Moment of Inertia: Integration Method

From Discrete Masses to Continuous Bodies

For a collection of discrete point masses $m_i$, each at a perpendicular distance $r_i$ from an axis of rotation, the total moment of inertia $I$ is a simple sum:

However, real objects like flywheels, planets, or spinning tops are continuous distributions of mass. We can't simply sum up the particles. This is where integration comes in. We imagine the body is composed of an infinite number of infinitesimal mass elements, dm, and we sum their contributions by integrating.

The fundamental formula for the moment of inertia of a continuous body is:

Here, $r$ is the perpendicular distance of the mass element $dm$ from the axis of rotation. Our entire task boils down to expressing $r^2$ and $dm$ in terms of a single integration variable (like $x$, $r$, or $\theta$) and evaluating the integral over the body's geometry.

The video below provides an excellent introduction to this concept and the general strategy.

The General Strategy

Calculating the moment of inertia is a systematic process:

1. Choose Coordinates: Define a coordinate system and specify the axis of rotation.
2. Select a Mass Element (dm): This is the most crucial step. Choose a small piece of the object whose moment of inertia is either known or easy to express. The choice should exploit the body's symmetry.
3. Express dm using Density: Relate the mass element dm to a geometric element (like dx, dA, or dV) using the appropriate mass density:
   • Linear mass density ($\lambda$): For 1D objects like a thin rod. $\lambda =M/L$, so $dm=\lambda dx$.
   • Surface mass density ($\sigma$): For 2D objects like a disc. $\sigma =M/A$, so $dm=\sigma dA$.
   • Volume mass density ($\rho$): For 3D objects like a sphere. $\rho =M/V$, so $dm=\rho dV$.
4. Set up the Integral: Substitute the expressions for $r^2$ and $dm$ into $I=\int r^{2}dm$. Determine the correct integration limits to cover the entire object.
5. Evaluate: Solve the integral.

Derivations for Common Shapes

Let's apply this strategy to derive the moment of inertia for several standard shapes. We'll use the same video as a guide.

1. Thin Ring (or Hollow Cylinder) about its Central Axis

This is the simplest case. For a ring of mass $M$ and radius $R$, every mass element $dm$ is at the same perpendicular distance $R$ from the central axis.

Since $\int dm$ is just the total mass $M$, the result is immediate:

This is also the formula for a thin-walled hollow cylinder, as it is just a stack of identical rings.

2. Thin Rod about its Center

Consider a uniform thin rod of mass $M$ and length $L$, rotating about a perpendicular axis through its center.

• Coordinates & Axis: Let the rod lie on the x-axis from $-L/2$ to $+L/2$. The axis of rotation is the y-axis.
• Mass Element: A small segment of length $dx$ at position $x$.
• dm: The linear mass density is $\lambda =M/L$. So, $dm=\lambda dx$.
• Distance: The perpendicular distance from dm at position x to the y-axis is simply $r=x$.
• Integral: $$I=\int r^{2}dm={\int }_{-L/2}^{+L/2}{x}^2(\lambda dx)=\lambda {\int }_{-L/2}^{+L/2}{x}^{2}dx$$
• Evaluation: $$I=\lambda {\left[\frac{x^3}{3}\right]}_{-L/2}^{+L/2}=\frac{M}{L}\left(\frac{(L/2{)}^3}{3}-\frac{(-L/2{)}^3}{3}\right)=\frac{M}{L}\left(\frac{L^3}{24}+\frac{L^3}{24}\right)=\frac{M}{L}\left(\frac{L^3}{12}\right)$$ $$I_{rod,center}=\frac{1}{12}M{L}^2$$

The video below walks through this exact derivation.

3. Solid Disc (or Solid Cylinder) about its Central Axis

For a uniform solid disc of mass $M$ and radius $R$, rotating about the axis perpendicular to its plane and passing through its center.

