Moment of Inertia
Solved Problems
Example 5.
Find the moment of inertia of a right circular cone of mass \(m,\) base radius \(R\) and height \(H\) with respect to its centroidal axis.
Solution.
Figure 9. We assume that the cone is a uniform solid, so its density is
\[\rho = \frac{m}{V} = \frac{m}{{\frac{1}{3}\pi {R^2}H}} = \frac{{3m}}{{\pi {R^2}H}}.\]
Consider a thin slice of thickness \(dz\) at height \(z.\) The mass of the slice is
\[dm = \pi\rho {r^2}dz.\]
By triangle similarity,
\[\frac{r}{{H - z}} = \frac{R}{H},\]
so
\[r = \frac{{R\left( {H - z} \right)}}{H} = R\left( {1 - \frac{z}{H}} \right),\]
and
\[dm = \pi\rho {r^2}dz = \pi \rho {R^2}{\left( {1 - \frac{z}{H}} \right)^2}dz.\]
The moment of inertia of the slice is
\[dI = \frac{{{r^2}dm}}{2} = \frac{{\pi\rho {R^4}{{\left( {1 - \frac{z}{H}} \right)}^4}dz}}{2} = \frac{{\pi\rho {R^4}{{\left( {H - z} \right)}^4}}}{{2{H^4}}}dz.\]
Integrating from \(z = 0\) to \(z = H\) gives the moment of inertia of the cone:
\[I = \int\limits_0^H {\frac{{\pi\rho {R^4}{{\left( {H - z} \right)}^4}}}{{2{H^4}}}dz} = \frac{{\pi\rho {R^4}}}{{2{H^4}}}\int\limits_0^H {{{\left( {H - z} \right)}^4}dz} .\]
Make the substitution:
\[H - z = u,\;\; \Rightarrow z = H - u,\;\; dz = - du.\]
When \(z = 0,\) \(u = H,\) and when \(z = H,\) \(u = 0.\) Hence,
\[\int\limits_0^H {{{\left( {H - z} \right)}^4}dz} = \int\limits_H^0 {{u^4}\left( { - du} \right)} = \int\limits_0^H {{u^4}du} = \left. {\frac{{{u^5}}}{5}} \right|_0^H = \frac{{{H^5}}}{5}.\]
Then the moment of inertia is written in the form
\[I = \frac{{\pi\rho {R^4}}}{{2{H^4}}} \times \frac{{{H^5}}}{5} = \frac{{\pi\rho {R^4}H}}{{10}}.\]
Substitute the expression for \(\rho:\)
\[I = \frac{{3m}}{{\pi {R^2}H}} \times \frac{{\pi {R^4}H}}{{10}} = \frac{{3m{R^2}}}{{10}}.\]
Example 6.
A right uniform triangle with legs \(a\) and \(b\) and mass \(m\) is rotated about the leg \(b.\) Calculate the moment of inertia of the triangle.
Solution.
Figure 10. The moment of inertia of the triangular lamina about the \(y-\)axis is given by the integral
\[{I_y} = \int\limits_0^a {{x^2}\rho \left( x \right)f\left( x \right)dx} .\]
The function \(f\left( x \right)\) is the hypotenuse \(AB\) of the triangle. We can easily derive its equation using the two-point form:
\[\frac{{x - {x_A}}}{{{x_B} - {x_A}}} = \frac{{y - {y_A}}}{{{y_B} - {y_A}}},\;\; \Rightarrow \frac{{x - a}}{{0 - a}} = \frac{{y - 0}}{{b - 0}},\;\; \Rightarrow 1 - \frac{x}{a} = \frac{y}{b},\;\; \Rightarrow y = f\left( x \right) = b - \frac{b}{a}x.\]
Assuming that the density \(\rho\) is constant, we have
\[I = {I_y} = \rho \int\limits_0^a {{x^2}\left( {b - \frac{b}{a}x} \right)dx} = \rho \int\limits_0^a {\left( {b{x^2} - \frac{b}{a}{x^3}} \right)dx} = \rho \left. {\left( {\frac{{b{x^3}}}{3} - \frac{{b{x^4}}}{{4a}}} \right)} \right|_0^a = \rho \left( {\frac{{b{a^3}}}{3} - \frac{{b{a^3}}}{4}} \right) = \frac{{\rho b{a^3}}}{{12}}.\]
Let's now recall that the mass of the triangular lamina is
\[{m = \rho A = \frac{{\rho ab}}{2}.}\]
Then the moment of inertia is expressed as
\[I = \frac{{\rho b{a^3}}}{{12}} = \frac{{\rho ab}}{2} \times \frac{{{a^2}}}{6} = \frac{{m{a^2}}}{6}.\]
Example 7.
