Question

State and prove theorem of parallel axes about moment of inertia.

Answer 1

Theorem of parallel axes : The moment of inertia of a body about any axis is equal to the sums of its moment of inertia about a parallel axis passing through its center of mass and the product of its mass and the square of the perpendicular distance between the two parallel axes.
Consider a rigid of mass ${}^{\prime }{M}^{\prime }$ rotating about an axis passing through a point 'O' and perpendicular to the plane of the figure.
Let the the moment of inertia of the body about an axis passing through point 'O'. Take another parallel axis of rotation passing through the center of mass of the body.
Let ${}^{\prime }I{c}^{\prime }$ be the moment of inertia of the body about point 'C'.
Let the distance between the two parallel axes be $OC=h$.
$OP=randCP=r_0$
Take a small element of body of mass 'dm' situated at a point P. Join OP and CP, then
$I_0=\int O{P}^{2}dm=\int r^{2}dm$
$I_c=\int C{P}^{2}dm=\int r_0^{2}dm$
From point P draw a perpendicular to OC produced.
Let $CD=X$
From the figure ,
$O{P}^2=O{D}^2+P{D}^2$
$\therefore O{P}^2=(h+CD{)}^2+P{D}^2$
$=h^2+C{D}^2+2hCD+P{D}^2$
$\therefore O{P}^2=C{P}^2+h^2+2hCD$ $(\because C{D}^2+P{D}^2=C{P}^2)$
$\therefore r^2=r_0^2+h^2+2hx$
Multiplying the above equation with 'dm' on the both side and intergrating, we get
$\int r^{2}dm=\int r_0^{2}dm+\int h^{2}dm+\int 2hxdm$
$r^{2}dm=\int r_0^{2}dm+\int h^{2}dm+2h\int xdm$
$\int xdm=0$ as 'C' is the center of mass and algebraic sum of moments of all the particles about the center amass is always zero, for body in equilibrium.
$\therefore \int r^{2}dm=\int r_0^{2}dm+h^2\int dm+0$ ...(1)
But $\int dm=M=$ Mass of the body.
$\int r^{2}dm=I_0$ and $\int r_0^{2}dm=I_c$
Sustituting in equation (1), we get
$I_o=I_c+M{h}^2$
This proves the theorem of parallel axes about moment of inertia.