# 16.3: Rotational Kinetic Energy and Moment of Inertia

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## Rotational Kinetic Energy and Moment of Inertia

We have already defined translational kinetic energy for a point object as $$K=(1 / 2) m v^{2}$$; we now define the rotational kinetic energy for a rigid body about its center of mass.

Choose the z-axis to lie along the axis of rotation passing through the center of mass. As discussed previously, divide the body into volume elements of mass $$\Delta m_{i}$$ (Figure $$\PageIndex{1}$$). Each individual mass element $$\Delta m_{i}$$ undergoes circular motion about the center of mass with z-component of angular velocity $$\omega_{\mathrm{cm}}$$ in a circle of radius $$r_{\mathrm{cm}, i}$$. Therefore the velocity of each element is given by $$\overrightarrow{\mathbf{v}}_{\mathrm{cm}, i}=r_{\mathrm{cm}, i} \omega_{\mathrm{cm}} \hat{\boldsymbol{\theta}}$$. The rotational kinetic energy is then

$K_{\mathrm{cm}, i}=\frac{1}{2} \Delta m_{i} v_{\mathrm{cm}, i}^{2}=\frac{1}{2} \Delta m r_{i} r_{\mathrm{cm}, i}^{2} \omega_{\mathrm{cm}}^{2} \nonumber$

We now add up the kinetic energy for all the mass elements,

\begin{aligned} K_{\mathrm{cm}} &=\lim _{i \rightarrow \infty \atop \Delta m_{i} \rightarrow 0} \sum_{i=1}^{i=N} K_{\mathrm{cm}, i}=\lim _{i \rightarrow \infty} \sum_{\Delta m_{i} \rightarrow 0}^{i=N}\left(\sum_{i} \frac{1}{2} \Delta m_{i} r_{\mathrm{cm}, i}^{2}\right) \omega_{\mathrm{cm}}^{2} \\ &=\left(\frac{1}{2} \int_{\mathrm{body}} d m r_{\mathrm{dm}}^{2}\right) \omega_{\mathrm{cm}}^{2} \end{aligned} \nonumber

where $$dm$$ is an infinitesimal mass element undergoing a circular orbit of radius $$r_{d m}$$ about the axis passing through the center of mass.

The quantity

$I_{c m}=\int_{body} r_{d m}^{2}\,dm \nonumber$

is called the moment of inertia of the rigid body about a fixed axis passing through the center of mass, and is a physical property of the body. The SI units for moments of inertia are $$\left[\mathrm{kg} \cdot \mathrm{m}^{2}\right]$$.

Thus

$K_{\mathrm{cm}}=\left(\frac{1}{2} \int_{\mathrm{body}} r_{\mathrm{dm}}^{2}\, dm \right) \omega_{\mathrm{cm}}^{2} \equiv \frac{1}{2} I_{c m} \omega_{\mathrm{cm}}^{2} \nonumber$

## Moment of Inertia of a Rod of Uniform Mass Density

Consider a thin uniform rod of length $$L$$ and mass $$m$$. In this problem, we will calculate the moment of inertia about an axis perpendicular to the rod that passes through the center of mass of the rod. A sketch of the rod, volume element, and axis is shown in Figure $$\PageIndex{2}$$. Choose Cartesian coordinates, with the origin at the center of mass of the rod, which is midway between the endpoints since the rod is uniform. Choose the x-axis to lie along the length of the rod, with the positive x-direction to the right, as in Figure $$\PageIndex{2}$$.

Identify an infinitesimal mass element $$d m=\lambda d x$$, located at a displacement x from the center of the rod, where the mass per unit length $$\lambda=m / L$$ is a constant, as we have assumed the rod to be uniform. When the rod rotates about an axis perpendicular to the rod that passes through the center of mass of the rod, the element traces out a circle of radius $$r_{d m}=x$$. We add together the contributions from each infinitesimal element as we go from $$x=-L / 2$$ to $$x=L / 2$$. The integral is then

\begin{align} I_{\mathrm{cm}} &=\int_{\text {body }} r_{d \mathrm{m}}^{2} d m \label{eq5} \\[4pt] &=\lambda \int_{-L / 2}^{L / 2}\left(x^{2}\right) d x=\left.\lambda \frac{x^{3}}{3}\right|_{-L / 2} ^{L / 2} \\[4pt] &=\frac{m}{L} \frac{(L / 2)^{3}}{3}-\frac{m}{L} \frac{(-L / 2)^{3}}{3}=\frac{1}{12} m L^{2} \end{align} \nonumber

