1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    [tex]\sum{r_i\sin{\theta_i}}[/tex]

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

    Thanks,
    Sean
     
  2. jcsd
  3. Feb 21, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Angular Momentum of a sliding disc about a point on the floor
Loading...