Questions: Four objects are held in position at the corners of a rectangle by light rods as shown in the figure below. (The mass values are given in the table.) m1(kg) m2(kg) m3(kg) mA(kg) 2,70 2,30 3,70 2,10 (a) Find the moment of inertia of the system about the x-axis. kg · m^2 (b) Find the moment of inertia of the system about the y-axis. kg · m^2 (6) Find the moment of inertia of the system about an axis through O and perpendicular to the page. kg · m^2

Four objects are held in position at the corners of a rectangle by light rods as shown in the figure below. (The mass values are given in the table.)
m1(kg) m2(kg) m3(kg) mA(kg)
2,70 2,30 3,70 2,10

(a) Find the moment of inertia of the system about the x-axis.
kg · m^2
(b) Find the moment of inertia of the system about the y-axis.
kg · m^2
(6) Find the moment of inertia of the system about an axis through O and perpendicular to the page.
kg · m^2

Transcript text: Four objects are held in position at the corners of a rectangle by light rods as shown in the figure below. (The mass values are given in the table.) \begin{tabular}{|c|c|c|c|} \hline$m_{1}(kg)$ & $m_{2}(kg)$ & $m_{3}(kg)$ & $m_{A}(kg)$ \\ \hline 2,70 & 2,30 & 3,70 & 2,10 \\ \hline \end{tabular} (a) Find the moment of inertia of the system about the x-axis. $\square$ $\mathrm{kg} \cdot \mathrm{m}^{2}$ (b) Find the moment of inertia of the system about the $y$-axis. $\square$ $\mathrm{kg} \cdot \mathrm{m}^{2}$ (6) Find the moment of inertia of the system about an axis through $O$ and perpendicular to the page. $\square$ $\mathrm{kg} \cdot m^{2}$

Solution

Solution Steps

Step 1: Moment of inertia about the x-axis

The moment of inertia about the x-axis is given by the sum of the products of each mass and the square of its distance from the x-axis. In this case, masses $m_1$ and $m_2$ are located at a distance of 0 m from the x-axis, while $m_3$ and $m_4$ are located at a distance of 4.00 m from the x-axis. Therefore, the moment of inertia about the x-axis is:

$I_x = m_1(0)^2 + m_2(0)^2 + m_3(4.00)^2 + m_4(4.00)^2$ $I_x = (3.70)(4.00)^2 + (2.10)(4.00)^2$ $I_x = 59.2 + 33.6$ $I_x = 92.8 \text{ kg} \cdot \text{m}^2$

Step 2: Moment of inertia about the y-axis

The moment of inertia about the y-axis is given by the sum of the products of each mass and the square of its distance from the y-axis. Masses $m_1$ and $m_3$ are located at a distance of 0 m from the y-axis, and $m_2$ and $m_4$ are located at a distance of 6.00 m from the y-axis. Thus, the moment of inertia about the y-axis is:

$I_y = m_1(0)^2 + m_3(0)^2 + m_2(6.00)^2 + m_4(6.00)^2$ $I_y = (2.30)(6.00)^2 + (2.10)(6.00)^2$ $I_y = 82.8 + 75.6$ $I_y = 158.4 \text{ kg} \cdot \text{m}^2$

Step 3: Moment of inertia about an axis through O and perpendicular to the page

The moment of inertia about an axis through O and perpendicular to the page can be found by summing the moments of inertia about the x and y axes. This is because the x and y axes are perpendicular and intersect at O. This is also known as the perpendicular axis theorem. $I_z = I_x + I_y$ $I_z = 92.8 + 158.4$ $I_z = 251.2 \text{ kg} \cdot \text{m}^2$