Solving Review Problem on Dance Floor

[Maximus24]

Hello,

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!

 

[HallsofIvy]

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?

 

[thepatientmental]

Hello!

Solving review problems can be tricky, but with some practice and understanding of the concepts, you'll be able to solve them easily. Let's break down this problem and see how we can solve it step by step.

A. To find the initial y coordinate of the center of gravity, we need to find the total mass and the individual moments of the dance floor and the three couples. The total mass is given as 150 kg + (3 x 150 kg) = 600 kg. The moment of the dance floor is calculated as Ml(x)/2 = (150 kg)(20 m)/2 = 1500 kgm. Similarly, the moments of the three couples are calculated as ml(x)/2 = (150 kg)(20 m)/2 = 1500 kgm, ml(y)/2 = (150 kg)(16 m)/2 = 1200 kgm, and ml(y)/2 = (150 kg)(16 m)/2 = 1200 kgm. Now, we can find the moment of the center of gravity by adding all the individual moments and dividing by the total mass. This gives us (1500 kgm + 1500 kgm + 1200 kgm + 1200 kgm)/600 kg = 10.5 m. Therefore, the initial y coordinate of the center of gravity is 10.5 m.

B. To find the new x coordinate of the center of gravity, we need to consider the movement of the couple in the bottom left corner. Since they move 7.3 m to the right, the moment of this couple changes from ml(x)/2 to m(l(x)+7.3)/2. Substituting the values, we get (150 kg)(27.3 m)/2 = 2050 kgm. Now, we can find the new moment of the center of gravity by adding all the individual moments and dividing by the total mass. This gives us (1500 kgm + 1500 kgm + 2050 kgm + 1200 kgm)/600 kg = 12.5 m. Therefore, the new x coordinate of the center of gravity is 12.5 m.

C. To find the speed of the center of gravity, we need to use the formula v = d/t, where v is the speed, d is the distance, and t is the time. In this case, the distance is