Question

Determine the location of the center of mass of an L-shaped object whose thin vertical and horizontal members have the same length Land the same mass M. Use the formal definition to find the x and ycoordinates, and check your result by doing the calculation with respect to two different origins, one in the lower left corner at the intersection of the horizontal and vertical members and at the right end of the horizontal member.

Short answer

• the coordinate of center of mass is $\left(L/4,L/4,0\right)$
• the coordinate of center of mass is$\left(-L/4,L/4,0\right)$

Step-by-step solution

Identification of given data

• The total mass is M
• The Length of each member (horizontal and vertical member) is L

Concept of the center of mass of the system

The center of mass of the system is calculated by considering the average positions of all the objects acting in the system.

Determination of x and y coordinates for the first origin

For the first origin (one in the lower left corner at the intersection of the horizontal and vertical members), the diagram will be the following,

The center of mass of the system,

$$\begin{gathered} \mathit{M}{\mathit{x}}_{\mathbf{C}\mathbf{M}}\mathbf{=}{\mathit{m}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{m}}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}} \\ \mathit{M}{\mathit{y}}_{\mathbf{C}\mathbf{M}}\mathbf{=}{\mathit{m}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{1}}\mathbf{+}{\mathit{m}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{2}} \end{gathered}$$

Here,

$$\begin{gathered} m_1=m_2=M \\ {M}_{total}=2M \\ \left(x_1,y_1\right)=\left(L/2,0\right) \\ \left(x_2,y_2\right)=\left(0,L/2\right) \end{gathered}$$

Substitute these values in Equation (1),

To find x coordinate.

$$\begin{gathered} 2M.x_{CM}=M.L/2+M.0 \\ 2M.x_{CM}=M.L/2 \\ {x}_{CM}=L/4 \end{gathered}$$

To find y coordinate,

$$\begin{gathered} M{y}_{CM}=m_1{y}_1+m_2{y}_2 \\ 2M.y_{CM}=M.0+M.L/2 \\ 2M.y_{CM}=M.L/2 \\ {y}_{CM}=L/4 \end{gathered}$$

Hence, the coordinate of center of mass is$\left(L/4,L/4,0\right)$

Determination of x and y coordinates for the second origin

For the second origin (at the right end of the horizontal member), the diagram will be the following,

$$\begin{gathered} \mathit{M}{\mathit{x}}_{\mathbf{CM}}\mathbf{=}{\mathit{m}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{m}}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}} \\ \mathit{M}{\mathit{y}}_{\mathbf{CM}}\mathbf{=}{\mathit{m}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{1}}\mathbf{+}{\mathit{m}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{2}} \end{gathered}$$

Here,

$$\begin{gathered} m_1=m_2=M \\ {M}_{total}=2M \\ \left(x_1,y_1\right)=\left(-L/2,0\right) \\ \left(x_2,y_2\right)=\left(0,L/2\right) \end{gathered}$$

Substitute these values in Equation (1),

To find x coordinate.

$$\begin{gathered} 2M.x_{CM}=M.\left(-L/2\right)+M.0 \\ 2M.x_{CM}=-M.L/2 \\ {x}_{CM}=-L/4 \end{gathered}$$

To find y coordinate,

$$\begin{gathered} M{y}_{CM}=m_1{y}_1+m_2{y}_2 \\ 2M.y_{CM}=M.0+M.L/2 \\ 2M.y_{CM}=M.L/2 \\ {y}_{CM}=L/4 \end{gathered}$$

Hence, the coordinate of center of mass is$\left(-L/4,L/4,0\right)$