Complete Complete
The complete question is shown on the first uploaded image
Answer:
The moment of inertia of the bar about the center of mass is
[tex] I_r = 1888.80 \ kg m^2 [/tex]
Explanation:
The free body diagram is shown on the second uploaded image
From the diagram we see that is
The mass of each segment is
[tex] m_1 = \rho_1 l_1 w = 1 * 6 * 2 = 12 [/tex]
[tex] m_1 = \rho_2 l_2 w = 8 * 6 * 2 = 96 [/tex]
[tex] m_1 = \rho_2 l_2 w = 5 * 5 * 2 = 50 [/tex]
The distance from the origin to the center of the segments i.e the center of masses for the individual segments
[tex] x_2 = \frac{6}{2} + 6 = 9 m [/tex]
[tex] x_3 = \frac{4}{2} + 12 = 14 m [/tex]
The resultant center of mass is mathematically evaluated as
[tex] x_r = \frac{m_1 * x_1 + m_2 *x_2 + m_3 *x_3}{m_1 + m_2 + m_3} [/tex]
[tex] = \frac{12 * 3 + 96 *9 + 50 *14}{12+ 96 + 50} [/tex]
[tex] x_r = 10.13m [/tex]
The moment of Inertia of each segment of the bar is mathematically evaluated
[tex] I_1 =\frac{m_1}{12}(l_1^2 + w^2) [/tex] = [tex] \frac{12}{12}(1^2 + 2^2) [/tex]
[tex] I_1 = 4 \ kgm^2 [/tex]
[tex] I_2 =\frac{m_2}{12}(l_2^2 + w^2) [/tex] = [tex] \frac{96}{12}(6^2 + 2^2) [/tex]
[tex] I_2 = 320 \ kgm^2 [/tex]
[tex] I_3 =\frac{m_3}{12}(l_3^2 + w^2) [/tex] = [tex] \frac{50}{12}(4^2 + 2^2) [/tex]
[tex] I_2 = 83.334 \ kgm^2 [/tex]
According to parallel axis theorem the moment of inertia about the center ([tex]x_r[/tex]) is mathematically evaluated as
[tex] I_r = (I_1 + m_1 r_1^2) + (I_2 + m_2 r_2^2) +(I_3 + m_3 r_3^2) [/tex]
[tex] I_r = (I_1 + m_1 |x_r - x_1|^2) + (I_2 + m_2 |x_r - x_2|^2) +(I_3 + m_3 |x_r - x_3|^2) [/tex]
[tex] I_r = (4 + 12 |10.13 - 3|^2) + (320 + 96 |10.13 - 9|^2) +(83.334 + 50 |10.13 - 14|^2) [/tex]
[tex] I_r = 1888.80 \ kg m^2 [/tex]
