Answer in Mechanics | Relativity for Valentine Agun #108070
March 12th, 2023
Solution
The moment of inertia of a mass m placed r meters from the axis of rotation is
“I=mr^2.”
This is is an additive quantity. Therefore, the moment of inertia around x-axis will be
“I_x=m_1r_1^2+m_2r^2_2+m_3r^2_3+m_4r^2_4,”
where
“I_x=m_1r_1^2+m_2r^2_2+m_3r^2_3+m_4r^2_4,\r_1=r_2=r_3=r_4=3text{ m}.\nI_x=(3.1+1.7+4.5+2.3)3^2=104text{ kg}cdottext{m}^2.”
Consider rotation around y-axis:
“I_y=m_1r_1^2+m_2r^2_2+m_3r^2_3+m_4r^2_4,\r_1=r_2=r_3=r_4=2text{ m}.\nI_y=(3.1+1.7+4.5+2.3)2^2=46.4text{ kg}cdottext{m}^2.”
Now consider rotation around z-axis:
“I_z=m_1r_1^2+m_2r^2_2+m_3r^2_3+m_4r^2_4,\r_1=r_2=r_3=r_4=sqrt{13}text{ m}.\nI_z=(3.1+1.7+4.5+2.3)sqrt{13^2}=151text{ kg}cdottext{m}^2.”
Conclusion: the further, the higher.
