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Angular Momentum by integration.

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose we have a yoyo (in the shape of a disk of radius R and mass M) set to be released. There is tension and gravity. Find the acceleration of the yoyo's center of mass. So far pretty easy.

    But instead of placing the origin at the yoyo's center of mass, I would like to place it anywhere in the plane of the disk, such that I would use [tex]\tau =\frac{dL}{dt}[/tex]

    and I would have to integrate for each particle in the disk, as they have different linear accelerations.

    3. The attempt at a solution
    Whenever I integrate angular momentums of each particle, I end up with the angular momentum of the center of mass of the disk. This means that vectorially the cross products sum to zero, and if I chose the origin (x=y=0) to be the center of mass, dL/dt becomes zero.

    I can imagine that when finding the rotational inertia of something and multiply it by the angular acceleration, you do not handle the acceleration of each particle as a vector, but add their tangential components as scalars.

    Now I have trouble establishing that connection between [tex]I\alpha[/tex] and [tex]\int \vec{r}\times d\vec{P}[/tex].

  2. jcsd
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