: L = rmv = (l/2)(mv) and you can solve this for v = solution.
     
  10. Jul 16, 2015 #9
    I forgot it was an inelastic collision. I will try again from an angular momentum perspective.

    Thanks
     
  11. Jul 16, 2015 #10
    Yup, I got it. Thanks guys. I really appreciate it.
     
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