Chapter 9, Problem 42P

To determine

The moment of inertia of a system of four particles, about an axis passing through its center of mass and parallel to the z axis.

Answer to Problem 42P

The moment of inertia of a system of four particles, one each at the corners of a square of edge $2.0\text{ m}$ about the axis passing through its center of mass and parallel to the z axis is found to be $28\text{ kg}\cdot {\text{m}}^2$ . The value calculated using the parallel axis theorem agrees with the value obtained by direct computation

Explanation of Solution

Given info:

The masses of the balls:

  $\begin{array}{l}{m}_1=m_3=3.0\text{ kg}\\ m_2=m_4=4.0\text{ kg}\end{array}$

The length of the edge of the cube

  $l=2.0\text{ m}$

The moment of inertia of the system about the z axis

  $I_z=56\text{ kg}\cdot {\text{m}}^2$

Formula used:

The moment of inertia of a particle of mass $m$ about an axis is given by,

  $I=m{r}^2$  .............(1)

Where, $r$ is the perpendicular distance between the particle and the axis of rotation.

The moment of inertia of a system of particles about an axis is equal to the sum of moments of inertia of the individual particles about the axis of rotation.

  $I_{\text{sys}}={\displaystyle \sum _{i=1}^{i=n}{I}_i}$  .............(2)

The coordinates $\left(x,y\right)$ of the center of mass of a system of particles are given by,

  $x=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$  .............(3)

  $y=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}$  .............(4)

Here, $\left(x_1,y_1\right),\left(x_2,y_2\right),\left(x_3,y_3\right)\text{ and }\left(x_4,y_4\right)$ are the coordinates of the particles of masses $m_1,m_2,m_3\text{ and }{m}_4$ respectively.

According to parallel axis theorem,

  $I_z=I_{\text{CG}}+M{r}^2$  .............(5)

Here, $I_{\text{CG}}$ is the moment of inertia of the system of particles about an axis passing through the center of gravity and parallel to the z axis, $r$ is the distance between the axis through the center of gravity and the z axis and $M$ is the sum of the all the masses comprising the system.

  $M=m_1+m_2+m_3+m_4$  .............(6)

Calculation:

Four particles of masses $m_1,m_2,m_3\text{ and }{m}_4$ are placed at the four corners A,B,C,D of a square of edge $l$ . The origin of the coordinate system lies at B, with the x axis along BC and the y axis along BA. The center of mass of the system of particles lies at O. This is shown in figure1.

  

  Figure 1

The mass $m_1$ is along the y axis, hence its coordinates $\left(x_1,y_1\right)$ are $\left(0,2.0\text{ m}\right)$ . The mass $m_2$ is at the origin, hence its coordinates $\left(x_2,y_2\right)$ are $\left(0,0\right)$ . The mass $m_3$ lies on the x axis, hence its coordinates $\left(x_3,y_3\right)$ are $\left(2.0\text{ m},0\right)$ . The mass $m_4$ at D has its coordinates $\left(x_4,y_4\right)$ as $\left(2.0\text{ m},2.0\text{ m}\right)$ .

Use equations (3) and (4) to determine the coordinates $\left(x,y\right)$ of the center of mass.

  $\begin{array}{c}x=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}\\ =\frac{\left(3.0\text{ kg}\right)\left(0\right)+\left(4.0\text{ kg}\right)\left(0\right)+\left(3.0\text{ kg}\right)\left(2.0\text{ m}\right)+\left(4.0\text{ kg}\right)\left(2.0\text{ m}\right)}{\left(3.0\text{ kg}\right)+\left(4.0\text{ kg}\right)+\left(3.0\text{ kg}\right)+\left(4.0\text{ kg}\right)}\\ =1.0\text{ m}\end{array}$

  $\begin{array}{c}y=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}\\ =\frac{\left(3.0\text{ kg}\right)\left(2.0\text{ m}\right)+\left(4.0\text{ kg}\right)\left(0\right)+\left(3.0\text{ kg}\right)\left(0\right)+\left(4.0\text{ kg}\right)\left(2.0\text{ m}\right)}{\left(3.0\text{ kg}\right)+\left(4.0\text{ kg}\right)+\left(3.0\text{ kg}\right)+\left(4.0\text{ kg}\right)}\\ =1.0\text{ m}\end{array}$

The center of mass of the system of 4 particles lies at $\left(1.0\text{ m},1.0\text{ m}\right)$ from the origin.

Determine the distance $r$ between the center of mass and the z axis using the expression,

  $r^2=x^2+y^2$  .............(7)

Substitute the values of $\left(x,y\right)$ in equation (7).

  $\begin{array}{c}{r}^2=x^2+y^2\\ ={\left(1.0\text{ m}\right)}^2+{\left(1.0\text{ m}\right)}^2\\ =2.0{\text{ m}}^2\end{array}$

Determine the value of the mass $M$ by substituting the values of the variables in equation (6).

  $\begin{array}{c}M=m_1+m_2+m_3+m_4\\ =\left(3.0\text{ kg}\right)+\left(4.0\text{ kg}\right)+\left(3.0\text{ kg}\right)+\left(4.0\text{ kg}\right)\\ =14\text{ kg}\end{array}$

Rewrite equation (5) for $I_{\text{CG}}$ .

  $I_{\text{CG}}=I_z-M{r}^2$  .............(8)

Substitute the values of the variables in equation (8) and calculate the moment of inertia of the system of particles about the axis passing through the center of gravity and parallel to the z axis.

  $\begin{array}{c}{I}_{\text{CG}}=I_z-M{r}^2\\ =\left(56\text{ kg}\cdot {\text{m}}^2\right)-\left(14\text{ kg}\right)\left(2.0{\text{ m}}^2\right)\\ =28\text{ kg}\cdot {\text{m}}^2\end{array}$

  

  Figure 2

The particles of masses $m_1,m_2,m_3\text{ and }{m}_4$ are all at a distance $r$ from the center of gravity O. The moment of inertia of the system of particles about the axis passing through the center of gravity and parallel to the z axis is obtained by using equations (1) and (2) as follows:

  $\begin{array}{c}{I}_{\text{CG}}=I_1+I_2+I_3+I_4\\ =m_1{r}^2+m_2{r}^2+m_3{r}^2+m_4{r}^2\\ =\left(m_1+m_2+m_3+m_4\right)r^2\end{array}$  .............(9)

Substitute the values of the given quantities in equation (9) and determine the value of $I_{\text{CG}}$ .

  $\begin{array}{c}{I}_{\text{CG}}=\left(m_1+m_2+m_3+m_4\right)r^2\\ =\left[\left(3.0\text{ kg}\right)+\left(4.0\text{ kg}\right)+\left(3.0\text{ kg}\right)+\left(4.0\text{ kg}\right)\right]\left(2.0\text{ m}\right)\\ =28\text{ kg}\cdot {\text{m}}^2\end{array}$

The calculated value agrees with the value calculated using the parallel axis theorem.

Conclusion:

Thus, the moment of inertia of a system of four particles, one each at the corners of a square of edge $2.0\text{ m}$ about the axis passing through its center of mass and parallel to the z axis is found to be $28\text{ kg}\cdot {\text{m}}^2$ .The value calculated using the parallel axis theorem agrees with the value obtained by direct computation