Moments of Inertia

The moment of inertia measures how mass is distributed relative to an axis of rotation. Physically, it describes the tendency of an object to resist changes in rotational motion about that axis.

Moment of Inertia for a Particle

For a single particle of mass [latex]m[/latex] located at distance [latex]r[/latex] from an axis, the moment of inertia is:

[latex]I = mr^2[/latex]

The distance [latex]r[/latex] is measured perpendicular to the axis of rotation.

Using the same partition approach from the previous section, we divide region [latex]R[/latex] into subrectangles [latex]R_{ij}[/latex] with sample points [latex](x_{ij}^*, y_{ij}^*)[/latex] and area [latex]\Delta A[/latex].

From Figure 3, we can see that:

• The moment of inertia of subrectangle [latex]R_{ij}[/latex] about the [latex]x[/latex]-axis is [latex]{({y}^{*}_{ij})^{2}}{\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\Delta}{A}[/latex]
• The moment of inertia of subrectangle [latex]R_{ij}[/latex] about the [latex]y[/latex]-axis is [latex]{({x}^{*}_{ij})^{2}}{\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\Delta}{A}[/latex]

Notice that we square the distance from the axis, not just use the distance itself. This is what distinguishes moments of inertia from ordinary moments.

The moment of inertia [latex]I_x[/latex] about the [latex]x[/latex]-axis for region [latex]R[/latex] is the limit of the sum of moments of inertia of all subrectangles:

[latex]{{I}_{x}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{\left ( {y}^{*}_{ij} \right )}^{2}{{m}_{ij}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{\left ( {y}^{*}_{ij} \right )}^{2}{\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\Delta}{A} = {\underset{R}{\displaystyle\iint}}{{y}^{2}}{\rho}{({x},{y})}{dA}.[/latex]

Similarly, the moment of inertia [latex]I_y[/latex] about the [latex]y[/latex]-axis is:

[latex]{{I}_{y}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{\left ( {x}^{*}_{ij} \right )}^{2}{{m}_{ij}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{\left ( {x}^{*}_{ij} \right )}^{2}{\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\Delta}{A} = {\underset{R}{\displaystyle\iint}}{{x}^{2}}{\rho}{({x},{y})}{dA}.[/latex]

moments of inertia about coordinate axes

For a lamina occupying region [latex]R[/latex] with density [latex]\rho(x,y)[/latex]:

• About the [latex]x[/latex]-axis: [latex]I_x = \iint_R y^2 \rho(x,y) dA[/latex]
• About the [latex]y[/latex]-axis: [latex]I_y = \iint_R x^2 \rho(x,y) dA[/latex]

Moments vs. Moments of Inertia

Compare these formulas carefully:

• Moment [latex]M_x = \iint_R y\rho(x,y) dA[/latex] (linear in [latex]y[/latex])
• Moment of inertia [latex]I_x = \iint_R y^2\rho(x,y) dA[/latex] (quadratic in [latex]y[/latex])

The squared distance term makes moments of inertia more sensitive to mass far from the axis.

Polar Moment of Inertia

Sometimes we need the moment of inertia about the origin rather than about one of the coordinate axes. This is called the polar moment of inertia, denoted [latex]I_0[/latex].

Since the distance from a point [latex](x,y)[/latex] to the origin is [latex]\sqrt{x^2 + y^2}[/latex], squaring this distance gives [latex]x^2 + y^2[/latex]. This leads to:

[latex]{{I}_{0}} = {{I}_{x}} + {{I}_{y}} = {\underset{R}{\displaystyle\iint}}{({x^{2}} + {y^{2}})}{\rho}{({x},{y})}{dA}.[/latex]

polar moment of inertia

The moment of inertia about the origin is:

[latex]I_0 = I_x + I_y = \iint_R (x^2 + y^2) \rho(x,y) dA[/latex]

These expressions can be converted to polar coordinates using the substitutions:

• [latex]x = r\cos\theta[/latex]
• [latex]y = r\sin\theta[/latex]
• [latex]dA = r \, dr \, d\theta[/latex]

For example, the polar moment of inertia becomes:

[latex]I_0 = \iint_R r^2 \rho(r\cos\theta, r\sin\theta) r \, dr \, d\theta[/latex]

This is particularly useful when the region or density function has circular symmetry.

