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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? Thank you.
  2. jcsd
  3. May 30, 2006 #2


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    I think you have got the nut and olive vectors mixed up. Here's the steps I would now take;

    1. Convert the intial positions into unit vectors (ai + bj)
    2. Convert the forces into accelerations
    3. Use kinematic equations to find the position vector of the nut and the olive
    4. Find the centre of mass

    Alternatively, as Doc Al mentioned on the other thread, you could treat it a single system.

    Last edited: May 31, 2006
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