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!

Total Angular Momentum

  1. Nov 9, 2009 #1
    The problem statement
    Icm=[tex]\frac{2}{5}[/tex]mr2
    [tex]\omega[/tex]=v/r
    m=mass of object

    There's some things I don't really get about total angular momentum in a rigid body. Suppose a perfectly spherical object is rolling in uniform circular motion (ie. in a loop). Find the total angular momentum.

    2. Relevant equations
    L=mvr for particle, and L=Icm[tex]\omega[/tex] for rigid body

    3. Attempt at solution

    Since it is rolling, it has angular momentum due to its own rotation around its axis, like a normal object that's just spinning right? So then L = Icm[tex]\omega[/tex] and I would use [tex]\omega[/tex]=v/r with r=radius of sphere

    But then, it also has angular momentum around the centre of the loop-the-loop that it rolls around. Here's where i get lost. Would i consider the object as a particle and use L=mvr, or do i use L = Icm[tex]\omega[/tex] again, and which radius do i use when i sub in for [tex]\omega[/tex]? the radius of the loop or the radius of the mass itself?

    I get that the total angular momentum for a situation like this would require adding together the seperate momentums but i don't get how exactly they are seperated.
     
    Last edited: Nov 9, 2009
  2. jcsd
  3. Nov 9, 2009 #2
    actually when u are asked to calculate the angular momentum of such system, what u need to do is to apply the mv X r (with v and r being vectors) to all the particles in the rigid body, but that will be complicated
    but thankfully, the calculation can be simplified (in most simple problems) by separating the angular mementum in the way you have described, they will give the same and the correct answer just like if u do it rigorously for each and every particles in the rigid body

    hope that helps :)
     
  4. Nov 10, 2009 #3
    Thanks, that makes it clear, i thought i was simplifying it too much, so its good to know that I don't have to do the less simplified way with all the rigorous work.
     
  5. Nov 10, 2009 #4

    ideasrule

    User Avatar
    Homework Helper

    For the sphere's angular momentum about the center of the loop, you use mvr, where r is the distance from the loop's center to the sphere's center. For the sphere's angular momentum about its own center of mass, you'd use 2/5MR^2. For the total angular momentum you'd add them together.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Total Angular Momentum
  1. Total Angular Momentum (Replies: 1)

Loading...