Solved Problem on Center of Mass
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From the opposite sides of a platform of length L, resting on frictionless rollers, an adult and a child are running toward each other. Determine how far the platform will slide when the adult moves from one end of the platform to the other. We know that the speed of the adult is twice the speed of the child, and the masses of the platform, the adult, and the child are m1, m2 and m3, respectively.


Problem data:
  • Platform Length:    L;
  • Platform mass:    m1;
  • Adult mass:    m2;
  • Adult speed:    v2;
  • Child mass:    m3;
  • Child Speed:    v3.
Solution

The position of the center of mass of a three-particle system is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}}} \tag{I} \end{gather} \]
In the first situation (Figure 1), we have a reference frame pointing to the right, with the origin at the point on the platform where the man is. Forgetting the platform, the man, and the child, and considering only their centers of mass, cp for the center of mass of the platform, cm for the center of mass of the man, and cc for the center of mass of the child, the center of mass of the man is at the origin of the reference frame x2 = 0, the center of mass of the platform, of length L, is at half its length \( x_{1}=\frac{L}{2} \), and the center of mass of the child is at the end of the platform opposite the man x3 = L. Thus, substituting these values in expression (I), we have for the center of mass of the platform-man-child system in the initial situation
\[ \begin{gather} x_{i}=\frac{m_{1}\dfrac{L}{2}+m_{2}\times 0+m_{3}L}{m_{1}+m_{2}+m_{3}}\\[5pt] x_{i}=\frac{m_{1}\dfrac{L}{2}+m_{3}L}{m_{1}+m_{2}+m_{3}} \end{gather} \]
multiplying and dividing the second term of the numerator on the right-hand side of the equality by 2
\[ \begin{gather} x_{i}=\frac{m_{1}\dfrac{L}{2}+\dfrac{2}{2}\times m_{3}L}{m_{1}+m_{2}+m_{3}}\\[5pt] x_{i}=\frac{m_{1}\dfrac{L}{2}+2m_{3}\dfrac{L}{2}}{m_{1}+m_{2}+m_{3}} \end{gather} \]
factoring \( \dfrac{L}{2} \) in the numerator
\[ \begin{gather} x_{i}=\frac{L}{2}\left(\frac{m_{1}+2m_{3}}{m_{1}+m_{2}+m_{3}}\right) \tag{II} \end{gather} \]
Figure 1

In the second part of Figure 1, we have the final situation, the man walked forward the length L of the platform, as the child's speed is half the speed of the man, he walked half the length of the platform, meanwhile, the platform moved by a certain distance D behind the man. So the center of mass of the man is in the position \( x_{2}=L-D \) and the center of mass of the platform and the child coincide in \( x_{1}=x_{3}=\dfrac{L}{2}-D \), again substituting these values in expression (I) for the final situation
\[ \begin{gather} x_{f}=\frac{m_{1}\left(\dfrac{L}{2}-D\right)+m_{2}(L-D)+m_{3}\left(\dfrac{L}{2}-D\right)}{m_{1}+m_{2}+m_{3}}\\[5pt] x_{f}=\frac{m_{1}\dfrac{L}{2}-m_{1}D+m_{2}L-m_{2}D+m_{3}\dfrac{L}{2}-m_{3}D}{m_{1}+m_{2}+m_{3}} \end{gather} \]
multiplying and dividing the third term of the numerator on the right-hand side of the equation by 2
\[ \begin{gather} x_{f}=\frac{m_{1}\dfrac{L}{2}-m_{1}D+\dfrac{2}{2}m_{2}L-m_{2}D+m_{3}\dfrac{L}{2}-m_{3}D}{m_{1}+m_{2}+m_{3}}\\[5pt] x_{f}=\frac{m_{1}\dfrac{L}{2}-m_{1}D+2m_{2}\dfrac{L}{2}-m_{2}D+m_{3}\dfrac{L}{2}-m_{3}D}{m_{1}+m_{2}+m_{3}} \end{gather} \]
factoring −D and \( \dfrac{L}{2} \) in the numerator on the right-hand side of the equation
\[ \begin{gather} x_{f}=\frac{\dfrac{L}{2}(m_{1}+2m_{2}+m_{3})-D(m_{1}+m_{2}+m_{3})}{m_{1}+m_{2}+m_{3}} \tag{III} \end{gather} \]
The Center of Mass remains in the same position, equating expressions (II) and (III)
\[ \begin{gather} x_{i}=x_{f}\\[5pt] \frac{L}{2}\left(\frac{m_{1}+2m_{3}}{\cancel{m_{1}+m_{2}+m_{3}}}\right)=\frac{\dfrac{L}{2}(m_{1}+2m_{2}+m_{3})-D(m_{1}+m_{2}+m_{3})}{\cancel{m_{1}+m_{2}+m_{3}}}\\[5pt] \frac{L}{2}(m_{1}+2m_{3})=\frac{L}{2}(m_{1}+2m_{2}+m_{3})-D(m_{1}+m_{2}+m_{3})\\[5pt] D(m_{1}+m_{2}+m_{3})=\frac{L}{2}(m_{1}+2m_{2}+m_{3})-\frac{L}{2}(m_{1}+2m_{3}) \end{gather} \]
factoring \( \dfrac{L}{2} \) on the right-hand side of the equation
\[ \begin{gather} D(m_{1}+m_{2}+m_{3})=\frac{L}{2}(m_{1}+2m_{2}+m_{3}-m_{1}-2m_{3})\\[5pt] D(m_{1}+m_{3}+m_{2})=\frac{L}{2}(2m_{2}-m_{3}) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {D=\frac{2m_{2}-m_{3}}{m_{1}+m_{3}+m_{2}}\frac{L}{2}} \end{gather} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .