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

Question

Given: A flat dance floor of dimensions ? x =17 m by ? y =17 m and has a mass of M=1500 kg. Use the bottom left corner of the dance floor as the origin. Three dance couples, each of mass m =140 kg start in the top left, top right, and bottom left corners. What is the initial y coordinate of the center of gravity of the dance floor and three couples? Answer in units of m

2. The couple in the bottom left corner moves ? ? =11 m to the right. What is the new x coordinate of the center of gravity? Answer in units of m

3.What was the speed of the center of gravity if it took that couple 7.5 s to change positions? Answer in units of m/s

Explanation / Answer

1)

Taking the centre of the floor as origin,

y coordinate of the COM = [2 *(17/2)* 140 + (-17/2) * 140]/ [1500+(3*140)]

=> y= 0.62 m above the centre of the floor

2). x coordinate = 140 * 2.5/ [140*3 + 1500] = 0.182 m to the right of the centre of the floor

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