Question

#198. Moment of inertia about center of a segmented bar A bar of width is formed of three uniform segments with lengths and areal densities given by: Matlab/Mathematica input: L1 = 6 rho1 = 1 L2 = 6 rho2 = 8 L3 = 4 rho3 = 5 What is the moment of inertia of the bar about the center of mass ?

Answer 1

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The complete question is shown on the first uploaded image

Answer:

The moment of inertia of the bar about the center of mass is

$I_r = 1888.80 \ kg m^2$

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

          $m_1 = \rho_1 l_1 w = 1 * 6 * 2 = 12$

          $m_1 = \rho_2 l_2 w = 8 * 6 * 2 = 96$

          $m_1 = \rho_2 l_2 w = 5 * 5 * 2 = 50$

The distance from the origin to the center of the segments i.e the center of masses for the individual segments

   $x_2 = \frac{6}{2} + 6 = 9 m$

   $x_3 = \frac{4}{2} + 12 = 14 m$

           

The  resultant center of mass is mathematically evaluated as

              $x_r = \frac{m_1 * x_1 + m_2 *x_2 + m_3 *x_3}{m_1 + m_2 + m_3}$    

        $= \frac{12 * 3 + 96 *9 + 50 *14}{12+ 96 + 50}$

                      $x_r = 10.13m$        

The moment of Inertia of each segment of the bar is mathematically evaluated

             $I_1 =\frac{m_1}{12}(l_1^2 + w^2)$ =    $\frac{12}{12}(1^2 + 2^2)$        

                   $I_1 = 4 \ kgm^2$

             $I_2 =\frac{m_2}{12}(l_2^2 + w^2)$  =    $\frac{96}{12}(6^2 + 2^2)$

                 $I_2 = 320 \ kgm^2$

             $I_3 =\frac{m_3}{12}(l_3^2 + w^2)$  =    $\frac{50}{12}(4^2 + 2^2)$        

                   $I_2 = 83.334 \ kgm^2$        

According to parallel axis theorem the moment of inertia about the center ($x_r$) is mathematically evaluated as

           $I_r = (I_1 + m_1 r_1^2) + (I_2 + m_2 r_2^2) +(I_3 + m_3 r_3^2)$

   $I_r = (I_1 + m_1 |x_r - x_1|^2) + (I_2 + m_2 |x_r - x_2|^2) +(I_3 + m_3 |x_r - x_3|^2)$

   $I_r = (4 + 12 |10.13 - 3|^2) + (320 + 96 |10.13 - 9|^2) +(83.334 + 50 |10.13 - 14|^2)$        

      $I_r = 1888.80 \ kg m^2$