A uniform lamina is bounded by the curves \[y = \sqrt x,\,y = {x^2}\] and has the mass \(m.\) Find the moment of inertia of the lamina about the \(x-\)axis.
Solution.
Figure 11. We can calculate the moment of inertia of the lamina by the formula
\[I = {I_x} = \rho \int\limits_0^1 {{y^2}\left[ {f\left( y \right) - g\left( y \right)} \right]dy} .\]
We rewrite the equations of both curves as functions of \(y:\)
\[y = \sqrt x ,;\; \Rightarrow x = g\left( y \right) = {y^2};\]
\[y = {x^2},\;\; \Rightarrow x = f\left( y \right) = \sqrt y .\]
Integration gives:
\[I = \rho \int\limits_0^1 {{y^2}\left( {\sqrt y - {y^2}} \right)dy} = \rho \int\limits_0^1 {\left( {{y^{\frac{5}{2}}} - {y^4}} \right)dy} = \rho \left. {\left( {\frac{{2{y^{\frac{7}{2}}}}}{7} - \frac{{{y^5}}}{5}} \right)} \right|_0^1 = \rho \left( {\frac{2}{7} - \frac{1}{5}} \right) = \frac{{3\rho }}{{35}}.\]
Now let's determine the mass of the lamina:
\[m = \rho A = \rho \int\limits_0^1 {\left( {\sqrt x - {x^2}} \right)dx} = \rho \left. {\left( {\frac{{2{x^{\frac{3}{2}}}}}{3} - \frac{{{x^3}}}{3}} \right)} \right|_0^1 = \rho \left( {\frac{2}{3} - \frac{1}{3}} \right) = \frac{\rho }{3}.\]
Hence, the moment of inertia is
\[I = \frac{{3\rho }}{{35}} = \frac{\rho }{3} \times \frac{9}{{35}} = \frac{{9m}}{{35}}.\]
Example 8.
Find the moment of inertia of a uniform ball of mass \(m\) and radius \(R\) with respect to a diameter.
Solution.
Figure 12. The density of the ball is
\[\rho = \frac{m}{V} = \frac{m}{{\frac{4}{3}\pi {R^3}}} = \frac{{3m}}{{4\pi {R^3}}}.\]
Consider a thin slice of thickness \(dz\) at height \(z.\) We can express its radius in terms of the coordinate \(z:\)
\[r = \sqrt {{R^2} - {z^2}} .\]
The mass of the elementary slice is
\[dm = \pi \rho {r^2}dz = \pi \rho \left( {{R^2} - {z^2}} \right)dz.\]
The moment of inertia of the disk is given by
\[dI = \frac{{{r^2}dm}}{2} = \frac{{\left( {{R^2} - {z^2}} \right)dm}}{2} = \frac{{\pi \rho {{\left( {{R^2} - {z^2}} \right)}^2}}}{2}dz.\]
To find the moment of inertia of the ball, we integrate from \(z = -R\) to \(z = R:\)
\[I = \int\limits_{ - R}^R {\frac{{\pi \rho {{\left( {{R^2} - {z^2}} \right)}^2}}}{2}dz} = \frac{{\pi \rho }}{2}\int\limits_{ - R}^R {{{\left( {{R^2} - {z^2}} \right)}^2}dz} = \pi \rho \int\limits_0^R {{{\left( {{R^2} - {z^2}} \right)}^2}dz} = \pi \rho \int\limits_0^R {\left( {{R^4} - 2{R^2}{z^2} + {z^4}} \right)dz} = \pi \rho \left. {\left( {{R^4}z - \frac{{2{R^2}{z^3}}}{3} + \frac{{{z^5}}}{5}} \right)} \right|_0^R = \pi \rho {R^5}\left( {1 - \frac{2}{3} + \frac{1}{5}} \right) = \frac{{8\pi \rho {R^5}}}{{15}}.\]
Substituting the equation for \(\rho,\) we have
\[I = \frac{{8\pi {R^5}}}{{15}} \times \frac{{3m}}{{4\pi {R^3}}} = \frac{{2m{R^2}}}{5}.\]
Example 9.
