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Conservation of angular momentum

  1. Aug 1, 2017 #1
    1. The problem statement, all variables and given/known data
    A 220-kg beam 2.8 m in length slides broadside down the ice with a speed of 23 m/s . A 68-kgman at rest grabs one end as it goes past and hangs on as both he and the beam go spinning down the ice. Assume frictionless motion. (Figure 1)

    2. Relevant equations
    L1=L2
    Iω=L GIANCOLI.ch11.p50.jpg

    3. The attempt at a solution
    With what angular velocity does the system rotate about its cm? For this question i tried to use the initial moment of inertia w.r.t the CM by....[ 1/12Ml^2 + Mh^2] ω------ where l=2.8m M=220kg and h=.33m ----- For some reason this doesn't work. I'm positive the other side of the equation i listed is correct, i had it checked out. I ended up getting the answer for this quesiton too see where i went wrong, it ended up being ω=6.8 rad/s--- And they used (Mhv)
    M=220kg h=.33m v=23 m/s. I'm not sure how that equates too being angular momentum (Iω), but it worked for the answer. Can someone explain to me why they used that and why it worked. Also how else can you approach the initial angular momentum, somewhere along the lines of what i was doing. Thank you,
     
  2. jcsd
  3. Aug 1, 2017 #2

    CWatters

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    OK so h=0.33m is the distance from the CM of the beam to the CM of the system.

    That's because before the collision the beam isn't rotating about the systems CoM. It's travelling in a straight line "past" the systems CoM.

    That's the correct equation for a particle mass m moving in a straight line past a point. See second paragraph..

    http://courses.ncssm.edu/apb11o/resources/guides/G11-4.angmom.htm

    In your case r=h and just before the collision the sinθ=1 so the angular momentum is Mhv.
     
  4. Aug 1, 2017 #3

    CWatters

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    If you want to do it "your way" you need to calculate the moment of inertia as if it's a particle of mass 220kg moving past the system CoM...

    https://en.wikipedia.org/wiki/List_of_moments_of_inertia

    For a particle mass m and radius h..

    I = mh2
    ω = V/h

    Angular Momentum
    = I*ω
    = mh2*V/h
    h cancels
    = mhv
    same as before
     
  5. Aug 1, 2017 #4
    Thank you i understand now. Also, how come they only want the component rsin? Even for particles that are far away from the axis of rotation. Is it because that there is no angular momenta till it hits that point?
     
  6. Aug 1, 2017 #5

    CWatters

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    No. rSinθ gives you the tangential offset shown on this this diagram (left hand diagram). It's the same rule for calculating the lever arm when working out the torque produced by a force.

    If the particle passes through the centre point then the radius r is unchanged but rSinθ=0 and the angular momentum equals zero as you would expect (Right hand diagram.
    Angular momentum.jpg
     
  7. Aug 1, 2017 #6

    haruspex

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    A tip: with such problems, there is nearly always an easier way than finding the common mass centre.
    Say the beam has mass M, length L and initial speed u. The man has mass m, and just after the impact the beam's mass centre has linear speed v and angular speed w.
    What is the man's linear speed just after impact?
    What equation do you get for linear momentum?
    For angular momentum, I suggest using as reference axis either the man's initial position or the position of the beam's mass centre at instant of impact. Either way, just find the before and after angular momenta of each component and write out the conservation equation.

    Fwiw, I get nearly double the answer you quote. 23m/s is very fast. The man will be seriously injured.
     
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