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Angular Momentum

  1. Nov 21, 2005 #1
    A 2.6 kg particle that is moving horizontally over a floor with velocity (-3.00 m/s)j undergoes a totally inelastic collision with a 4.00 kg particle that is moving horizontally over the floor with velocity (4.20 m/s)i. The collision occurs at xy coordinates (-0.50 m, -0.10 m). After the collision, what is the angular momentum of the stuck-together particles with respect to the origin using vector notation?

    I first tried using conservation of momentum, which gives me 9 as the x-momentum. This is wrong. I know that angular momentum (L) = Iw, but I'm not sure how to calculate I or w.

    I'm also stuck on a similar problem:
    At the instant the displacement of a 2.00 kg object relative to the origin is d = (2.00 m)i + (4.00 m)j - (3.00 m)k, its velocity is v = - (6.40 m/s)i + (2.90 m/s)j + (2.70 m/s)k, and it is subject to a force F = (6.40 N)i - (7.80 N)j + (4.10 N)k. Find the angular momentum of the object and the torque acting on the object about the origin using vector notation.

    To find momentum, I again didn't know how to find Iw, so I just tried multiplying the given mass times the velocity vector. I'm guessing this is wrong because thats for linear momentum.

    Any help would greatly be appreciated!
  2. jcsd
  3. Nov 22, 2005 #2


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    The angular momentum of a particle about an origin is given by,

    L = rp

    where r is position vector of the particle and p is the linear momentum
  4. Nov 22, 2005 #3
    how do you determine which way the velocity of stuck-together particles are going? would i just use pythagorean theorem to find the hypotenuse of the 4.2 velocity and the -3 velocity? and from there, how do i figure out which way the momentum is going? how can there be momentum in the 'k' (vector notation) direction?
  5. Nov 22, 2005 #4


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    You need to specify (and keep track of) the direction of the velocity and momentum.
    For particles you can calculate the angular momentum as:
    [tex]\vec{L}=\vec{r} \times \vec{p}[/tex]
    where [itex]\times[/itex] is the cross product, [itex]\vec{r}[/itex] is the vector from the axis of rotation, and [itex]\vec{p}[/itex] is the momentum of the particle.
    The magnitude of [itex]\vec{L}[/itex] will be equal to
    [tex]|r p \sin \theta|[/tex]
    Where [itex]\theta[/itex] is the angle between [itex]\vec{r}[/itex] and [itex]\vec{p}[/itex].
    For forces that apply at a single point, torque can be calculated by
    [tex]\vec{\tau}=\vec{r} \times \vec{F}[/tex]
  6. Nov 22, 2005 #5
    how do i determine r? it isn't just the displacement from the origin, right? because that doesnt work
  7. Nov 22, 2005 #6

    Doc Al

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    Sure it works. (Use the coordinates of the collision to figure out the perpendicular distance between the momentum vectors and the origin.)
  8. Nov 22, 2005 #7
    sorry i was referring to the second problem. but on the first problem, how do i know how fast the stuck-together objects are going in the x or y directions? would they both be moving -3j down and 4.2i across? if so, would i determine the angular momentum in the j direction by doing

    (-.5)(2.6 kg)(-3.00 m/s) = 3.9 kg*m^2/s?
  9. Nov 22, 2005 #8

    Doc Al

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    For the first problem, realize that angular momentum is conserved during the collision. (Find the total angular momentum of the particles before the collision.)

    For the second problem, calculate [itex]\vec{L}=\vec{r} \times \vec{p}[/itex]. (I assume you know how to take the cross product of vectors in cartesian coordinates.)
  10. Nov 25, 2005 #9
    i.. actually don't know how to take the cross product of vectors in cartesian coordinates :( but anyhow, after i find the angular momentum, how do i figure out which direction it's going? because i have to give the angular momentum in vector notation.
  11. Nov 28, 2005 #10

    Doc Al

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