A uniform lamina of mass \(m\) is bounded by the exponential curve \[y = \exp \left({-x}\right)\] and the coordinate axes. Find the moment of inertia of the lamina about the \(y-\)axis.
Solution.
Figure 13. The moment of inertia of the infinite lamina about the \(y-\)axis is determined by the improper integral
\[I = {I_y} = \int\limits_0^\infty {{x^2}\rho \left( x \right)f\left( x \right)dx} = \rho \int\limits_0^\infty {{x^2}{e^{ - x}}dx} .\]
Integrating by parts twice, we can write the indefinite integral in the form:
\[\int {{x^2}{e^{ - x}}dx} = \left[ {\begin{array}{*{20}{l}}
{u = {x^2}}\\
{dv = {e^{ - x}}dx}\\
{du = 2xdx}\\
{v = - {e^{ - x}}}
\end{array}} \right] = - {x^2}{e^{ - x}} - \int {\left( { - 2x{e^{ - x}}} \right)dx} = - {x^2}{e^{ - x}} + 2\int {x{e^{ - x}}dx} = \left[ {\begin{array}{*{20}{l}}
{u = x}\\
{dv = {e^{ - x}}dx}\\
{du = dx}\\
{v = - {e^{ - x}}}
\end{array}} \right] = - {x^2}{e^{ - x}} + 2\left[ { - x{e^{ - x}} - \int {\left( { - {e^{ - x}}} \right)dx} } \right] = - {x^2}{e^{ - x}} + 2\left[ { - x{e^{ - x}} + \int {{e^{ - x}}dx} } \right] = - {x^2}{e^{ - x}} + 2\left[ { - x{e^{ - x}} - {e^{ - x}}} \right] = - {x^2}{e^{ - x}} - 2x{e^{ - x}} - 2{e^{ - x}} = - {e^{ - x}}\left( {{x^2} + 2x + 2} \right).\]
Returning back to the improper integral, we have
\[I = \rho \int\limits_0^\infty {{x^2}{e^{ - x}}dx} = - \rho \lim \limits_{b \to \infty } \left. {\left[ {{e^{ - x}}\left( {{x^2} + 2x + 2} \right)} \right]} \right|_0^b = - \rho \lim \limits_{b \to \infty } \left[ {{e^{ - b}}\left( {{b^2} + 2b + 2} \right) - 2} \right] = - \rho \lim \limits_{b \to \infty } \left[ {\frac{{{b^2} + 2b + 2}}{{{e^b}}} - 2} \right] = 2\rho - \rho \lim \limits_{b \to \infty } \frac{{{b^2} + 2b + 2}}{{{e^b}}}.\]
By L'Hopital's Rule ,
\[\lim \limits_{b \to \infty } \frac{{{b^2} + 2b + 2}}{{{e^b}}} = \left[ {\frac{\infty }{\infty }} \right] = \lim \limits_{b \to \infty } \frac{{\left( {{b^2} + 2b + 2} \right)^\prime}}{{\left( {{e^b}} \right)^\prime}} = \lim \limits_{b \to \infty } \frac{{2b + 2}}{{{e^b}}} = \left[ {\frac{\infty }{\infty }} \right] = \lim \limits_{b \to \infty } \frac{{\left( {2b + 2} \right)^\prime}}{{\left( {{e^b}} \right)^\prime}} = \lim \limits_{b \to \infty } \frac{2}{{{e^b}}} = 0.\]
Thus, we see that the moment of inertia of the infinite lamina has the finite value equal to
\[I = 2\rho .\]
We can express the moment of inertia of the lamina in terms of its mass \(m.\) Given that
\[m = \rho A = \rho \int\limits_0^\infty {f\left( x \right)dx} = \rho \int\limits_0^\infty {{e^{ - x}}dx} = - \left. {\rho {e^{ - x}}} \right|_0^\infty = - \rho \lim \limits_{b \to \infty } \left. {\left[ {{e^{ - x}}} \right]} \right|_0^b = - \rho \lim \limits_{b \to \infty } \left[ {{e^{ - b}} - 1} \right] = \rho ,\]
the final answer is written as
\[I = 2m.\]
Example 10.
