{"id":1286,"date":"2025-10-17T13:14:38","date_gmt":"2025-10-17T13:14:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1286"},"modified":"2025-11-03T15:09:05","modified_gmt":"2025-11-03T15:09:05","slug":"center-of-mass-and-moments-of-inertia-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/center-of-mass-and-moments-of-inertia-learn-it-2\/","title":{"raw":"Center of Mass and Moments of Inertia: Learn It 2","rendered":"Center of Mass and Moments of Inertia: Learn It 2"},"content":{"raw":"

Moments of Inertia<\/h2>\r\n

The moment of inertia<\/strong> 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.<\/p>\r\n\r\n

\r\n

Moment of Inertia for a Particle<\/strong><\/p>\r\n

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

[latex]I = mr^2[\/latex]<\/p>\r\n

The distance [latex]r[\/latex] is measured perpendicular to the axis of rotation.<\/p>\r\n\r\n<\/section>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].\r\n\r\n[caption id=\"attachment_1410\" align=\"aligncenter\" width=\"454\"]\"A Figure 3. Subdividing the lamina into tiny rectangles\u00a0[latex]R_{ij}[\/latex], each containing a sample point\u00a0[latex](x_{ij}^{\\ast},y_{ij}^{\\ast})[\/latex].[\/caption]\r\n

From Figure 3, we can see that:<\/p>\r\n\r\n

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

    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.<\/p>\r\n

    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:<\/p>\r\n

    [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]<\/p>\r\n

    Similarly, the moment of inertia [latex]I_y[\/latex] about the [latex]y[\/latex]-axis is:<\/p>\r\n

    [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]<\/p>\r\n\r\n

    \r\n

    moments of inertia about coordinate axes<\/h3>\r\n

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

      \r\n \t
    • About the [latex]x[\/latex]-axis:<\/strong> [latex]I_x = \\iint_R y^2 \\rho(x,y) dA[\/latex]<\/li>\r\n \t
    • About the [latex]y[\/latex]-axis:<\/strong> [latex]I_y = \\iint_R x^2 \\rho(x,y) dA[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>
      \r\n

      Moments vs. Moments of Inertia<\/strong><\/p>\r\n

      Compare these formulas carefully:<\/p>\r\n\r\n

        \r\n \t
      • Moment<\/strong> [latex]M_x = \\iint_R y\\rho(x,y) dA[\/latex] (linear in [latex]y[\/latex])<\/li>\r\n \t
      • Moment of inertia<\/strong> [latex]I_x = \\iint_R y^2\\rho(x,y) dA[\/latex] (quadratic in [latex]y[\/latex])<\/li>\r\n<\/ul>\r\n

        The squared distance term makes moments of inertia more sensitive to mass far from the axis.<\/p>\r\n\r\n<\/section>\r\n

        Polar Moment of Inertia<\/h3>\r\nSometimes 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<\/strong>, denoted [latex]I_0[\/latex].\r\n\r\nSince 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:\r\n

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

        \r\n

        polar moment of inertia<\/h3>\r\n

        The moment of inertia about the origin is:<\/p>\r\n

        [latex]I_0 = I_x + I_y = \\iint_R (x^2 + y^2) \\rho(x,y) dA[\/latex]<\/p>\r\n\r\n<\/section>

        \r\n

        These expressions can be converted to polar coordinates using the substitutions:<\/p>\r\n\r\n

          \r\n \t
        • [latex]x = r\\cos\\theta[\/latex]<\/li>\r\n \t
        • [latex]y = r\\sin\\theta[\/latex]<\/li>\r\n \t
        • [latex]dA = r \\, dr \\, d\\theta[\/latex]<\/li>\r\n<\/ul>\r\n

          For example, the polar moment of inertia becomes:<\/p>\r\n

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

          This is particularly useful when the region or density function has circular symmetry.<\/p>\r\n\r\n<\/section>

          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.[reveal-answer q=\"567926450\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"567926450\"]\r\n

          Using the expressions established above for the moments of inertia, we have<\/p>\r\n

          [latex]\\begin{aligned}\r\n{{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}}, \\\\\r\n{{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}}, \\\\\r\n{{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} \\\\\r\n= & {{I}_{x}} + {{I}_{y}} = {8}.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

          Radius of Gyration<\/h3>\r\n

          The radius of gyration<\/strong> 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.<\/p>\r\n

          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.<\/p>\r\n

          The radii of gyration with respect to the [latex]x[\/latex]-axis, [latex]y[\/latex]-axis, and origin are:<\/p>\r\n

