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Angular momentum help

  1. Nov 28, 2007 #1
    Two small objects each of mass m = 0.6 kg are connected by a lightweight rod of length d = 1.3 m (Figure 9.60). At a particular instant they have velocities whose magnitudes are v1 = 27 m/s and v2 = 67 m/s and are subjected to external forces whose magnitudes are F1 = 51 N and F2 = 32 N. The distance h = 0.3 m, and the distance w = 0.5 m. The system is moving in outer space.


    (a) What is the total (linear) momentum total of this system?

    (b) What is the velocity cm of the center of mass?

    (c) What is the total angular momentum A of the system relative to point A?

    (d) What is the rotational angular momentum rot of the system?

    (e) What is the translational angular momentum trans of the system relative to point A?

    (f) After a short time interval t = 0.16 s, what is the total (linear) momentum total of the system?

    Solved So far....

    a)linear momentum = <.6 kg * 27 m/s + 0.6 kg * 27 m/s, 0, 0>
    = <56.4, 0,0>

    b) Vcm = Ptotal/Mtotal = <56.4, 0,0>/1.2 kg = 47.0

    c) stuck on this one.....
    I'm having trouble understanding what exactly to use for the angular momentum.
    LA = r X p (both vectors) but i don't know what to use for r, the position of the center of mass??

    Thanks :-)
  2. jcsd
  3. Nov 28, 2007 #2

    Doc Al

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    Staff: Mentor


    Find the angular momentum of each mass separately and add them up.
  4. Nov 28, 2007 #3
    ahh...thats a lot easier to understand. thank you . The direction of both angular momentums would be in the negative direction correct? I got the answer to be 38 kg*m^2/s and I'm trying to apply the right hand rule to find the sign. Usually the momentum is given in the picture so I'm just trying to figure out which direction it is going.
  5. Nov 28, 2007 #4
    I'm having trouble with part d now though....

    I have Lrot = cm*m*v2 + cm*m*v2

    center of mass = cm = d/2

    (1.3/2)*0.6*27 + (1.3/2)*0.6 * 67 = 37 ...which the answer is 15.6.
  6. Nov 28, 2007 #5
    Well i found my problem but I don't know why its the case....the first d/2 is negative so that makes it work out to 15.6...but why is it negative?
  7. Nov 29, 2007 #6

    Doc Al

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    Staff: Mentor

    That's not it. To find the "rotational angular momentum" of the system, look at the system in the center of mass frame. In that frame the top mass moves to the left and the bottom mass moves to the right.

    Total angular momentum is the angular momentum about the center of mass plus the angular momentum of the center of mass (which is part e).
  8. Nov 29, 2007 #7
    Do you mean its not negative or the 15.6 is wrong? The answer key said -15.6
  9. Nov 29, 2007 #8

    Doc Al

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    Staff: Mentor

    I was objecting to your reasoning, not your answer. Looks to me that the system is rotating counterclockwise about the center of mass, which I would call positive. (In the + z direction.)
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