Question

Three point mass particles are located in a plane: 4.34 kg located at the origin, 8.029 kg at [(6.44 cm),(12.88 cm)], and 2.9078 kg at [(15.8424 cm),(0 cm)]. How far is the center of mass of the three particles from the origin? Answer in units of cm.

Answer 1

Answer:

9.3155 cm

Explanation:

Let

$m_1=4.34 kg$

$x_1=0$

$y_1=0$

$m_2=8.029 kg$

$x_2=6.44 cm,y_2=12.88 cm$

$m_3=2.9078 kg$

$x_3=15.8424 cm,y_3=0$

We have to find the center of mass of the three particles from the origin.

$X=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$

$X=\frac{0+8.029\times 6.44+15.8424\times 2.9078}{4.34+8.029+2.9078}$

$X=6.400 cm$

$Y=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$

$Y=\frac{0+8.029\times 12.88+2.9078\times 0}{4.34+8.029+2.9078}$

$Y=6.769 cm$

$d=\sqrt{X^2+Y^2}=\sqrt{(6.400)^2+(6.769)^2}=9.3155 cm$

Hence, the distance of center of mass of the three particles from the origin=9.3155 cm