Why Measure Moment of Inertia?
Mass tells us how much matter a body contains, but it does not describe how that mass is distributed relative to an axis. For rotation about the z-axis, a small mass far from the axis contributes more to rotational inertia than the same mass close to it. The appropriate measure weights each piece of mass by the square of its distance from the axis.
For a body described in cylindrical coordinates, that distance is especially simple: the distance from \((r,\theta,z)\) to the z-axis is \(r\). The calculation combines this squared distance with the density and the cylindrical volume element.
Writing the Integral in Cylindrical Coordinates
In cylindrical coordinates, \(x=r\cos\theta\) and \(y=r\sin\theta\), so
The cylindrical volume element is \(dV=r\,dz\,dr\,d\theta\), with the order of integration chosen to suit the bounds. Therefore, the integrand contains two factors involving \(r\): \(r^2\) from the squared distance and \(r\) from the volume element. For density written as \(\rho(r,\theta,z)\), the integral is
The displayed order indicates the factors in the integrand, not a requirement that \(z\) be integrated first. The bounds must describe the whole solid without omitting or counting any region twice. If both the solid and its density are rotationally symmetric about the z-axis, the density and the radial and vertical bounds do not depend on \(\theta\), and a full revolution has \(0\leq\theta\leq2\pi\).
Why the Formula Works
For a small piece of mass \(dm\), its contribution to the moment of inertia about the z-axis is its mass multiplied by the square of its distance from that axis. If the distance is \(d\), that contribution is \(d^2\,dm\). Adding the contributions of all the pieces gives the continuous integral.
A point \((x,y,z)\) is at perpendicular distance \(d=\sqrt{x^2+y^2}\) from the z-axis.
A mass element with density \(\rho\) and volume \(dV\) has mass \(dm=\rho\,dV\), so its contribution is \(d^2\,dm=(x^2+y^2)\rho\,dV\).
Integrating those contributions over the solid gives \(I_z=\iiint_E(x^2+y^2)\rho\,dV\). In cylindrical coordinates, \(x^2+y^2=r^2\) and \(dV=r\,dz\,dr\,d\theta\), giving the factor \(\rho r^3\).
This argument is the continuous version of summing \(d_i^2m_i\) for many small masses \(m_i\) located at distances \(d_i\). For a density that is integrable over the solid, the triple integral adds their contributions throughout the body.
Moment of inertia has units of mass times length squared. If density is measured in kilograms per cubic metre and lengths in metres, then \(\rho\,dV\) has units of kilograms, and multiplying by \(r^2\) gives \(\mathrm{kg\,m^2}\).
Worked Example: A Uniform Solid Cylinder
Worked Example: A Uniform Solid Cylinder
A solid cylinder has radius \(2\,\mathrm{m}\), height \(3\,\mathrm{m}\), and constant density \(5\,\mathrm{kg/m^3}\). Find its moment of inertia about its central z-axis.
Use \(0\leq r\leq2\,\mathrm{m}\), \(0\leq\theta\leq2\pi\), and \(0\leq z\leq3\,\mathrm{m}\). The density is constant, so
Integrate first with respect to \(z\):
Then integrate with respect to \(r\):
Finally, integrate over the full revolution:
As a check, the cylinder’s mass is its density times its volume: \(M=5\pi(2\,\mathrm{m})^2(3\,\mathrm{m})=60\pi\,\mathrm{kg}\). For a uniform disk at each height, the radial integral gives \(I_z=\tfrac12MR^2\); here this is \(\tfrac12(60\pi)(2^2)=120\pi\,\mathrm{kg\,m^2}\), agreeing with the triple integral.
Worked Example: A Cylinder with Radially Varying Density
Worked Example: A Cylinder with Radially Varying Density
Consider a cylinder with radius \(1\,\mathrm{m}\) and height \(2\,\mathrm{m}\). Its density is \(\rho(r)=2\,\mathrm{kg/m^3}\left(1+(r/(1\,\mathrm{m}))^2\right)\). The density increases toward the outer edge. Find \(I_z\).
