Center of Mass!

a) Use the equation for the center of mass.

b) Use the equation of center of mass, but remember that the origin is now centered at the top mass.

a) The equation for the center of mass formula gives:

\begin{equation*}
\vec{r}_{\text{cm}}=\frac{\sum_i m_i\vec{r}_i}{\sum_i m_i}.
\end{equation*}

Using a coordinate system centered on the lower left mass, we have:

\begin{equation*}
\vec{r}_{\text{cm}}=\frac{m(b\cos(60^{\circ})\,\hat{\textbf{i}}+b\sin(60^{\circ})\,\hat{\textbf{j}})+2m(b\,\hat{\textbf{i}})+m\vec{0}}{m+2m+m},
\end{equation*}

which, with numerical values, yields:

\begin{equation*}
\vec{r}_{\text{cm}}\approx4.38\,\text{cm}\,\hat{\textbf{i}}+1.52\,\text{cm}\,\hat{\textbf{j}}.
\end{equation*}

b) We now need to center our coordinate system at the top mass. The equation for the center of mass can now be written as:

\begin{equation*}
\vec{r}_{\text{cm}}^{\prime}=\frac{m\vec{0}+2m(b\cos(60^{\circ})\,\hat{\textbf{i}}-b\sin(60^{\circ})\,\hat{\textbf{j}})+m(-b\cos(60^{\circ})\,\hat{\textbf{i}}-b\sin(60^{\circ})\,\hat{\textbf{j}})}{m+2m+m},
\end{equation*}

which, with numerical values, yields:

\begin{equation*}
\vec{r}_{\text{cm}}^{\prime}\approx0.88\,\text{cm}\,\hat{\textbf{i}}-4.54\,\text{cm}\,\hat{\textbf{j}}.
\end{equation*}

The difference between the location of Particle 1 and the center of mass \(cm\), is significant enough to warrant a comparison:

\begin{equation*}
\vec{r}_{\text{cm}} – \vec{r}_1  \approx0.88\,\text{cm}\,\hat{\textbf{i}}-4.54\,\text{cm}\,\hat{\textbf{j}},
\end{equation*}

which is the same result as \(\vec{r}_{\text{cm}}^{\prime}\).

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) To find the center of mass position \(\vec{r}_{\text{cm}}\) for an array of point particles, we must use the following expression

\begin{equation}
\label{rcm}
\vec{r}_{\text{cm}}=\frac{\sum_i m_i\vec{r}_i}{\sum_i m_i},
\end{equation}

where \(m_i\) is the mass of the \(i\)-th particle and \(\vec{r}_i\) is the position vector of the particle with respect to some coordinate system.

We’ll label the particles with numbers: particle 1 will be the one on the top, particle 2 will be the one on the right, and particle 3 will be the particle on the left of the triangle. Then \(m_1=m\), \(m_2=2m\) and \(m_3=m\).

If the origin of the coordinate system is located in particle 3, then

\begin{equation}
\label{r3}
\vec{r}_{3}=\vec{0}.
\end{equation}

The position for particle 1 and 2 is given by the geometry of the equilateral triangle as

\begin{equation}
\label{r1}
\vec{r}_1=b\cos(60^{\circ})\,\hat{\textbf{i}}+b\sin(60^{\circ})\,\hat{\textbf{j}}.
\end{equation}

The position vector for particle 2 will be

\begin{equation}
\label{r2}
\vec{r}_2=b\,\hat{\textbf{i}}.
\end{equation}

This can be seen in figure 1.

Figure 1: We place our coordinate system on the mass on the left. The position vectors \(\vec{r}_1\), \(\vec{r}_2\) for the other two masses in the triangular array are also shown.

Explicitly writing down the expression for the center of mass given in equation \eqref{rcm}, we obtain

\begin{equation}
\vec{r}_{\text{cm}}=\frac{m_1\vec{r}_1+m_2\vec{r}_2+m_3\vec{r}_3}{m_1+m_2+m_3},
\end{equation}

which, after using the explicit values for the position of the particles given by equations \eqref{r3}, \eqref{r1} and \eqref{r2}, becomes

\begin{equation}
\vec{r}_{\text{cm}}=\frac{m(b\cos(60^{\circ})\,\hat{\textbf{i}}+b\sin(60^{\circ})\,\hat{\textbf{j}})+2m(b\,\hat{\textbf{i}})+m\vec{0}}{m+2m+m}.
\end{equation}

After grouping terms in the same direction and factorizing \(mb\), we have

\begin{equation}
\vec{r}_{\text{cm}}=\frac{mb\left(\left(\cos(60^{\circ})+2\right)\,\hat{\textbf{i}}+\sin(60^{\circ})\,\hat{\textbf{j}}\right)}{4m}.
\end{equation}

