In Fig. 9-37, three uniform thin rods, each of length L=22, form an inverted U. The vertical rods each have a mass of $14g$; the horizontal rod has a mass of$14g$. What is (a) the xcoordinate and (b) the ycoordinate of the system’s center of mass?

1. Length of the three uniform rods,$L=22m$
2. The rods form an inverted U.
3. Mass of the vertical rods,$m=14kg$
4. Mass of the horizontal rod,role="math" localid="1663215003780" $M=42kg$

The center of mass of a body is defined as the relative position of the mass of the system at which all the distributed mass sums to zero or at which the forces acting on the system sums up to zero value. It can also be defined as the average mass of all the particles relative to their position in the system considering a common measuring point.

Formulae:

The x-coordinate of the center of mass, $x_{cm}=\frac{{\sum }_i^n{M}_i{x}_i}{{\sum }_i^n{M}_i}$ (i)

The y-coordinate of the center of mass, $y_{cm}=\frac{{\sum }_i^n{M}_i{y}_i}{{\sum }_i^n{M}_i}$ (ii)

From the given figure and coordinate system, the position of the center of mass of the left vertical leg can be given as:$(x_1,y_1)=\left(0,-L/2\right)$

Similarly, the position of the center of mass of the horizontal leg can be given as:$(x_2,y_2)=\left(L/2,0\right)$

Similarly, the position of the center of mass of the right vertical leg can be given as:$(x_3,y_3)=\left(+L,-L/2\right)$

Now, the x-coordinate of the center of mass of the system can be given using equation (i) as follows:

$$\begin{gathered} x_{cm}=\frac{m{x}_1+M{x}_2+m{x}_3}{m+M+m} \\ =\frac{\left(14\text{kg}\right)\left(0\right)+\left(\text{42kg}\right)\left(L/2\right)+\left(14\text{kg}\right)\left(L\right)}{\left(14+42+14\right)\text{kg}} \\ =\frac{L\left(\text{35kg}\right)}{70\text{kg}} \\ =\frac{L}{2} \\ =\frac{22\text{cm}}{2} \\ =11\text{cm} \end{gathered}$$

Hence, the value of the x-coordinate is$11\text{cm}$.

Now, the y-coordinate of the center of mass of the system can be given using equation (ii) as follows:

$$\begin{gathered} y_{cm}=\frac{m{y}_1+M{y}_2+m{y}_3}{m+M+m} \\ =\frac{\left(14\text{kg}\right)\left(-L/2\right)+\left(\text{42kg}\right)\left(0\right)+\left(14\text{kg}\right)\left(-L/2\right)}{\left(14+42+14\right)\text{kg}} \\ =\frac{-L\left(\text{14kg}\right)}{70\text{kg}} \\ =\frac{-L}{5} \\ =\frac{-22\text{cm}}{5} \\ =-4.4\text{cm} \end{gathered}$$

Hence, the value of the x-coordinate is $-4.4\text{cm}$.