Given: A flat dance floor of dimensions x = 21 m by y = 21 m and has a mass of M

Question

Given: A flat dance floor of dimensions x = 21 m by y = 21 m and has a mass of M = 1300 kg. Use the bottom left corner of the dance floor as the origin. Three dance couples, each of mass m = 120 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.

(part 2 of 3)

The couple in the bottom left corner moves _ = 7.4 m to the right. What is the new x coordinate of the center of gravity?

Answer in units of m.

(part 3 of 3)

What was the speed of the center of gravity if it took that couple 12 s to change positions?

Answer in units of m/s.

Explanation / Answer

Here ,

lx = 21 m

ly = 21 m

M = 1300 Kg

m = 120 kg

part 1) for the initial y - coordinate ,

ycom = (m*y1 + m *y2 + m * y3 + M * y )/(3 * m + M)

ycom = (120 * 0 + 120 * 21 + 120*21 + 1300 * 21/2)/(120 * 3 + 1300)

ycom = 11.26 m

the y- coordinate of the centre of mass is 11.26 m

Xcom = (120 * 0 + 120 * 21 + 120*0 + 1300 * 21/2)/(120 * 3 + 1300)

Xcom = 9.741 m

the x -coordinate of the centre of mass is 9.741 m

part 2)

for the couple to move 7.4 m to the left ,

Xcom = (120 * 7.1 + 120 * 21 + 120*0 + 1300 * 21/2)/(120 * 3 + 1300)

Xcom = 10.25 m

the position of centre of mass is 10.25 m

part 3)

speed = distance moved /time

speed = (10.25 - 9.741)/12

speed = 0.043 m/s

the speed of centre of mass is 0.043 m/s

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