          [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]<\/p>\r\n\r\n

          \r\n

          radius of gyration<\/h3>\r\n

          The radius of gyration measures the effective distance from an axis where the entire mass could be concentrated:<\/p>\r\n

          [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]<\/p>\r\n

          where [latex]m[\/latex] is the total mass of the object.<\/p>\r\n\r\n<\/section>

          \r\n

          In engineering and physics, the radius of gyration helps describe:<\/p>\r\n\r\n

            \r\n \t
          • How mass is distributed around a centroidal axis<\/li>\r\n \t
          • Structural stability and resistance to bending<\/li>\r\n \t
          • Rotational dynamics of machinery and moving parts<\/li>\r\n<\/ul>\r\n

            While moments of inertia are useful for calculating torque and rotational motion, radii of gyration provide insight into the spatial distribution of mass.<\/p>\r\n\r\n<\/section>

            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.[reveal-answer q=\"634752734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"634752734\"]\r\n

            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\u00a0Example \"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,<\/p>\r\n

            [latex]\\begin{aligned}\r\n{{R}_{x}} = & {\\sqrt{\\frac{{I}_{x}}{m}}} = {\\sqrt{\\frac{8\/3}{2}}} = {\\sqrt{\\frac{8}{6}}} = {\\frac{{2}{\\sqrt{3}}}{3}}, \\\\\r\n{{R}_{y}} = & {\\sqrt{\\frac{{I}_{y}}{m}}} = {\\sqrt{\\frac{16\/3}{2}}} = {\\sqrt{\\frac{8}{3}}} = {\\frac{{2}{\\sqrt{6}}}{3}}, \\\\\r\n{{R}_{0}} = & {\\sqrt{\\frac{{I}_{0}}{m}}} = {\\sqrt{\\frac{8}{2}}} = {\\sqrt{4}} = {2}.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

            Moments of Inertia<\/h2>\n

            The moment of inertia<\/strong> 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.<\/p>\n

            \n

            Moment of Inertia for a Particle<\/strong><\/p>\n

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

            [latex]I = mr^2[\/latex]<\/p>\n

            The distance [latex]r[\/latex] is measured perpendicular to the axis of rotation.<\/p>\n<\/section>\n

            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].<\/p>\n

            \"A
            Figure 3. Subdividing the lamina into tiny rectangles\u00a0[latex]R_{ij}[\/latex], each containing a sample point\u00a0[latex](x_{ij}^{\\ast},y_{ij}^{\\ast})[\/latex].<\/figcaption><\/figure>\n

            From Figure 3, we can see that:<\/p>\n

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

              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.<\/p>\n

              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:<\/p>\n

              [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]<\/p>\n

              Similarly, the moment of inertia [latex]I_y[\/latex] about the [latex]y[\/latex]-axis is:<\/p>\n

              [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]<\/p>\n

              \n

              moments of inertia about coordinate axes<\/h3>\n

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

                \n
              • About the [latex]x[\/latex]-axis:<\/strong> [latex]I_x = \\iint_R y^2 \\rho(x,y) dA[\/latex]<\/li>\n
              • About the [latex]y[\/latex]-axis:<\/strong> [latex]I_y = \\iint_R x^2 \\rho(x,y) dA[\/latex]<\/li>\n<\/ul>\n<\/section>\n
                \n

                Moments vs. Moments of Inertia<\/strong><\/p>\n

                Compare these formulas carefully:<\/p>\n

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

                  The squared distance term makes moments of inertia more sensitive to mass far from the axis.<\/p>\n<\/section>\n

                  Polar Moment of Inertia<\/h3>\n

                  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<\/strong>, denoted [latex]I_0[\/latex].<\/p>\n

                  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:<\/p>\n

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

                  \n

                  polar moment of inertia<\/h3>\n

                  The moment of inertia about the origin is:<\/p>\n

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

                  \n

                  These expressions can be converted to polar coordinates using the substitutions:<\/p>\n

                    \n
                  • [latex]x = r\\cos\\theta[\/latex]<\/li>\n
                  • [latex]y = r\\sin\\theta[\/latex]<\/li>\n
                  • [latex]dA = r \\, dr \\, d\\theta[\/latex]<\/li>\n<\/ul>\n

                    For example, the polar moment of inertia becomes:<\/p>\n

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

                    This is particularly useful when the region or density function has circular symmetry.<\/p>\n<\/section>\n

                    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.<\/p>\n