The equation above can be further simplified by cancelling out the mass \(m\), namely

\begin{equation}
\vec{r}_{\text{cm}}=\frac{b\left(\left(\cos(60^{\circ})+2\right)\,\hat{\textbf{i}}+\sin(60^{\circ})\,\hat{\textbf{j}}\right)}{4}.
\end{equation}

Using the numerical values, we get

\begin{equation}
\vec{r}_{\text{cm}}=\frac{(7\,\text{cm})\left(\left(\cos(60^{\circ})+2\right)\,\hat{\textbf{i}}+\sin(60^{\circ})\,\hat{\textbf{j}}\right)}{4},
\end{equation}

\begin{equation}
\vec{r}_{\text{cm}}\approx4.38\,\text{cm}\,\hat{\textbf{i}}+1.52\,\text{cm}\,\hat{\textbf{j}}.
\end{equation}

b) Now, if the origin of the coordinate system is located in particle 1, then

\begin{equation}
\label{r1p}
\vec{r}_{1}’=\vec{0}.
\end{equation}

The position for particle 2 and 3 is given by the geometry of the equilateral triangle, namely

\begin{equation}
\label{r2p}
\vec{r}_2’=b\cos(60^{\circ})\,\hat{\textbf{i}}-b\sin(60^{\circ})\,\hat{\textbf{j}}.
\end{equation}

The position vector for particle 3 will be

\begin{equation}
\label{r3p}
\vec{r}_3’=-b\cos(60^{\circ})\,\hat{\textbf{i}}-b\sin(60^{\circ})\,\hat{\textbf{j}},
\end{equation}

as seen in figure 2.

Figure 2: We place our coordinate system ot the mass on the top of the triangular array. The position vectors \(\vec{r}_2’\), \(\vec{r}_3’\) for the other two masses in the triangular array are also shown.

Explicitly writing down the expression for the center of mass given in equation \eqref{rcm}, we obtain

\begin{equation}
\vec{r}_{\text{cm}}^{\prime}=\frac{m_1\vec{r}_1^{\prime}+m_2\vec{r}_2^{\prime}+m_3\vec{r}_3^{\prime}}{m_1+m_2+m_3},
\end{equation}

which, after using the explicit values for the position of the particles in this second case given by equations \eqref{r1p}, \eqref{r2p} and \eqref{r3p}, becomes

\begin{equation}
\vec{r}_{\text{cm}}^{\prime}=\frac{m\vec{0}+2m(b\cos(60^{\circ})\,\hat{\textbf{i}}-b\sin(60^{\circ})\,\hat{\textbf{j}})+m(-b\cos(60^{\circ})\,\hat{\textbf{i}}-b\sin(60^{\circ})\,\hat{\textbf{j}})}{m+2m+m}.
\end{equation}

After grouping terms in the same direction and factorizing \(mb\), we have

\begin{equation}
\vec{r}_{\text{cm}}^{\prime}=\frac{mb\left(\left(2\cos(60^{\circ})-\cos(60^{\circ})\right)\,\hat{\textbf{i}}-3\sin(60^{\circ})\,\hat{\textbf{j}}\right)}{4m}.
\end{equation}

The equation above can be further simplified by cancelling out the mass \(m\) and the term along the X axis, namely

\begin{equation}
\vec{r}_{\text{cm}}^{\prime}=\frac{b\cos(60^{\circ})}{4}\,\hat{\textbf{i}}-\frac{3b\sin(60^{\circ})}{4}\,\hat{\textbf{j}}.
\end{equation}

Using the numerical values, we get

\begin{equation}
\vec{r}_{\text{cm}}^{\prime}=\frac{(7\,\text{cm})\cos(60^{\circ})}{4}\,\hat{\textbf{i}}-\frac{(3)(7\,\text{cm})\sin(60^{\circ})}{4}\,\hat{\textbf{j}},
\end{equation}

\begin{equation}
\vec{r}_{\text{cm}}^{\prime}\approx0.88\,\text{cm}\,\hat{\textbf{i}}-4.54\,\text{cm}\,\hat{\textbf{j}}.
\end{equation}

To compare the positions, let’s calculate the difference between particle 1 and the center of mass \(cm\), which is:

\begin{equation}
\vec{r}_{\text{cm}} – \vec{r}_1 = \frac{b( (\cos (60^\circ) + 2) \, \hat{\textbf{i}} + \sin (60^\circ) \, \hat{\textbf{j}} )}{4} – ( b \cos (60^\circ) \, \hat{\textbf{i}} + b \sin (60^\circ) \, \hat{\textbf{j}},
\end{equation}

or

\begin{equation}
\vec{r}_{\text{cm}} – \vec{r}_1  \approx0.88\,\text{cm}\,\hat{\textbf{i}}-4.54\,\text{cm}\,\hat{\textbf{j}},
\end{equation}

which is exactly the center of mass \( \vec{r}_{\text{cm}}^{\prime} \) as expected. The location of the center of mass for any object never changes. The references’ points or coordinate system can change, from where the center of mass was measured, but that does not mean the center of mass is different.

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