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) m4(kg) 2.70 1.70 3.50 2.50 (a) Find the moment of inertia of the system about the x-axis. 4.0 kg ⋅ m^2 (b) Find the moment of inertia of the system about the y-axis. 4.0 kg ⋅ m^2 (c) Find the moment of inertia of the system about an axis through O and perpendicular to the page. 4.0 kg ⋅ m^2

Solution Steps

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. The distances of $m_1$ and $m_2$ from the x-axis are 6.00 m, and the distances of $m_3$ and $m_4$ from the x-axis are 0 m. Therefore, the moment of inertia about the x-axis is:

$I_x = m_1(6.00)^2 + m_2(6.00)^2 + m_3(0)^2 + m_4(0)^2$ $I_x = (2.70)(36.00) + (1.70)(36.00) = 97.2 + 61.2 = 158.4$ kg⋅m²

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. The distances of $m_3$ and $m_4$ from the y-axis are 4.00 m, and the distances of $m_1$ and $m_2$ from the y-axis are 0 m. Therefore, the moment of inertia about the y-axis is:

$I_y = m_1(0)^2 + m_2(0)^2 + m_3(4.00)^2 + m_4(4.00)^2$ $I_y = (3.50)(16.00) + (2.50)(16.00) = 56.0 + 40.0 = 96.0$ kg⋅m²

The moment of inertia about an axis through O perpendicular to the page (let's call it the z-axis) is given by the sum of the products of each mass and the square of its distance from point O. The distances can be calculated using the Pythagorean theorem. For $m_1$, the distance is $\sqrt{(6.00)^2 + (0)^2} = 6.00$ m. For $m_2$, the distance is $\sqrt{(6.00)^2 + (4.00)^2} = \sqrt{36+16} = \sqrt{52}$ m. For $m_3$, the distance is 4.00 m and for $m_4$ the distance is $\sqrt{52}$ m. Therefore, the moment of inertia about the z-axis is:

$I_z = m_1(6.00)^2 + m_2(\sqrt{52})^2 + m_3(4.00)^2 + m_4(\sqrt{52})^2$ $I_z = 2.70(36) + 1.70(52) + 3.50(16) + 2.50(52) = 97.2+88.4+56+130 = 371.6$ kg⋅m²

Final Answer: