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!

Homework Help: Center of mass for rigid body and point particle

  1. May 6, 2012 #1
    Hi, this is a problem on center mass.

    1. The problem statement, all variables and given/known data
    A beam hangs straight down from a point O(O is placed at x=0 and y = 0, aka origo). The beam is attached to the point O. Beam has length L and mass M. The density of the beam is uniform, so the centermass of the beam is [itex]-\frac{L}{2}\boldsymbol{\hat{\jmath}}[/itex]. A point particle with mass m is shot into the beam at (0,-L,0) and latches itself onto the beam in (0,-L,0) so it becomes a part of the beam at that point.

    What is the center of mass for the system consisting of the beam and the particle?
    (I first thought the center of mass was still at [itex]-\frac{L}{2}[/itex], but the task does not state anywhere that m<< M).
    2. Relevant equations


    3. The attempt at a solution

    I was thinking of using center of mass for a particle system, but since one of the "particles" is a beam, I assume I cannot use [itex]\vec{R} = \frac{1}{M}\Sigma m_i \vec{r}_i [/itex]. And since I dont know the densitydifference for the point particle nor the beam, I don't think I can use [itex]\vec{R} = \frac{1}{M} \int \int \int \vec{r} \rho dV[/itex]

    Not sure how to approach finding the total center of mass, though maybe the answer is right there and I dont see it. Any ideas?

    Edit; Is the subdivision principle applicable here?


    [itex]\vec{R} = \frac{-M\frac{L}{2} -mL}{M+m}\boldsymbol{\hat{\jmath}}[/itex]
    Last edited: May 6, 2012
  2. jcsd
  3. May 6, 2012 #2
    That looks good to me at least, and you do see that in the limit m<<M the stick goes back to L/2 CM.
  4. May 6, 2012 #3
    Yeah, that's a good point! Thanks=)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook