Question

a rod with uniform density is centered on $x = 0$, as shown in the figure. spheres 1 and 2 with masses $m_1$ and $m_2$ respectively, are connected to the ends of the rod. the center of mass of the rod - spheres system is located at $x_{cm}$. if the mass of each of the two spheres is doubled, but the mass of the rod does not change, where will the new center of mass of the spheres - rod system be located?
a between 0 and $\frac{x_{cm}}{2}$
b at $\frac{x_{cm}}{2}$
c at $x_{cm}$
d between $x_{cm}$ and $2x_{cm}$

Explanation:

Step1: Recall the formula for center of mass

The formula for the \(x\)-coordinate of the center of mass of a system of particles is \(x_{cm}=\frac{\sum_{i = 1}^{n}m_ix_i}{M_{total}}\), where \(m_i\) is the mass of the \(i -\)th particle and \(x_i\) is its \(x\)-coordinate, and \(M_{total}=\sum_{i = 1}^{n}m_i\). Let the mass of the rod be \(M_{rod}\), the position of sphere 1 be \(x_1\) and sphere 2 be \(x_2\). Initially, \(x_{cm}=\frac{M_1x_1 + M_2x_2+M_{rod}x_{rod}}{M_1 + M_2+M_{rod}}\).

Step2: Calculate the new center of mass

When the mass of each sphere is doubled (\(M_1'\) = \(2M_1\), \(M_2'\)= \(2M_2\)), the new center of mass \(x_{cm}'=\frac{2M_1x_1 + 2M_2x_2+M_{rod}x_{rod}}{2M_1 + 2M_2+M_{rod}}\). Factor out 2 from the numerator and denominator (for the terms involving \(M_1\) and \(M_2\)): \(x_{cm}'=\frac{2(M_1x_1 + M_2x_2)+M_{rod}x_{rod}}{2(M_1 + M_2)+M_{rod}}\). Let \(a = M_1x_1 + M_2x_2\) and \(b = M_1 + M_2\). Then \(x_{cm}=\frac{a + M_{rod}x_{rod}}{b + M_{rod}}\) and \(x_{cm}'=\frac{2a+M_{rod}x_{rod}}{2b + M_{rod}}\).

We can also use the property of the center of mass formula. If we consider the two - sphere - rod system, the center of mass formula is a weighted average. Let \(M_{s}=M_1 + M_2\) (initial total mass of spheres) and \(M_{s}'=2M_1 + 2M_2\) (new total mass of spheres). The contribution of the rod (\(M_{rod}\)) to the center of mass formula:

$$x_{cm}=\frac{M_{s}x_{s}+M_{rod}x_{rod}}{M_{s}+M_{rod}},x_{cm}'=\frac{2M_{s}x_{s}+M_{rod}x_{rod}}{2M_{s}+M_{rod}}$$

Let \(k=\frac{M_{s}}{M_{s}+M_{rod}}\), then \(x_{cm}=kx_{s}+(1 - k)x_{rod}\). And \(x_{cm}'=\frac{2M_{s}}{2M_{s}+M_{rod}}x_{s}+\frac{M_{rod}}{2M_{s}+M_{rod}}x_{rod}\). Let \(M_{s}+M_{rod}=N\), \(2M_{s}+M_{rod}=P\). We know that \(\frac{2M_{s}}{P}=\frac{2(M_{s})}{2M_{s}+M_{rod}}\) and \(\frac{M_{rod}}{P}=\frac{M_{rod}}{2M_{s}+M_{rod}}\).

Another way: Assume \(x_{rod} = 0\) (since the rod is uniform and centered at \(x = 0\)), \(x_{cm}=\frac{M_1x_1+M_2x_2}{M_1 + M_2}\). When we double \(M_1\) and \(M_2\), \(x_{cm}'=\frac{2M_1x_1 + 2M_2x_2}{2M_1+2M_2}=\frac{M_1x_1 + M_2x_2}{M_1 + M_2}\)

Answer:

C. At \(x_{cm}\)