- Prove the theorem of perpendicular axes.
- Prove the theorem of parallel axes.
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- According to the theorem of perpendicular axes the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its plane and intersecting each other at the point where the perpendicular axis passes through it. let us consider a physical body with center O and a point mass m,in the x-yplane at (x, y) is shown in the following figure:
Moment of inertia about x-axis, $I_x = mx^2$
Moment of inertia about y-axis, $I_y = my^2$
Moment of inertia about z-axis, $I_z = m(x^2 + y^2)^{1/2}$
$I_x + I_y = mx^2 + my^2$
$= m(x^2 + y^2)$
$= m [(x^2 + y^2)^{1/2}]^{1/2}$
$I_x + I_y = I_z$
Thus, the theorem is verified.
- According to the theorem of parallel axes the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its center of mass and the product of its mass and the square of the distance between the two parallel axes.
Suppose a rigid body is made up of n number of particles, having masses $m_1, m_2, m_3, … ,m_n$, at perpendicular distances $r_1, r_2, r_3, … , r_n$ respectively from the center of mass O of the rigid body.
The moment of inertia about axis RS passing through the point O,
$\text{I}_{\text{RS}}=\sum\limits^\text{n}_{\text{i}=1}=\text{m}_{\text{i}}\text{r}_{\text{i}}^2$
The perpendicular distance of mass $m_i$ from the axis $QP = a+ r_i$
$\text{I}_{\text{QR}}=\sum\limits^\text{n}_{\text{i}=1}\text{m}_{\text{i}}(\text{a}+\text{r}_\text{i})^2$
$\text{I}_{\text{QR}}=\sum\limits^\text{n}_{\text{i}=1}\text{m}_{\text{i}}(\text{a}^2+\text{r}_\text{i}^2+2\text{ar}_{\text{i}})$
$\text{I}_{\text{QR}}=\sum\limits^\text{n}_{\text{i}=1}\text{m}_{\text{i}}\text{a}^2+\sum\limits^\text{n}_{\text{i}=1}\text{m}_{\text{i}}\text{r}_\text{i}^2+\sum\limits^\text{n}_{\text{i}=1}\text{m}_{\text{i}}2\text{ar}_{\text{i}}$
$\text{I}_{\text{QR}}=\text{I}_{\text{RS}}=\sum\limits^\text{n}_{\text{i}=1}\text{m}_{\text{i}}\text{a}^2+2\sum\limits^\text{n}_{\text{i}=1}\text{m}_{\text{i}}\text{ar}^2_{\text{i}}$
We know, the moment of inertia of all particles about the axis passing through the center of mass is zero.
$2\sum\limits^\text{n}_{\text{i}=1}\text{m}_{\text{i}}\text{ar}_{\text{i}}=0$
as $\text{a}\neq0$
Therefore, $\sum\text{m}_{\text{i}}\text{r}_{\text{i}}=0$
Also,
Therefore, $\sum\text{m}_{\text{i}}$
= M; M = Total mass of the rigid body.
Therefore, $IQP = IRS + Ma^2$
Therefore the theorem is verified.
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