Chapter 8, Problem 37P

Four objects are held in position at the corners of a rectangle by light rods as shown in Figure P8.37. Find the moment of inertia of the system about (a) the x-axis, (b) they-axis, and (c) an axis through O and perpendicular to the page.

To determine

The moment of inertia of the system about $x$-axis.

Answer to Problem 37P

The moment of inertia of the system about $x$-axis is $99.0\text{kg}\cdot {\text{m}}^{\text{2}}$.

Explanation of Solution

In general, the moment of inertia of the objects is defined as $I=m_i{r}_i^2$ and in the same way, this problem is solved.

Given info The masses $m_1$, $m_2$, $m_3$, and $m_4$ of the objects  are $2.00\text{kg}$, $3.00\text{kg}$, $4.00\text{kg}$, and $2.00\text{kg}$ respectively and the distance of these objects from the $x$-axis is $3.00\text{m}$.

The formula for the moment of inertia of these objects about $x$-axis is,

$I_x=m_1{y}^2+m_2{y}^2+m_3{y}^2+m_4{y}^2$

• $m_1$, $m_2$, $m_3$, and $m_4$ are masses of the objects.
• $y$ is the distance of the object from $x$-axis.

Substitute $2.00\text{kg}$ for $m_1$, $3.00\text{kg}$ for $m_2$, $4.00\text{kg}$ for $m_3$, $2.00\text{kg}$ for $m_4$, and $3.00\text{m}$ for $y$ to find $I_x$.

$\begin{array}{c}{I}_x=\left(2.00\text{kg}+3.00\text{kg}+4.00\text{kg}+2.00\text{kg}\right){\left(3.00\text{m}\right)}^2\\ =99.0\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Thus, the moment of inertia of the system about $x$-axis is $99.0\text{kg}\cdot {\text{m}}^{\text{2}}$.

Conclusion:

Therefore, the moment of inertia of the system about $x$-axis is $99.0\text{kg}\cdot {\text{m}}^{\text{2}}$.

To determine

The moment of inertia of the system about $y$-axis.

Answer to Problem 37P

The moment of inertia of the system about $y$-axis is $44.0\text{kg}\cdot {\text{m}}^{\text{2}}$.

Explanation of Solution

Given info: The masses $m_1$, $m_2$, $m_3$, and $m_4$ of the objects  are $2.00\text{kg}$, $3.00\text{kg}$, $4.00\text{kg}$, and $2.00\text{kg}$ respectively and the distance of these objects from the $y$-axis is $2.00\text{m}$.

The formula for the moment of inertia of these objects about $y$-axis is,

$I_y=m_1{x}^2+m_2{x}^2+m_3{x}^2+m_4{x}^2$

• $m_1$, $m_2$, $m_3$, and $m_4$ are masses of the objects.
• $x$ is the distance of the object from $y$-axis.

Substitute $2.00\text{kg}$ for $m_1$, $3.00\text{kg}$ for $m_2$, $4.00\text{kg}$ for $m_3$, $2.00\text{kg}$ for $m_4$, and $2.00\text{m}$ for $x$ to find $I_y$.

$\begin{array}{c}{I}_y=\left(2.00\text{kg}+3.00\text{kg}+4.00\text{kg}+2.00\text{kg}\right){\left(2.00\text{m}\right)}^2\\ =44.0\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Thus, the moment of inertia of the system about $y$-axis is $44.0\text{kg}\cdot {\text{m}}^{\text{2}}$.

Conclusion:

Therefore, the moment of inertia of the system about $x$-axis is $44.0\text{kg}\cdot {\text{m}}^{\text{2}}$.

To determine

The moment of inertia of the system about an axis through $O$ -and perpendicular to the page.

Answer to Problem 37P

The moment of inertia of the system about an axis through $O$ -and perpendicular to the page is $143\text{kg}\cdot {\text{m}}^{\text{2}}$.

Explanation of Solution

The distance of each object from the point $O$ is $r=\sqrt{x^2+y^2}$.

Given info: The masses $m_1$, $m_2$, $m_3$, and $m_4$ of the objects  are $2.00\text{kg}$, $3.00\text{kg}$, $4.00\text{kg}$, and $2.00\text{kg}$ respectively and the distances of these objects from $x$ and $y$-axes are $3.00\text{m}$ and $2.00\text{m}$.

The formula for the moment of inertia of these objects about $O$ is,

$I_y=\left(m_1+m_2+m_3+m_4\right)\sqrt{x^2+y^2}$

• $m_1$, $m_2$, $m_3$, and $m_4$ are masses of the objects.
• $x$ is the distance of the object from $y$-axis.
• $y$ is the distance of the object from $x$-axis

Substitute $2.00\text{kg}$ for $m_1$, $3.00\text{kg}$ for $m_2$, $4.00\text{kg}$ for $m_3$, $2.00\text{kg}$ for $m_4$, $2.00\text{m}$ for $x$, $3.00\text{m}$ for $y$ to find $I_O$.

$\begin{array}{c}{I}_y=\left(2.00\text{kg}+3.00\text{kg}+4.00\text{kg}+2.00\text{kg}\right)\sqrt{{\left(3.00\text{m}\right)}^2+{\left(2.00\text{m}\right)}^2}\\ =143\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Thus, the moment of inertia of the system about an axis through $O$ -and perpendicular to the page is $143\text{kg}\cdot {\text{m}}^{\text{2}}$.

Conclusion:

Therefore, the moment of inertia of the system about an axis through $O$ -and perpendicular to the page is $143\text{kg}\cdot {\text{m}}^{\text{2}}$.