Question

A flat dance floor of dimensions l(x) = 17m by l(y) = 19 m has a mass M = 1400 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.

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

PART TWO
B. The couple in the bottom left corner moves 7.6 m to the right. What is the new x coordinate of the center of gravity? Answer in units of m.

Answer 1

Answer:

A. 10.1 m

B. 8.44 m

Explanation:

Part One:

The coordinates of the three couples are (0, 19), (17, 19), and (0, 0).  The center of the dance floor is (8.5, 9.5).

To find the center of gravity, we multiply each coordinate by its mass, sum the results, then divide by the total mass.

$\left\begin{array}{ccccc}x&y&m&xm&ym\\0&19&110&0&2090\\17&19&110&1870&2090\\0&0&110&0&0\\8.5&9.5&1400&11,900&13,300\\&\sum&1730&13,770&17,480\\&&Avg&7.96&10.1\end{array}\right$

The initial y-coordinate is 10.1 m.

Part Two:

Repeat part one, but this time, (0, 0) becomes (7.6, 0).

$\left\begin{array}{ccccc}x&y&m&xm&ym\\0&19&110&0&2090\\17&19&110&1870&2090\\7.6&0&110&836&0\\8.5&9.5&1400&11,900&13,300\\&\sum&1730&14,606&17,480\\&&Avg&8.44&10.1\end{array}\right$

The new x-coordinate is 8.44 m.

Answer 2

(a) The initial y-coordinate of the centre of gravity of the dance floor and three couples is 10.10 m.

When the couple in the bottom left corner moves 7.6 m to the right, the new x-coordinate of the centre of gravity is 8.44 m.

Centre of Mass

(a) Here as the dance floor is flat, the centre of gravity is the same as that of the centre of mass.

We know that the centre of mass of a system of particles is given by;

$x_{CM}=\frac{\sum m_ix_i }{M}$

$y_{CM}=\frac{\sum m_iy_i }{M}$

The centre of mass without the dancers is the centre of the dance floor,

$(x_0 , y_0) = (8.5, 9.5)$

The mass at this point is 1400 kg.

The coordinates of the dance couples are;

$(x_1 , y_1)= (0, 19)$

$(x_2, y_2)=(17, 19)$

$(x_3, y_3)=(0,0)$

So, the y-coordinate of the centre of mass with the dancers is given by,

$y_{CM}=\frac{(19\times 110)+(19\times 110)+(0\times 110)+(9.5\times 1400) }{(3\times 110)+1400}=10.10\,m$

(b) When the couple in the bottom left corner moves 7.6 m to the right;

$(x_3, y_3)=(7.6,0)$

Therefore, the y-coordinate of the centre of mass with the dancers is given by:

$x_{CM}=\frac{(0\times 110)+(17\times 110)+(7.6\times 110)+(8.5\times 1400) }{(3\times 110)+1400}=8.44\,m$

Learn more about the concept of centre of mass here:

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