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Conservation of Angular Momentum problem

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data
    A thin, uniform metal bar, 2.00 m long and weighing
    90.0 N, is hanging vertically from the ceiling by a frictionless
    pivot. Suddenly it is struck 1.50 m below the ceiling by a small
    3.00-kg ball, initially traveling horizontally at 10 m/s The ball
    rebounds in the opposite direction with a speed of 6 m/s
    (a) Find the angular speed of the bar just after the collision.
    (b) During the collision, why is the angular momentum conserved
    but not the linear momentum?


    2. Relevant equations

    L = Iw
    F = dP/dt

    3. The attempt at a solution

    I tried various ways to equate the momentum of the ball to the angular momentum of the rod but it just won't work. I've been at this problem for 8 hours straight and I can barely think anymore. Please help me.
     
    Last edited: Mar 13, 2014
  2. jcsd
  3. Mar 13, 2014 #2

    BvU

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    You missed typing the rebound speed of the ball...
    What about the angular momentum of the ball when it hits and rebounds ?
    Any idea what b) is about ?
    Get some sleep first 8h is too much...
     
  4. Mar 13, 2014 #3
    Sorry, I edited the post and added the speeds. I only understand how angular momentum affects something that is already rotating...Like an ice skater pulling his arms in to increase his rotational velocity. I don't understand how an object travelling in a horizontal vector can even be affected by angular momentum :(
     
  5. Mar 13, 2014 #4

    BvU

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    The ball bounces back from the bar. That's a change in momentum. Also a trivial change in angular momentum wrt the center of the rotation of the bar (at a fixed distance) and, as you say more or less, not very interesting.

    So we concentrate on the bar. F = dp/dt is interesting but we only know Δp, not Δt. However, the rotation analogue is ##\tau = d/dt (I\omega)## and that is nice because we do know the integral. Does this make sense ? Otherwise we assume a short Δt and it will cancel.

    Action = - reaction helps us translate F into torque ##\tau##.

    Reason I drew attention to b was that you have to realize that (horizontal) force of ball on bar is not the only (horizontal) force on the bar.
     
  6. Mar 14, 2014 #5
    I'm back at this problem...

    So I=1/3*9,1*4 according to my book, I for a slender rod around an end-point = 1/3*ML^2.

    I see that there is a change in momentum of the ball. Do you mean to say that the change in momentum of the ball = momentum of the rod?
     
  7. Mar 14, 2014 #6

    BvU

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    Nope. If you look at (b) with one eye, you should get that message !
     
  8. Mar 14, 2014 #7
    I see. The linear momentum of the ball is NOT conserved (where does it go, though? Isn't this an elastic collision?) but the angular momentum of both the ball AND the rod are conserved. Does this mean that the angular momentum of the ball gets transferred to the rod? How do I get a feeling for the angular momentum of the ball?
     
  9. Mar 14, 2014 #8
    Can you just make it obvious what I am supposed to do, I've been sitting here for a few hours again and there's just something that I'm missing. Please, I got this due for monday and I need to work on other assignments during the weekend :'(
     
  10. Mar 14, 2014 #9
    I've got the impulse at the point of collision to be 16. If I could convert this to force I could use torque to solve my problem, but I don't have a time value.
     
  11. Mar 14, 2014 #10

    BvU

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    What formula can give "the" impulse and lead to an answer of 16 without any units?


    For the ball you can write linear momentum before = 10 m/s x 3 kg = 30 kgm/s
    and momentum after -6 m/s x 3 kg = -18 kgm/s, so a momentum change of 48 kgm/s.
    So during the collision it has experienced a force F = Δp/Δt = 48 kgm/s / Δt.

    The analogue of ##\vec F = d\vec p/dt## for rotations is ## \vec \tau = d\vec L/dt## in vector notation, with
    ##\vec \tau ## the torque and ##\vec L## the angular momentum

    And if things aren't too complicated , like in F = Δp/Δt this becomes ##\tau = \Delta L/\Delta t##.

    In the absence of external forces, and making use of action = -reaction, one can state for a system of e.g. two colliding masses ## \vec F=0 \Rightarrow d\vec p/dt = 0 \Rightarrow \vec p## is conserved.

    Likewise, in the absence of external torques, we get ## \tau \equiv \vec r \times \vec F=0 \Rightarrow d(\vec r \times \vec p)/dt \equiv d\vec L/dt = 0 \Rightarrow \vec L## is conserved (r is constant during the collision).

    If we can combine: FΔt and ##\tau##Δt via ##\tau## = rF we have rΔp = ΔL and no more Δt!


    Now take the pivot point of the bar as the center for rotation. And, just for the fun of it, consider the change in angular momentum of the ball:
    before: ##\vec L = \vec r \times \vec p ## and since ## \vec v \bot \vec r## this becomes L = rp
    after: idem, so ΔL = r Δp
    We have all we need to calculate it. Do so, you need it later on.​

    Wrt this same point the bar initially has zero angular momentum and for the system of bar + ball there are no external torques wrt this center of rotation, which means that ΔLball + ΔLbar=0 !

    (for the ball r Δp = I Δω :
    ## \vec v = \vec \omega \times \vec r## and since ## \vec v \bot \vec r## this becomes ## |\vec \omega | = {|\vec v|\over | \vec r|}## so ω = v/r and Δω = Δ(v/r) = Δv /r. Moment of inertia: I = mr2, so Angular momentum L = I ω= mr2 v/r = mvr and ΔL = r Δp). You already calculated it.​

    For the bar we now have ΔL = ΔIω = I Δω = I ωafter, which is the one we were after !


    Reason I drew attention to b was that you have to realize that (horizontal) force of ball on bar is not the only (horizontal) force on the bar, but the torque from the collsion with the ball ball is the only non-zero torque wrt the pivot point!

    Your questions [quote The linear momentum of the ball is NOT conserved (where does it go, though? Isn't this an elastic collision?) [/quote] can now be answered with ease, right ? Try me!
     
  12. Mar 15, 2014 #11

    haruspex

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    Is that the exact wording? Angular momentum of a system is usually specified as being in relation to some fixed point. (The exception is where there is zero linear momentum.) In this case, the angular momentum is only conserved if the pivot point of the bar is taken as the reference point.
    As BvU notes, when the ball hits the bar there is also an impulse from the pivot. If you hold a bat vertically and a ball strikes the far end horizontally you feel an impulse on your hand.
    Can you now see why the angular momentum of the ball+bar about the pivot point is conserved?
     
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