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Dumbell movement in space

  1. Apr 6, 2006 #1
    Suppose you have 2 iron balls that are connected to one another by a bar. The balls have equal mass.
    The bar has a fixed length, has no mass, and cannot be compressed or stretched.
    The object is in a total void with no gravity or wind resistance.

    The balls can move on a 2D plain, the X,Y plain.

    The inputs are ball A and ball B's (x,y), mass, and the force, and direction of force applied on each.

    How do you find the outputs of the new (x,y) position for each iron ball?


    You can assume the iron balls have no volume just a weight, and the bar has no volume or weight, just a length.
    Outside motion is only applied on the point for each ball. (Outside motion isn't applied on the bar, but each ball affects the position of the other via the bar.)

    Basically the only function of the bar is keeping the two dumbells a fixed length apart. So any force in one end of the bar should effect the other end in someway.


    For example, we could start with a dumbell with it's first weight, A, on (0,0) and it's 2nd wieght, B, on (10,10). We apply upward force to B. If B weren't attached to A, B would go straight up. Given that B is attachedto A, when upward force is applied to B it should curve upward toward the y axis, and A should curve away from it's start point.

    I'd really like an equation or a set of equations that describe this. I imagine this is just could be done with just an application of Newton's laws.
     
    Last edited: Apr 6, 2006
  2. jcsd
  3. Apr 6, 2006 #2

    Astronuc

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    Staff: Mentor

    Depending on the forces, one has to track the motion of the center-of-mass (CM) and orientation of the dumbell axis.

    So there is both translation and rotation in the X,Y plane.

    So one needs to calculate the force on the CM and the torque (moment) on the dumbells.
     
  4. Apr 6, 2006 #3
    A = first ball; B = 2nd ball
    A_x,A_y= A's x,y
    B_x,B_y = B's x,y
    CM = Center of Mass

    At rest the CM will be
    x = (A_x+B_x)/2; y = (A_y+B_y)/2

    So the CM isn't always in the middle of the bar, as it changes based on the force and direction of each ball?

    So you're saying that the forces on each ball can be combined and placed on the center of mass to get the new (x,y) of the CM, and then only the rotation remains to be calculated?

    Is the "torque" some derivitive of the speed for an instant?
     
    Last edited: Apr 6, 2006
  5. Apr 6, 2006 #4

    Doc Al

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    Why would you think that? The CM is always in the middle of the bar.

    The net force on the object determines the acceleration of its CM; the net torque about the CM determines the angular acceleration of the object about its CM. (All per Newton's 2nd law.)
     
  6. Apr 6, 2006 #5
    I thought that speed could change the mass or something.

    -
    Hm that's nice and simple.

    So if I applied 1 foot/second of force upward on B, then I'm applying it at 90 degrees relative to A, so then 1 second later the CM would be 1 foot up, and the angle of the bar would be 45?
     
    Last edited: Apr 6, 2006
  7. Apr 6, 2006 #6

    Doc Al

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    I don't know what that means, since "foot/sec" is not a unit of force.

    Here's what you could say: If you apply a force F upward on B, then the CM of the system (assuming each ball has a mass M) will have an acceleration equal to F/(2M) upward.
     
  8. Apr 6, 2006 #7

    Astronuc

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    The applied forces (vectors) will impart a net displacement of the CM and a net moment about the CM.

    One will have a change in both translational momentum and rotational momentum.

    CM stays the same as long as the mass distribution does not change. How does one determine the moment of inertia?

    Note quite.

    F = dp/dt = d/dt (mv) = m dv/dt if m not equal m(t) or m(t) = constant.

    Torque = dL/dt = I d[itex]\omega[/itex]/dt,

    where L = angular momentum, [itex]\omega[/itex] = angular velocity

    See this - http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html#am

    and

    http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mi
     
    Last edited: Apr 6, 2006
  9. Jun 17, 2006 #8
    Thank you all for the replies so far.

    When determining the angle of net force, is it necessary to convert the vectors of ball A and B to horizontal and vertical vectors, add the horizontal and vertical vectors, and then convert the horizontal and vertical vector sums into the vector of net force? Or what is the shorter trigonometry method that I'm forgetting?
     
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