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Homework Help: Problem Help

  1. Mar 8, 2005 #1

    I have been looking at various review problems to prep for an exam and there is one I could not solve at all. Here it is:

    A flat dance floor of dimensions l(x) = 20m by l(y) = 16 m has a mass M of 150 kg. Use the bottom left corner of the dance floor as the origin. Three dance couples, each of mass m = 150 kg start in the top left, top right, and bottom left corners.

    A. What is the initial y coordinate of the center of gravity of the dance floor and three couples? Answer in units of m.

    B. The couple in the bottom left corner moves l = 7.3 m to the right. What is the new x coordinate of the center of gravity? Answer in units of m.

    C. What was the speed of the center of gravity if it took the couple 9.2 seconds to change positions? Amswer in units of m/s.

    Any help would be greatly appreciated. Thanks!
  2. jcsd
  3. Mar 8, 2005 #2


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    A is pretty straight forward. Assuming the dance floor is uniform, its center of gravity (not including the dancers) is at its center: (10, 8). The dance couples are at (0,16), (20,16), and (0,0). If we take (x,y) as the center of gravity, the x- moment of "torque" is 150(x-10)+ 150(x-0)+ 150(x-20)+ 150(x- 0) and that must be equal to 0 since there is, in fact, no "twisting" of the floor. That is:
    150(x-10)+ 150x+ 150(x-20)+ 150x= 0 . We can immediately divide by 150 to get
    x-10+ x+ x-20+ x= 0 (since all couples and the floor have the same mass, it cancels out) or 4x= 30 so x= 30/4= 15/2. Do the same thing with y (and the y coordinates of each mass).

    B. Move that couple from (0,0) to (7.3, 0) and recalculate.

    C. What is the distance between the center of gravity in A and the center of gravity in B? Since the center of gravity takes 9.2 seconds to move that far, what was its speed?
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