The full cylindrical bounds are \(0\leq r\leq1\,\mathrm{m}\), \(0\leq\theta\leq2\pi\), and \(0\leq z\leq2\,\mathrm{m}\). Thus
The integrand does not depend on \(z\), so the vertical integration contributes the height \(2\,\mathrm{m}\):
Integrate each radial term:
Therefore, the radial integral, including its factor \(4\), is \(4(\tfrac14+\tfrac16)\,\mathrm{m^4}=\tfrac53\,\mathrm{m^4}\) before accounting for the density units in the original expression. Integrating over \(\theta\) gives
The positive radial density term increases the result: compared with the same cylinder at constant density \(2\,\mathrm{kg/m^3}\), the added outer density contributes \(4\pi/3\,\mathrm{kg\,m^2}\). This agrees with the calculation, since the constant-density contribution is \(2\pi\,\mathrm{kg\,m^2}\) and \(2\pi+4\pi/3=10\pi/3\).
Worked Example: A Paraboloid-Shaped Body
Worked Example: A Paraboloid-Shaped Body
A body of constant density \(3\,\mathrm{kg/m^3}\) fills the region above \(z=0\) and below the paraboloid \(z=4\,\mathrm{m}-r^2/(1\,\mathrm{m})\). The paraboloid meets the plane at \(r=2\,\mathrm{m}\). Find the moment of inertia about the z-axis.
At each radius from \(0\) to \(2\,\mathrm{m}\), the vertical coordinate runs from \(0\) to \(4\,\mathrm{m}-r^2/(1\,\mathrm{m})\), and \(\theta\) runs from \(0\) to \(2\pi\). Hence
Integrating in \(z\) replaces the inner integral by the height of the body at radius \(r\):
Evaluate the two radial terms separately:
The difference is \(16-\tfrac{32}{3}=\tfrac{16}{3}\,\mathrm{m^5}\). Therefore,
For a check, the volume is \(2\pi\int_0^{2\,\mathrm{m}}(4\,\mathrm{m}-r^2/(1\,\mathrm{m}))r\,dr=8\pi\,\mathrm{m^3}\), so the mass is \(24\pi\,\mathrm{kg}\). The computed inertia divided by this mass is \(\tfrac43\,\mathrm{m^2}\), a plausible average squared distance for a body whose radius ranges from \(0\) to \(2\,\mathrm{m}\).
A Common Mistake: Using Distance Instead of Squared Distance
The definition weights mass by the square of the distance to the axis. Replacing \(r^2\) with \(r\) calculates a different quantity and also gives the wrong units for moment of inertia.
For the uniform cylinder in the first example, the incorrect use of distance \(r\) would give
The first \(r\) in the integrand above was incorrectly used as the distance, and the second comes from the cylindrical volume element. The correct squared distance is \(r^2\), so the actual integrand is \(5r^2r=5r^3\), giving \(120\pi\,\mathrm{kg\,m^2}\), not \(80\pi\). Checking that the distance has been squared, and that the final units are mass times length squared, helps catch this error.
Setting Up an Inertia Calculation
For a new body, organize the calculation before integrating:
- State the axis. Here it is the z-axis, so the squared distance is \(x^2+y^2=r^2\).
- Write the density in cylindrical coordinates, including any dependence on \(r\), \(\theta\), or \(z\).
- Choose radial, angular, and vertical bounds that cover the solid.
- Include both the squared-distance factor and the volume-element factor, giving \(\rho r^3\).
- Evaluate the iterated integral and check the units and, when practical, an independent quantity such as the mass.
Moment of inertia is useful whenever the distribution of mass about a rotational axis matters. Two bodies can have equal mass but different moments of inertia if one places more of its mass farther from the axis. In the integral, that difference is captured by the factor \(r^2\): mass farther out receives a larger weight.
Check Your Understanding
Use the definition and cylindrical-coordinate setup to answer these questions.
- What is the perpendicular distance from \((x,y,z)\) to the z-axis, and what is its square?
- Why does the cylindrical-coordinate inertia integrand contain \(r^3\) when the density is written as \(\rho(r,\theta,z)\)?
- For a uniform cylinder, which bounds correspond to radius, height, and a full revolution?
- For a cylinder with density that increases with \(r\), how should that density appear in the inertia integral?
- What units should the moment of inertia have if mass is measured in kilograms and length in metres?