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Acceleration of center of mass of cart-block system

  1. Nov 26, 2006 #1
    The figure above shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block.
    The cart has mass m1 = 0.600 kg, and its center is initially at xy coordinates (-0.500 m, 0 m).
    The block has mass m2 = 0.400 kg, and its center is initially at xy coordinates (0, 0.100 m).
    The cart is released from rest, and both cart and block move until the cart hits the pulley.
    The friction between the cart and air track and between the pulley and its axle is negligible as is the mass of the cord and pully.

    • In unit-vector notation, what is the acceleration of the center of mass of the cart-block system?
    • What is the velocity of the com as a function of time t?

    so..i know that acom = Fnet/M and that i probably have to find acom,x and acom,y, but i'm not sure and if i do need to solve for those, how do i find the Fnet?
  2. jcsd
  3. Nov 26, 2006 #2
    I think you are making it too complicated. The internal forces will cancel, by the 3rd law, right? Fnet would just be the external forces.

  4. Nov 27, 2006 #3
    Still not sure what to do, sorry :(.
  5. Nov 27, 2006 #4
    The only forces are the tension on the cord, and gravity, as far as I can tell. The tension on the cord is an internal force, which just cancels out by Newton's third law. So Fnet has got to have something to do with gravity, it would seem.
  6. Nov 27, 2006 #5
    so if the only external force acting on the system is gravity, then
    Fnet,system = Mg = (m1 + m2)g = 9.8?
  7. Nov 27, 2006 #6


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    There is another external force besides gravity. You don't need to find it to do the problem, but you could. It is the force of the pulley acting on the string. You can continue to find the motions of the individual masses and then find the motion of the CM from those. But you have too many g in your equation, and something is missing.
  8. Nov 27, 2006 #7
    Yeah, I guess I'm pretty lost. For some reason, the more I try to think about this problem, the more confused I seem to become :(.
    I hate to ask this, but might you be willing to give me a step-by-step explanation of how to solve it?

    I don't know if they have any significance (or if they are even correct), but here are some values I've come up with so far:
    • x1 = -0.5, y1 = 0
    • x2 = 0, y2 = -0.1
    • xcom,sys = (m1x1 + m2x2)/(M) = -0.3
    • ycom,sys = (m1y1 + m2y2)/(M) = -0.04
    and a couple of the equations I've been trying to work with:
    • Fnet = Macom
    • Macom = m1a1 + m2a2
  9. Nov 27, 2006 #8


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    The initial position and CM equations are a good start. You don't have Fnet given, adn you will not be able to find it unless you find the tension in the string. You can do that, but you don't have to. The two masses share a common distance of motion, which means their velocities and accelerations must have the same magnitude. Those magnitudes can be found without actually calculating the tension. You know the directions. From your CM position equations you can come up with the CM velocity and CM acceleration equations.
  10. Nov 27, 2006 #9
    so what you're saying is:
    (change in d1) = (change in d2), v1 = v2, and a1 = a2?

    the equations for CM of velocity and acceleration:
    vcom = (m1v1 + m2v2)/(M)
    acom = (m1a1 + m2a2)/(M)

    i don't know what to do...if i set the velocities and accelerations equal to eachother, all i end up with is vcom = v and acom = a.
  11. Nov 27, 2006 #10


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    The vectors are not equal. The magnitudes of the vectors are equal. The horizontal distance m1 moves is equal to the vertical distance m2 falls
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