Answer:
[tex]153.9 kgm^2[/tex]
Explanation:
We are given that
[tex]m_r=6.85 kg[/tex]
Length,L=5.76 m
[tex]m_s=34.25 kg[/tex]
Radius,R=1.44 m
We have to find the moment of inertia of the object about an axis at the center of mass of the object.
Moment of inertia of the object about an axis at the center of mass of the object,I=[tex]I_r+I_s=\frac{1}{12}m_rL^2+m_r(\frac{L}{2}+\frac{R}{2})^2+\frac{13}{20}m_sR^2[/tex]
Substitute the values
[tex]I=\frac{1}{12}(6.85)(5.76)^2+6.85(\frac{5.76}{2}+\frac{1.44}{2})^2+\frac{13}{20}(34.25)(1.44)^2[/tex]
[tex]I=153.9 kgm^2[/tex]