Use the triangular region [latex]R[/latex] with vertices [latex](0, 0)[/latex], [latex](2, 2)[/latex], and [latex](2, 0)[/latex] and with density [latex]{\rho}{({x},{y})} = {x}{y}[/latex]. Find the moments of inertia.

Show Solution

Using the expressions established above for the moments of inertia, we have

[latex]\begin{aligned} {{I}_{x}} = & {\underset{R}{\displaystyle\iint}}{{y}^{2}}{\rho}{({x},{y})}{dA} = {\displaystyle\int^{{x} = {2}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {x}}_{{y} = {0}}}{x}{{y}^{3}}{dy} \ {dx} = {\frac{8}{3}}, \\ {{I}_{y}} = & {\underset{R}{\displaystyle\iint}}{{x}^{2}}{\rho}{({x},{y})}{dA} = {\displaystyle\int^{{x} = {2}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {x}}_{{y} = {0}}}{{x}^{3}}{y} \ {dy} \ {dx} = {\frac{16}{3}}, \\ {{I}_{0}} = & {\underset{R}{\displaystyle\iint}}{({{x}^{2}} + {{y}^{2}})} \ {\rho}{({x},{y})}{dA} = {\displaystyle\int^{2}_{0}}{\displaystyle\int^{x}_{0}}{({{x}^{2}} + {{y}^{2}})} \ {xy} \ {dy} \ {dx} \\ = & {{I}_{x}} + {{I}_{y}} = {8}. \end{aligned}[/latex]

Radius of Gyration

The radius of gyration tells us how far from an axis we could concentrate all the mass of an object while maintaining the same moment of inertia. It’s a useful measure of how mass is distributed around an axis.

For a particle of mass [latex]m[/latex] at distance [latex]r[/latex] from an axis, we have [latex]I = mr^2[/latex]. Solving for [latex]r[/latex] gives [latex]r = \sqrt{I/m}[/latex], which motivates the definition of radius of gyration.

The radii of gyration with respect to the [latex]x[/latex]-axis, [latex]y[/latex]-axis, and origin are:

[latex]R_x = \sqrt{\frac{I_x}{m}}, \quad R_y = \sqrt{\frac{I_y}{m}}, \quad \text{and} \quad R_0 = \sqrt{\frac{I_0}{m}}[/latex]

radius of gyration

The radius of gyration measures the effective distance from an axis where the entire mass could be concentrated:

[latex]R_x = \sqrt{\frac{I_x}{m}}, \quad R_y = \sqrt{\frac{I_y}{m}}, \quad R_0 = \sqrt{\frac{I_0}{m}}[/latex]

where [latex]m[/latex] is the total mass of the object.

In engineering and physics, the radius of gyration helps describe:

• How mass is distributed around a centroidal axis
• Structural stability and resistance to bending
• Rotational dynamics of machinery and moving parts

While moments of inertia are useful for calculating torque and rotational motion, radii of gyration provide insight into the spatial distribution of mass.

Consider the same triangular lamina [latex]R[/latex] with vertices [latex](0, 0)[/latex], [latex](2, 2)[/latex], and [latex](2, 0)[/latex] and with density [latex]{\rho}{({x},{y})} = {x}{y}[/latex]. Find the radii of gyration with respect to the [latex]x[/latex]-axis, the [latex]y[/latex]-axis, and the origin.

Show Solution

If we compute the mass of this region we find that [latex]m=2[/latex]. We found the moments of inertia of this lamina in Example “Finding Mass, Moments, and Center of Mass”. From these data, the radii of gyration with respect to the [latex]x[/latex]-axis, the [latex]y[/latex]-axis, and the origin are, respectively,

[latex]\begin{aligned} {{R}_{x}} = & {\sqrt{\frac{{I}_{x}}{m}}} = {\sqrt{\frac{8/3}{2}}} = {\sqrt{\frac{8}{6}}} = {\frac{{2}{\sqrt{3}}}{3}}, \\ {{R}_{y}} = & {\sqrt{\frac{{I}_{y}}{m}}} = {\sqrt{\frac{16/3}{2}}} = {\sqrt{\frac{8}{3}}} = {\frac{{2}{\sqrt{6}}}{3}}, \\ {{R}_{0}} = & {\sqrt{\frac{{I}_{0}}{m}}} = {\sqrt{\frac{8}{2}}} = {\sqrt{4}} = {2}. \end{aligned}[/latex]