By using a constant mass per unit length along the rod, we need not consider variations in the mass density in any direction other than the x-axis. We also assume that the width is the rod is negligible. (Technically we should treat the rod as a cylinder or a rectangle in the xy-plane if the axis is along the z-axis. The calculation of the moment of inertia in these cases would be more complicated.)

Example $$\PageIndex{1}$$: Moment of Inertia of a Uniform Disk

A thin uniform disk of mass $$M$$ and radius $$R$$ is mounted on an axle passing through the center of the disk, perpendicular to the plane of the disk. Calculate the moment of inertia about an axis that passes perpendicular to the disk through the center of mass of the disk

Solution

As a starting point, consider the contribution to the moment of inertia from the mass element $$dm$$ show in $$\PageIndex{3}$$. Let $$r$$ denote the distance form the center of mass of the disk to the mass element.

Choose cylindrical coordinates with the coordinates $$(r, \theta)$$ in the plane and the z-axis perpendicular to the plane. The area element

$d a=r dr d\theta \nonumber$

may be thought of as the product of arc length $$r d \theta$$ and the radial width $$dr$$. Since the disk is uniform, the mass per unit area is a constant,

$\sigma=\frac{d m}{d a}=\frac{m_{\text {total }}}{\text { Area }}=\frac{M}{\pi R^{2}} \nonumber$

Therefore the mass in the infinitesimal area element as given in Equation \ref{16.2.6}, a distance $$r$$ from the axis of rotation, is given by

$d m=\sigma r d r d \theta=\frac{M}{\pi R^{2}} r\, dr\, d\theta \label{16.2.6}$

When the disk rotates, the mass element traces out a circle of radius $$r_{d m}=r$$; that is, the distance from the center is the perpendicular distance from the axis of rotation. The moment of inertia integral is now an integral in two dimensions; the angle $$θ$$ varies from $$\theta=0 \text { to } \theta=2 \pi$$, and the radial coordinate r varies from $$r = 0$$ to $$r = R$$. Thus the limits of the integral are

$I_{\mathrm{cm}}=\int_{\mathrm{body}} r_{d m}^{2} d m=\frac{M}{\pi R^{2}} \int_{r=0}^{r=R} \int_{\theta=0}^{\theta=2 \pi} r^{3} d \theta d r \nonumber$

The integral can now be explicitly calculated by first integrating the $$θ$$-coordinate

$I_{\mathrm{cm}}=\frac{M}{\pi R^{2}} \int_{r=0}^{r=R}\left(\int_{\theta=0}^{\theta=2 \pi} d \theta\right) r^{3} d r=\frac{M}{\pi R^{2}} \int_{r=0}^{r=R} 2 \pi r^{3} d r=\frac{2 M}{R^{2}} \int_{r=0}^{r=R} r^{3} d r \nonumber$

and then integrating the $$r$$-coordinate,

$I_{\mathrm{cm}}=\frac{2 M}{R^{2}} \int_{r=0}^{r=R} r^{3} d r=\left.\frac{2 M}{R^{2}} \frac{r^{4}}{4}\right|_{r=0} ^{r=R}=\frac{2 M}{R^{2}} \frac{R^{4}}{4}=\frac{1}{2} M R^{2} \nonumber$

Remark: Instead of taking the area element as a small patch $$d a=r d r d \theta$$, choose a ring of radius $$r$$ and width $$dr$$. Then the area of this ring is given by

$d a_{\text {ring }}=\pi(r+d r)^{2}-\pi r^{2}=\pi r^{2}+2 \pi r d r+\pi(d r)^{2}-\pi r^{2}=2 \pi r d r+\pi(d r)^{2} \nonumber$

