Respuesta :
Answer:
Explanation:
m1 = 3.77 kg (0, 0 )
m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)
m3 = 2.46181 kg (16.7024 cm, 0 cm )
Let x and y be the coordinates of centre of mass.
[tex]x = \frac{m_{1}x_{1}+ m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}}[/tex]
[tex]x = \frac{3.77\times 0+ 6.7106\times 5.72 + 2.46181\times 16.7024}{3.77+6.7106+2.46181}[/tex]
x = 6.1428 cm
[tex]y = \frac{m_{1}y_{1}+ m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}}[/tex]
[tex]y= \frac{3.77\times 0+ 6.7106\times 11.44 + 2.46181\times 0}{3.77+6.7106+2.46181}[/tex]
y = 5.9316 cm
The distance of the center of mass of the three particles from the origin is 6.1428 cm and 5.9316 cm.
Calculation of the distance:
Since
m1 = 3.77 kg (0, 0 )
m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)
m3 = 2.46181 kg (16.7024 cm, 0 cm )
Now here we assume x and y be the coordinates with respect to the centre of mass.
So,
We know that
[tex]x = \frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}\\\\ = \frac{3.77\times 0 + 6.7106 \times 5.72 + 2.46181 \times 16.7024}{3.77 + 6.7106 + 2.46181}[/tex]
= 6.1428 cm
Now
[tex]y = \frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}\\\\ = \frac{3.77\times 0 + 6.7106 \times 11.44 + 2.46181 \times 0}{3.77 + 6.7106 + 2.46181}[/tex]
= 5.9316 cm
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