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Angular Momentum of a sliding disc about a point on the floor

  1. Feb 20, 2013 #1
    Hi everybody,

    A seemingly straightforward example from lecture is causing me some confusion. The example was about calculating the angular momentum of a sliding disk (not rolling) about a point on the floor. The result given in lecture says the distance to the point on the floor is unrelated to the angular momentum:
    [tex]\vec{L}=\vec{r} \times \vec{p}=\sum{\vec{r}_i \times \vec{p}_i}=\sum{\vec{r}_i \times m_i\vec{v}_i}=MRv [/tex]

    where M is the total mass of the disc, R is the radius of the disc, and v is the translational velocity of the sliding disc. Now my confusion comes in at the last equal sign. I think it should read:
    [tex]\sum{\vec{r}_i \times m_i \vec{v}_i}=\sum{m_i r_i v_i \sin{\theta_i}}[/tex]

    where [itex]\theta_i[/itex] is the angle between each particles radius vector and the constant velocity vector. I don't see how this sum ends up as [itex]MRv[/itex] as was claimed in lecture.

    I tried to write [itex] \theta_i [/itex] as a function of each ri and integrate over the disc, but didn't make progress. I know I can take the mi and vi out of the sum because they are the same for each i, but I still can't deal with the


    Any help would be greatly appreciated! Maybe I'm just missing something really obvious, I don't know.

  2. jcsd
  3. Feb 21, 2013 #2


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    Homework Helper

    Hi Sean! :smile:
    No, R is the perpendicular distance from the point to the line of motion of the centre of mass.

    ri x mi vi

    = (∑ miri) x v since vi = a constant, v

    and then use ∑ mi(ri - ro) = 0 by definition, where ro is the centre of mass :wink:
  4. Feb 21, 2013 #3
    Ok, I see. That makes sense. In my situation it just so happens that the perpendicular distance is equal to the radius of the disc. In a more general scenario this would be different. Thank you!
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