Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Conservation of linear momentum when wheels are included

  1. Nov 21, 2016 #1
    I am not a student. This is not a homework question. I am a 64 year old man who wants to understand a principle of physics. Suppose there is a large block of matter, designated as m2, which lies on a flat surface that initially is at rest. At some point in time a small block of matter, designated as m1, moves at an initial velocity vi and makes an inelastic collision with m2. I know from the conservation of linear momentum that the final momentum of the combined m1-m2 system must be equal to the initial momentum m1vi. What I am not sure of is what happens if there are 4 wheels attached to m2. Assume the moment of inertia of each wheel is 1/2 m3 r sq, where r is the radius of the wheel and m3 is the mass of each wheel.

    Suppose that m1 collides with the same initial velocity vi as before. Intuitively, it would seem that the translational momentum of the center of mass of the system would still be equal to the initial momentum m1vi per the conservation of linear momentum. But am I wrong and missing something having to do with overcoming the rotational inertia of the wheels?
     
  2. jcsd
  3. Nov 21, 2016 #2

    Dale

    Staff: Mentor

    Good question. Your intuition is correct. The wheels rotation does not contribute to its momentum. The linear momentum of the wheel is the mass of the wheel times the speed of the center of mass of the wheel, regardless of the rotation.
     
  4. Nov 21, 2016 #3

    jbriggs444

    User Avatar
    Science Advisor

    As @Dale certainly is aware, the moment of inertia of the wheels can play a role. If the objects are rolling without slipping then it is almost certainly friction with the floor that is speeding them up or slowing them down so that their rotation rate matches the movement of the carts. The frictional force to accomplish that contributes to the momentum of the system.
     
  5. Nov 21, 2016 #4

    Dale

    Staff: Mentor

    I agree, I am assuming all other things held constant, eg total mass, friction force, etc. If the wheels change the mass or the friction then there may be changes.
     
  6. Nov 21, 2016 #5
    I want to be more explicit regarding what jbriggs444 said and in so doing pretty much disagree with Dale. Our system is the two carts. If there is no external force (so no impulse) between the carts and the rest of the universe, then momentum must be conserved. The corollary is also true. If there is a net force, momentum will not be conserved.

    In the case as described let's say that there is no slipping. The only thing that starts and stops the wheels is torque FROM THE TABLE delivered through static friction. Suppose we have two carts with equal mass, but one has high rotational inertia wheels and the other has low rotational inertia wheels of the same diameter. (Lead vs plastic, more weight at the rim, whatever and weight the carts to make up any difference) If we imagine that the linear momentum of the carts is conserved, then the torque from the table stopped the high inertia wheels and spun up the low inertia wheels to the same angular speed. The torque cannot be equal, the force cannot be equal, and linear momentum of the two carts cannot be conserved.

    This is just an obvious counter example. It is very easy to construct lots of scenarios where the linear momentum is not conserved. Situations where it is conserved are more contrived. (Equal mas identical wheels). Bottom line, as soon as you have wheels whose motion is constrained by contact with the table you have an external force of constraint, the system is no longer closed, and the conservation laws are lost.
     
  7. Nov 21, 2016 #6

    Dale

    Staff: Mentor

    That is a good point. If all things are equal then the momentum will be the same. But your point is that it may not be the case that it is possible to make all things equal. Your logic seems reasonable.

    To the OP. Angular and linear momentum are separately conserved for an isolated system. For a non isolated system the rate of change in linear momentum is equal to the net force and the rate of change of angular momentum is equal to the net torque.
     
  8. Nov 21, 2016 #7
    Your answer makes sense. If we assumed no rolling friction, then the moment of inertia of the wheels would not matter because there would be no rotation of the wheels and the linear momentum of the system would be conserved. However, is there a generalized formula that would give me the final momentum of the system given the initial velocity of m1, the mass of m1,m2,m3,the moment of inertia of the wheels, and the rolling friction of the wheels? Not sure how I could derive this.
     
  9. Nov 21, 2016 #8
    I think that could be done, say, assuming the wheels don't slip or similar caveats. I think I would at least start without rolling friction. (I mean dissipation. Of course it takes friction for the wheels to roll, you just don't necessarily have to lose too much energy in that process (particularly without slipping). I'll think about it and see what I can come up with.
     
  10. Nov 21, 2016 #9
    thanks.
     
  11. Nov 22, 2016 #10
    Ok, I have some time let's give this a try

    Cart 1 has m1 wheels with diameter d1 and collective rotational inertia I1 (as long as they all do the same thing lumping them together will be fine)

    Cart 2 the same with 2s

    Before and after the collision cart 1 goes from v11 to v12 and cart 2 v21 and v22 respectively

    The wheels on cart 1 go from rotational velocity w11 = 2 v11 / d1 to w12 = 2 v12 / d1
    And similar for cart 2

    The net rotational momentum changed by

    Delta L1 = (2 I1 / d1) (v12 - v11)
    Delta L2 = (2 I2 / d2)(v22 - v12)

    The impulse to do this had to come from the table. Labeling torque T

    T1 t = F1 d1 t = (2 I1 / d1) (v12 - v11)
    T2 t = F2 d2 t = (2 I2 / d2)(v22 - v12)

    Or taking the ds to the other side

    F1 t = (2 I1 / d1^2) (v12 - v11)
    F2 t = (2 I2 / d2^2)(v22 - v21)

    But those are the linear impulses. Together that is the degree to which momentum was NOT conserved

    m1 v11 + m2 v21 = m1 v12 + m2 v22 - F1 t - F2 t

    And so

    m1 v11 + m2 v21 = m1 v12 + m2 v22 - (2 I1 / d1^2) (v12 - v11) - (2 I2 / d2^2)(v22 - v21)

    Is your new "momentum conservation" equation.

    Cases: m1 = m2, v21 = 0 , so
    m v11 = m v12 + m v22 - (2 I1 / d1^2) (v12 - v11) - (2 I2 / d2^2) v21

    So something like that. I'm too sleepy to go much further
     
  12. Nov 22, 2016 #11
    Thanks for your analysis. I will use it as a starting point. Appreciate your time.
     
  13. Nov 22, 2016 #12
    You are very welcome
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Conservation of linear momentum when wheels are included
Loading...