Answer:
L= 1.468 m
Explanation:
The moment of inertia of the rod about its center is (1/12)m_RL^2
The moment of inertia of each of the two bodies about the described axes is m_B(L/2)2
Hence, the moment of inertia of three body system is (1/12)m_RL^2+ 2×m_B(L/2)^2 which is given to be equal to I_T
=> L^2[(1/12)m_R+m_B/2] = I_T
=> L2 = IT/(mR/12+ mB/2)
=> L = sqrt( 12I_T/(m_R+6m_B))
now putting the value of m_R= 3.47 kg
m_B= 0.263 kg
I_T = 0.97 kg/m^2
[tex]L= \sqrt{\frac{12I_T}{m_R+6m_B} }[/tex]
[tex]L= \sqrt{\frac{12\times0.907}{3.47+6\times0.263} }[/tex]
L= 1.468 m is the length of the rod be so that the moment of inertia of the three-body system
