Given: A flat dance floor of dimensions x =17 m by y =19 m and has a mass of M=1

Question

                    Given: A flat dance floor of dimensions x =17 m by y =19 m and has a mass of M=1100 kg. Use the bottom left corner of the dance floor as the origin. Three                     dance couples, each of mass m =110 kg start in the top left, top right, and bottom left corners.

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

                    2. The couple in the bottom left corner moves x = 6.6 m to the right. What is the new x coordinate of the center of gravity? Answer in units of m -- ALREADY SOLVED --

                    3.What was the speed of the center of gravity if it took that couple 8.4 s to change positions? Answer in units of m/s

My thought is that you find the x coordinate from part 1 and subtract off the x coordinate in part 2 and divide by the 8.4 s, but the x coordinate I calculate from part 1 is negative, and to me that doesn't make sense.

Show me the way and I will give nice points.

Explanation / Answer

1) Xcg1 = ((1100*(17/2)) + 110*17)/(1100+110+110+110)
Xcg1 = 7.846 m

Ycg1 = ((1100*(19/2)) + 110*19 + 110*19)/(1100+110+110+110)
Xcg1 = 10.23 m

r1cm = Xcg1 i + Ycg1 j = 7.846 i + 10.23 j

2) Xcg2 = ((1100*(17/2)) + (110*17) + (110*6.6))/(1100+110+110+110)= 8.353 m

Ycg2 = 10.23

r2cm = 8.353i + 10.23j

3) v = r2-r1/t = (8.353i + 10.23j) -(7.846 i + 10.23 j) / 8.4

v = (8.353-7.846)j/8.4 = 0.060 m/s

speed = 0.06 m/s

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.