In the limit that $$d r \rightarrow 0$$, the term proportional to $$(d r)^{2}$$ can be ignored and the area is $$d a=2 \pi r d r$$. This equivalent to first integrating the $$dθ$$ variable

$d a_{\mathrm{ring}}=r d r\left(\int_{\theta=0}^{\theta=2 \pi} d \theta\right)=2 \pi r d r \nonumber$

Then the mass element is

$d m_{\text {ring }}=\sigma d a_{\text {ring }}=\frac{M}{\pi R^{2}} 2 \pi r d r \nonumber$

The moment of inertia integral is just an integral in the variable $$r$$,

$I_{\mathrm{cm}}=\int_{\mathrm{body}}\left(r_{\perp}\right)^{2} d m=\frac{2 \pi M}{\pi R^{2}} \int_{r=0}^{r=R} r^{3} d r=\frac{1}{2} M R^{2} \nonumber$

## Parallel Axis Theorem

Consider a rigid body of mass m undergoing fixed-axis rotation. Consider two parallel axes. The first axis passes through the center of mass of the body, and the moment of inertia about this first axis is $$I_{\mathrm{cm}}$$ The second axis passes through some other point $$S$$ in the body. Let $$d_{S, \mathrm{cm}}$$ denote the perpendicular distance between the two parallel axes (Figure $$\PageIndex{4}$$).

Then the moment of inertia $$I_{S}$$ about an axis passing through a point $$S$$ is related to $$I_{\mathrm{cm}}$$ by

$I_{S}=I_{\mathrm{cm}}+m d_{S, \mathrm{cm}}^{2} \label{16.2.16}$

## Parallel Axis Theorem Applied to a Uniform Rod

Let point $$S$$ be the left end of the rod of Figure $$\PageIndex{4}$$. Then the distance from the center of mass to the end of the rod is $$d_{S, \mathrm{cm}}=L / 2$$ The moment of inertia $$I_{S}=I_{\mathrm{end}}$$ about an axis passing through the endpoint is related to the moment of inertia about an axis passing through the center of mass, $$I_{\mathrm{cm}}=(1 / 12) m L^{2}$$, according to Equation \ref{16.2.16},

$I_{S}=\frac{1}{12} m L^{2}+\frac{1}{4} m L^{2}=\frac{1}{3} m L^{2} \nonumber$

In this case it is easy and useful to check by direct calculation. Use Equation \ref{eq5} but with the limits changed to $$x^{\prime}=0$$ and $$x^{\prime}=L$$, where $$x^{\prime}=x+L / 2$$,

\begin{aligned} I_{\text {end }} &=\int_{\text {body }} r_{\perp}^{2} d m=\lambda \int_{0}^{L} x^{\prime 2} d x^{\prime} \\ &=\left.\lambda \frac{x^{\prime 3}}{3}\right|_{0} ^{L}=\frac{m}{L} \frac{(L)^{3}}{3}-\frac{m}{L} \frac{(0)^{3}}{3}=\frac{1}{3} m L^{2} \end{aligned} \nonumber

Example $$\PageIndex{2}$$: Rotational Kinetic Energy of Disk

A disk with mass $$M$$ and radius $$R$$ is spinning with angular speed ω about an axis that passes through the rim of the disk perpendicular to its plane. The moment of inertia about cm is $$I_{c m}=(1 / 2) m R^{2}$$. What is the kinetic energy of the disk?

Solution

The parallel axis theorem states the moment of inertia about an axis passing perpendicular to the plane of the disk and passing through a point on the edge of the disk is equal to

$I_{e d g e}=I_{c m}+m R^{2} \nonumber$

The moment of inertia about an axis passing perpendicular to the plane of the disk and passing through the center of mass of the disk is equal to $$I_{c m}=(1 / 2) m R^{2}$$. Therefore

$I_{e d g e}=(3 / 2) m R^{2} \nonumber$

The kinetic energy is then

$K=(1 / 2) I_{e d g e} \omega^{2}=(3 / 4) m R^{2} \omega^{2} \nonumber$

This page titled 16.3: Rotational Kinetic Energy and Moment of Inertia is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Peter Dourmashkin (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.