Question

Prove the “parallel axis theorem”: The moment of inertia of a body about a given axis is $\mathbf{l}\mathbf{=}{\mathbf{l}}_{\mathbf{m}}\mathbf{+}{\mathbf{Md}}^{\mathbf{2}}$, where M is the mass of the body,${\mathrm{l}}_{\mathrm{m}}$is the moment of inertia of the body about an axis through the center of mass and parallel to the given axis, and dis the distance between the two axes.

Short answer

The Parallel Axis Theorem $\mathrm{l}={\mathrm{l}}_{\mathrm{m}}+{\mathrm{Md}}^2$is proved using the integral calculation.

Step-by-step solution

Parallel Axis Theorem

The sum of the moment of inertia of the body about the axis passing through its center and the product of the mass of the body times the square of the distance between the two axes equals the moment of inertia of the body about the axis passing through its center.

Plotting graph for “Parallel Axis Theorem”

The graph for the parallel axis theorem is plotted and is referred to for finding the integrals.

Deducing the first integral

The definition of the moment of inertia about the bottom axis is –

$\mathrm{l}={\int }_{\mathrm{v}}\mathrm{p}({\mathrm{r}}^2-{\mathrm{s}}^2)\mathrm{dV}$

The first integral is recognized as the total mass of the body and the second one as the moment of inertia about the axis going through the center of mass:

$$\begin{gathered} \mathrm{l}={\int }_{\mathrm{v}}\mathrm{p}({\mathrm{d}}^2+2\mathrm{d}.\mathrm{r}\text{'}+{\mathrm{r}}^2-{\mathrm{s}}^2)\mathrm{dV} \\ ={\mathrm{d}}^2{\int }_{\mathrm{v}}\mathrm{p}\mathrm{dV}+{\int }_{\mathrm{v}}\mathrm{p}({\mathrm{r}}^2-{\mathrm{s}}^2)\mathrm{dV}+2{\int }_{\mathrm{v}}\mathrm{p}\mathrm{d}.\mathrm{r}\text{'}\mathrm{dV} \end{gathered}$$

Deducing the final integral

The final integral is

$$\begin{gathered} 2{\int }_{\mathrm{v}}\mathrm{pd}.\mathrm{r}\text{'}\mathrm{dV}=2{\mathrm{d}}_{\mathrm{x}}{\int }_{\mathrm{v}}\mathrm{x}\text{'}\mathrm{p}\mathrm{dV}+2{\mathrm{d}}_{\mathrm{y}}{\int }_{\mathrm{v}}\mathrm{y}\text{'}\mathrm{pdV}+2{\mathrm{d}}_{\mathrm{z}}{\int }_{\mathrm{v}}\mathrm{z}\text{'}\mathrm{pdV} \\ =2\mathrm{d}.{\int }_{\mathrm{v}}\mathrm{pr}\text{'}\mathrm{dV} \end{gathered}$$

This integral is zero since it is just the position of the center of mass with respect to the center of mass.

Thus, the parallel axis theorem $\mathrm{l}={\mathrm{l}}_{\mathrm{m}}+{\mathrm{Md}}^2$is proven.