Respuesta :
The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
What is the center of mass of a system of particles?
- A place at which the entire mass of the body or all the masses of a system of particles appears to be concentrated is known as the center of mass of a body or system of particles. According to physics, the center of mass is a location where the total of the weighted relative positions of the distributed mass's points in space equals zero.
a)
The center of mass of a three-particle system is expressed as
[tex]r_{cm}=[/tex] [tex]$r_{c m}=\frac{\sum_{i=1}^{3} m_{i} r_{i}}{\sum_{i=1}^{3} m_{i}} \Rightarrow \frac{m_{1} r_{1}+m_{2} r_{2}+m_{3} r_{3}}{m_{1}+m_{2}+m_{3}}$[/tex]
When the system is only on [tex]$\mathrm{x}$[/tex]-axis (i.e. [tex]$\mathrm{y}=0$[/tex] )
[tex]$\begin{aligned}x_{c m} &=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\x_{c m} &=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\x_{c m} &=0.5909 \mathrm{~m} \\\text { Therefore } r_{c m}=(0.5909 \mathrm{~m}, 0)\end{aligned}$[/tex]
b)
When the two particles are shifted
[tex]\begin{aligned}&x_{c m}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\&x_{c m}=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\&x_{c m}=0.5909 \mathrm{~m} \\&y_{c m}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}} \\&y_{c m}=\frac{(5.0 \times 1.0)+(2.0 \times 0)+(4.0 \times 0.5)}{5.0+2.0+4.0} \\&y_{c m}=0.6364 \mathrm{~m}\end{aligned}[/tex]
Therefore [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
To learn more about the mass of a system of particles, refer to:
https://brainly.com/question/15572016
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