• Coordinates & Axis: Use polar coordinates in the plane of the disc. The axis is the z-axis.
• Mass Element: This is a key example of a clever choice. We choose dm to be a thin ring of radius $r$ and infinitesimal thickness $dr$. We already know the MOI of a ring is its mass times its radius squared.
• dm: The surface mass density is $\sigma =M/A=M/(\pi R^2)$. The area of our elemental ring is its circumference times its thickness: $dA=2\pi rdr$. Thus, $dm=\sigma dA=\sigma (2\pi rdr)$.
• Integral: The moment of inertia of this elemental ring is $dI=r^{2}dm=r^2(\sigma 2\pi rdr)$. To find the total moment of inertia, we integrate these rings from the center ($r=0$) to the edge ($r=R$): $$I=\int dI={\int }_0^R{r}^2(\sigma 2\pi rdr)=2\pi \sigma {\int }_0^R{r}^{3}dr$$
• Evaluation: $$I=2\pi \sigma {\left[\frac{r^4}{4}\right]}_0^R=2\pi \left(\frac{M}{\pi R^2}\right)\left(\frac{R^4}{4}\right)=\frac{2\pi M{R}^4}{4\pi R^2}$$ $$I_{disc}=\frac{1}{2}M{R}^2$$

This result also holds for a solid cylinder rotating about its long central axis, as a solid cylinder is just a stack of discs.

The Solid Sphere: A 3D Challenge

Calculating the moment of inertia for a solid sphere is a more advanced application of integration, but it's an excellent test of your skills. We'll explore two distinct methods.

Method 1: Slicing into Discs

This is the most direct integration approach. We slice the sphere into a series of thin discs perpendicular to the axis of rotation.

1. Setup: Let the sphere be centered at the origin, rotating about the z-axis. Consider a thin disc at height $z$ with thickness $dz$.
2. Disc Properties: From the sphere's geometry, $R^2=z^2+r_{disc}^2$, so the disc's radius is $r_{disc}=\sqrt{R^2-z^2}$.
3. Mass of Disc dm: The volume density is $\rho =M/(\frac{4}{3}\pi R^3)$. The disc's volume is $dV=(\pi r_{disc}^2)dz=\pi (R^2-z^2)dz$. So, $dm=\rho \pi (R^2-z^2)dz$.
4. MOI of Disc dI: From our previous result, the MOI of this elemental disc about the z-axis is $dI=\frac{1}{2}(dm)r_{disc}^2$. $$dI=\frac{1}{2}\left[\rho \pi (R^2-z^2)dz\right](R^2-z^2)=\frac{\pi \rho }{2}(R^2-z^2{)}^{2}dz$$
5. Integrate: We sum the contributions from all discs, from $z=-R$ to $z=+R$. $$I={\int }_{-R}^R\frac{\pi \rho }{2}(R^2-z^2{)}^{2}dz=\frac{\pi \rho }{2}{\int }_{-R}^R(R^4-2R^2{z}^2+z^4)dz$$ This is a standard polynomial integral. Evaluating it and substituting back $\rho$ gives the final result.

Method 2: The Elegance of Symmetry

There is a more clever and physically insightful method that uses the sphere's symmetry, which is a technique you will encounter often in theoretical physics.

Both methods lead to the same famous result:

Comparing the two methods is instructive. The first is a direct, brute-force calculation. The second requires a deeper physical insight about symmetry but results in a far simpler integral. Developing the ability to recognize such symmetries is a hallmark of an expert physicist.

Conclusion and Next Steps

In this lesson, you've learned the fundamental skill of calculating the moment of inertia for continuous rigid bodies using integration. This is the rotational equivalent of finding the mass of an object and is the first crucial parameter needed to describe its rotation.

Key Takeaways:

• The moment of inertia $I=\int r^{2}dm$ is the rotational analogue of mass.
• The value of $I$ depends critically on the mass distribution relative to the axis of rotation.
• The general strategy involves choosing a suitable differential mass element ($dm$) and integrating its contribution over the entire body.
• Clever choices of $dm$ (like using rings for a disc) and exploiting symmetry can significantly simplify calculations.
• We have derived the standard results for several key shapes:
  • Ring/Hollow Cylinder: $I=M{R}^2$
  • Rod (center): $I=\frac{1}{12}M{L}^2$
  • Disc/Solid Cylinder: $I=\frac{1}{2}M{R}^2$
  • Solid Sphere: $I=\frac{2}{5}M{R}^2$

Preview of the Next Lesson:
Now that we know how to calculate rotational inertia ($I$), we are ready to build a theory of rotational dynamics. In the next lesson, we will introduce the rotational analogue of force—torque ($\tau$)—and the rotational analogue of Newton's second law, $\tau =I\alpha$. We will then use these new tools to solve problems involving the angular acceleration of rigid bodies.