Given an ellipse with semi-major axis \(a\) and semi-minor axis \(b.\) Find the moment of inertia of the ellipse about the \(y-\)axis.
Solution.
Figure 14. The moment moment of inertia of the ellipse about the \(y-\)axis is given by the integral
\[{I_y} = \rho \int\limits_{ - a}^a {{x^2}\left[ {f\left( x \right) - g\left( x \right)} \right]dx} ,\]
where \(\rho\) is the surface density of the ellipse, and
\[f\left( x \right) = \frac{b}{a}\sqrt {{a^2} - {x^2}} ,\;\; g\left( x \right) = - \frac{b}{a}\sqrt {{a^2} - {x^2}}. \]
This yields:
\[I_y = \frac{{2\rho b}}{a}\int\limits_{ - a}^a {{x^2}\sqrt {{a^2} - {x^2}} dx} .\]
As the function in the integrand is even, we can rewrite the integral in the form
\[I_y = \frac{{4b\rho }}{a}\int\limits_0^a {{x^2}\sqrt {{a^2} - {x^2}} dx} .\]
Now we make the substitution
\[x = a\cos t,\;\; dx = - a\sin tdt.\]
When \(x = 0,\) then \(t = \frac{\pi }{2},\) and when \(x = a,\) then \(t = 0.\) So we have
\[{I_y} = \frac{{4\rho b{a^4}}}{a}\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}t\,{{\sin }^2}t\,dt} = \rho b{a^3}\int\limits_0^{\frac{\pi }{2}} {{{\left( {2\cos t\sin t} \right)}^2}dt} = \rho b{a^3}\int\limits_0^{\frac{\pi }{2}} {{{\left( {\sin 2t} \right)}^2}dt} = \frac{{\rho b{a^3}}}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 - \cos 4t} \right)dt} = \frac{{\rho b{a^3}}}{2}\left. {\left[ {t - \frac{{\sin 4t}}{4}} \right]} \right|_0^{\frac{\pi }{2}} = \frac{{\rho b{a^3}}}{2} \times \frac{\pi }{2} = \frac{{\pi \rho b{a^3}}}{4}.\]
Assuming that the density is \(\rho,\) we find the mass of the ellipse:
\[m = \rho A = \pi \rho ab,\]
where \(A\) denotes the area of the ellipse.
Hence,
\[{I_y} = \frac{{\pi \rho b{a^3}}}{4} = \pi \rho ab \times \frac{{{a^2}}}{4} = \frac{{m{a^2}}}{4}.\]
Notice that changing \(b \to a,\) \(a \to b\) gives us the moment of inertia of the ellipse about the \(x-\)axis:
\[{I_x} = \frac{{\pi \rho a{b^3}}}{4} = \frac{{m{b^2}}}{4}.\]
