Question

State and prove principle of parallel axes.

Answer 1

moment of inertia of system about an axis parallel to axis passes through centre of mass is summation of moment of inertia about C.M axis and product of mass and square of separation between axis.

see figure for arrangement,
Let $M=m_1+m_2+m_3$
so, we have to prove $I_{PQ}=I_{C.M}+Mx^2$

actually, $I_{PQ}=m_1(r_1+x)^2+m_2(r_2-x)^2+m_3(r_3+x)^2$
= $(m_1+m_2+m_3)x^2+(m_1r_1^2+m_2r_2^2+m_3r_3^2)+2x(m_1r_1-m_2r_2+m_3r_3)$

=$Mx^2+I_{C.M}+2x(m_1r_1-m_2r_2+m_3r_3)$.....(1)

now, Torque about C.M axis,
$\tau=m_1g\times r_1(\textbf{clockwise})+m_2g\times r_2(\textbf{anticlockwise})+m_3g\times r_3(\textbf{clockwise})$
because here no angular motion. so, $\tau=0$
so, $m_1r_1-m_2r_2+m_3r_3=0$...(2)

put equation (2) in equation (1),
$I_{PQ}=Mx^2+I_{C.M}$
hence proved