Moment of Inertia

Solved Problems

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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.

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.

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.

Example 8

Find the moment of inertia of a uniform ball of mass $$m$$ and radius $$R$$ with respect to a diameter.

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.

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.

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.

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.

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.

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.

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.

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

$I = \rho \int\limits_0^\infty {{x^2}{e^{ - x}}dx} = - \rho \lim \limits_{b \to \infty } \left. {\left[ {{e^{ - x}}\left( {{x^2} + 2x + 2} \right)} \right]} \right|_0^b = - \rho \lim \limits_{b \to \infty } \left[ {{e^{ - b}}\left( {{b^2} + 2b + 2} \right) - 2} \right] = - \rho \lim \limits_{b \to \infty } \left[ {\frac{{{b^2} + 2b + 2}}{{{e^b}}} - 2} \right] = 2\rho - \rho \lim \limits_{b \to \infty } \frac{{{b^2} + 2b + 2}}{{{e^b}}}.$
$\lim \limits_{b \to \infty } \frac{{{b^2} + 2b + 2}}{{{e^b}}} = \left[ {\frac{\infty }{\infty }} \right] = \lim \limits_{b \to \infty } \frac{{\left( {{b^2} + 2b + 2} \right)^\prime}}{{\left( {{e^b}} \right)^\prime}} = \lim \limits_{b \to \infty } \frac{{2b + 2}}{{{e^b}}} = \left[ {\frac{\infty }{\infty }} \right] = \lim \limits_{b \to \infty } \frac{{\left( {2b + 2} \right)^\prime}}{{\left( {{e^b}} \right)^\prime}} = \lim \limits_{b \to \infty } \frac{2}{{{e^b}}} = 0.$

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.

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:

${I_x} = \frac{{\pi \rho a{b^3}}}{4} = \frac{{m{b^2}}}{4}.$