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Center of mass (vectors)

  1. May 30, 2006 #1
    A big olive (m = 0.11 kg) lies at the origin of an xy coordinate system, and a big Brazil nut (M = 0.82 kg) lies at the point (0.99, 2.1) m. At t = 0, a force Fo = (4i + 4j) N begins to act on the olive, and a force Fn = (-4i -3j) N begins to act on the nut. What is the (a)x and (b)y displacement of the center of mass of the olive-nut system at t = 4.6 s, with respect to its position at t = 0?

    I first started approaching the problem by doing E(sigma)mixi/Emi, and the same for the y-direction. So, for x-direction, it would be:

    (.99molive + 0mnut)/(.82kg + .11kg)

    for the y-direction, it would be:

    (2.1molive + 0mnut)/(.82kg + .11kg)

    I don't even know if I did those correctly.

    For the rest, they give you the force in both directions and the duration time (4.6 sec). I have to find the displaceent, which means I first have to find the center of mass for 0 seconds and then for 4.6 seconds.

    Can someone help me with how to approach this problem, especially how I can use the vector forces? Thank you.
    Last edited: May 30, 2006
  2. jcsd
  3. May 30, 2006 #2

    Doc Al

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    Staff: Mentor

    Looks to me like you got the olive and nut mixed up; it's the olive that is at the origin.

    There are two ways to approach this. One way is to treat each "particle" separately: Given the force, find its acceleration, then it's displacement. (Treat each component independently.) Then find the new center of mass at t = 4.6 sec.

    Another way, a bit easier, is to treat the nut and olive as a single system. Find the net force on the system (just add the forces). Then, treating the system as a single "particle" (with mass equal to the total mass of both), you can find the acceleration--and then the displacement--of the center of mass directly.
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