Solved Problem on Center of Mass

A semicircular arc of radius R has the center of mass at a distance \( \frac{2R}{\pi} \) from the center. Determine the position of the center of mass, of a wire homogeneous and a constant cross-section with the shape shown in the figure, relative to the cartesian system given.

Problem data:

Length of straight wire segment: L = 8 cm;
Radius of the semicircular wire segment: R = 4 cm.

Solution

For the segment of the wire bent into a semicircle, the problem tells us that the center of mass is at a distance of \( \frac{2R}{\pi} \) from the center, and the x coordinate of the semicircle will be (Figure 1)

\[ \begin{gather} x_{s}=\frac{2R}{\pi}\\[5pt] x_{s}=\frac{-{2\times 4}}{\pi }\\[5pt] x_{s}=\frac{-{8}}{\pi} \end{gather} \]

where the minus sign indicates that the point is to the left of the origin.

Figure 1

Due to the symmetry of the semicircle, the y coordinate will be equal to zero ( y_s = 0, as the wire, is homogeneous and of a constant cross-section, there is the same mass above and below the x-axis). The point where the center of mass of the semicircle is located in

\[ \begin{gather} (x_{s},y_{s})=\left(-\frac{8}{\pi },\;0\right) \end{gather} \]

The linear mass density is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {\lambda =\frac{m}{\ell}} \end{gather} \]

Note: From Geometry, the circumference length

\[ \begin{gather} C=2\pi R \end{gather} \]

length of semicircle

\[ \begin{gather} \ell=\frac{C}{2}=\frac{\cancel{2}\pi R}{\cancel{2}}=\pi R \end{gather} \]

The mass of the semicircular segment will be

\[ \begin{gather} m_{s}=\lambda \ell\\[5pt] m_{s}=\lambda \pi R\\[5pt] m_{s}=4\pi \lambda \end{gather} \]

For the straight segment of the wire, which is also homogeneous and of a constant cross-section, we have that the x coordinate is in the middle of the wire, as its length is 8 cm (Figure 2)

\[ \begin{gather} x_{w}=\frac{L}{2}\\[5pt] x_{w}=\frac{8}{2}\\[5pt] x_{w}=4\;\text{cm} \end{gather} \]

Figure 2

Since the semicircular segment has a radius of 4 cm, it joins the straight segment at the point where the y coordinate has this value.

\[ \begin{gather} y_{w}=4\;\text{cm} \end{gather} \]

Thus the coordinate of the center of mass of the straight segment is

\[ \begin{gather} (x_{w},y_{w})=(4,\;4) \end{gather} \]

Since the material from which the straight segment is made is the same as the semicircle, its linear density is calculated using the same expression, and the mass of the straight segment will

\[ \begin{gather} m_{w}=\lambda L\\[5pt] m_{w}=8\lambda \end{gather} \]

The system behaves as if all the mass of the semicircular segment of the wire were at point \( (x_{s},y_{s})=\left(-{\frac{8}{\pi }},\;0\right) \) and all the mass of the straight segment were at point \( (x_{w},y_{w})=(4,\;4) \), the coordinates of the center of mass of the whole wire are at a point \( (x_{cm},y_{cm}) \) located on the line connecting the two previous points (Figure 3).

Figure 3

The coordinates of the center of mass will be given by the expressions

\[ \begin{gather} \bbox[#99CCFF,10px] {x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}} \end{gather} \]

\[ \begin{gather} \bbox[#99CCFF,10px] {y_{cm}=\frac{m_{1}y_{1}+m_{2}y_{2}}{m_{1}+m_{2}}} \end{gather} \]

Substituting the data obtained above for x_cm

\[ \begin{gather} x_{cm}=\frac{m_{s}x_{s}+m_{w}x_{w}}{m_{s}+m_{w}}\\[5pt] x_{cm}=\frac{4\cancel{\pi}\lambda \times\left(-{\frac{8}{\cancel{\pi}}}\right)+8\lambda \times4}{4\pi \lambda+8\lambda }\\[5pt] x_{cm}=\frac{\cancel{4\lambda}}{\cancel{4\lambda}}\frac{(-8+8)}{(\pi +2)}\\[5pt] x_{cm}=\frac{0}{\pi+2}\\[5pt] x_{cm}=0 \end{gather} \]

For y_cm

\[ \begin{gather} y_{cm}=\frac{m_{s}y_{s}+m_{w}y_{w}}{m_{s}+m_{w}}\\[5pt] y_{cm}=\frac{4\pi\lambda \times 0+8\lambda \times 4}{4\pi \lambda +8\lambda}\\[5pt] y_{cm}=\frac{0+32\lambda }{4\pi \lambda +8\lambda}\\[5pt] y_{cm}=\frac{32\lambda }{4\pi \lambda +8\lambda}\\[5pt] y_{cm}=\frac{\cancelto{8}{32\lambda} }{\cancel{4\lambda} (\pi+2)}\\[5pt] y_{cm}=\frac{8}{\pi +2} \end{gather} \]

assuming π = 3.14

\[ \begin{gather} y_{cm}=\frac{8}{3.14+2}\\[5pt] y_{cm}=\frac{8}{5.14}\\[5pt] y_{cm}=1.56\;\text{cm} \end{gather} \]

The coordinate of the center of mass will be

\[ \begin{gather} \bbox[#FFCCCC,10px] {(x_{cm},y_{cm})=(0,\;1.56)} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .