# [SOLVED]Calculating moments of inertia

#### dwsmith

##### Well-known member
What formula do I need here?

Starting from the sum $I = \sum\limits_{\alpha = 1}^Nm_{\alpha}\rho_{\alpha}^2$ and replacing it by the appropriate integral, find the moment of inertia of a uniform thin square of side $2b$, rotating about an axis perpendicular to the square and passing through the center.

#### MarkFL

Staff member
Re: I no longer own a Calculus book what formula is this that I need

This may or not be what you're after, but here is one method we could use to derive the result rather than simply look it up. We can deconstruct the sheet or slab into rods, each of which rotates about the axis of rotation of the sheet.

Calculate the moment of inertia of a uniform rigid rod of length $l$ and mass $m$ about an axis perpendicular to the rod (the $y$-axis) and passing through its center of mass.

The length element $dx$ has a mass $dm$ equal to the mass per unit length multiplied by $dx$:

$\displaystyle dm=\frac{m}{l}dx$

Substituting this expression for $dm$ with $r=x$, we get:

$\displaystyle I_y=\int r^2\,dm=\int_{-\frac{l}{2}}^{\frac{l}{2}} x^2\frac{m}{l}\,dx=\frac{m}{l}\int_{-\frac{l}{2}}^{\frac{l}{2}} x^2\,dx=\frac{m}{l}\left[\frac{x^3}{3}\right]^{\frac{l}{2}}_{-\frac{l}{2}}=\frac{1}{12}ml^2$

Now we may use the parallel-axis theorem and the even function rule to sum up the rods to form a sheet.

$\displaystyle I=2\int_{0}^{\frac{w}{2}} \frac{1}{12}ml^2+mw^2\,dw=\frac{2}{12}ml^2\left( \left(\frac{w}{2} \right)-0 \right)+\frac{2m}{3}\left( \left(\frac{w}{2} \right)^3-0^3 \right)$

$\displaystyle I=\frac{1}{12}ml^2w+\frac{1}{12}mw^3$

Let $M$ be the total mass of the sheet, thus the mass $m$ of a rod is $\displaystyle m=\frac{M}{w}$ and we have:

$\displaystyle I=\frac{1}{12}Ml^2+\frac{1}{12}Mw^2=\frac{1}{12}M(l^2+w^2)$

Now, let $l=w=2b$ to obtain:

$\displaystyle I=\frac{1}{12}M((2b)^2+(2b)^2)=\frac{2}{3}Mb^2$

#### dwsmith

##### Well-known member
Re: I no longer own a Calculus book what formula is this that I need

The r squared term is that just due to the fact we have a square and we want to integrate across the area?

For example, if we had a circle, it would be \pi r^2 instead?

#### MarkFL

Staff member
Re: I no longer own a Calculus book what formula is this that I need

If we had a circular disk, we could deconstruct it into a series of concentric rings. I'll post this later.

#### MarkFL

Staff member
Re: I no longer own a Calculus book what formula is this that I need

I posted the following some time ago at another forum, and it is essentially a paraphrasing of information from my old physics textbook.

Rotational Energy

Let us consider a rigid object as a collection of small particles and let us assume that the object rotates about the fixed $z$-axis with an angular speed $\omega$. Each particle has kinetic energy determined by its mass and speed. If the mass of the $i$th particle is $m_i$ and its speed is $v_i$, its kinetic energy is $\displaystyle K_i=\frac{1}{2}m_iv_i^2$.

To proceed further, we must recall that although every particle in the rigid object has the same angular speed, $\omega$, the individual linear speeds depend on the perpendicular distance $r_i$ from the axis of rotation according to the expression $v_i=r_i\omega$. The total energy of the rotating rigid object is the sum of the kinetic energies of the individual particles:

$\displaystyle K_R=\sum K_i=\sum\frac{1}{2}m_iv_i^2=\frac{1}{2}\sum m_ir_i^2\omega^2$

$\displaystyle K_R=\frac{1}{2}\left(\sum m_ir_i^2\right)\omega^2$

where we have factored $\omega^2$ from the sum since it is common to every particle. The quantity in parentheses is called the moment of inertia, $I$:

$\displaystyle I=\sum m_ir_i^2$

Using this notation, we can express the rotational energy of the rotating rigid object as

(1) $\displaystyle K_R=\frac{1}{2}I\omega^2$

Although we commonly refer to the quantity $\displaystyle \frac{1}{2}I\omega^2$ as the rotational energy, it is not a new form of energy. It is ordinary kinetic energy, because it was derived from a sum over individual kinetic energies of the particles contained in the rigid object. However, the form of the energy given by (1) is a convenient one when dealing with rotational motion, providing we know how to calculate $I$. It is important to recognize the analogy between kinetic energy associated with linear motion, $\displaystyle \frac{1}{2}mv^2$, and rotational energy, $\displaystyle \frac{1}{2}I\omega^2$. The quantities $I$ and $\omega$ in rotational motion are analogous to $m$ and $v$ in linear motion, respectively. In fact, $I$ takes the place of $m$ every time we compare a linear-motion equation to its rotational counterpart.

Calculation of Moments of Inertia

We can evaluate the moment of inertia of an extended object by imagining that the object is divided into many small volume elements, each of mass $\Delta m$. We use the definition $\displaystyle I=\sum r_i^2\Delta m$ and take the limit of this sum as $\displaystyle \Delta m\to0$. In this limit, the sum becomes an integral over the whole object:

(2) $\displaystyle I=\lim_{\Delta m\to0} \sum r_i^2\Delta m=\int r^2\,dm$

To evaluate the moment of inertia using (2), it is necessary to express each volume element (of mass $dm$) in terms of its coordinates. It is common to define a mass density in various forms. For a three-dimensional object, it is appropriate to use the volume density, that is mass per unit volume:

$\displaystyle \rho=\lim_{\Delta V\to0} \frac{\Delta m}{\Delta V}=\frac{dm}{dV}$

$\displaystyle dm=\rho\cdot dV$

Therefore, the moment of inertia for a three-dimensional object can be expressed in the form

$\displaystyle I=\int \rho r^2\,dV$

If the object is homogeneous, then $\rho$ is constant and the integral can be evaluated for a known geometry. If $\rho$ is not constant, then its variation with position must be specified.

When dealing with a sheet of uniform thickness $t$, rather than with a three dimensional object, it is convenient to define a surface density $\sigma=\rho t$, which signifies mass per unit area.

Finally, when mass is distributed along a uniform rod of cross-sectional area $A$, we sometimes use linear density, $\lambda=\rho A$, where $\lambda$ is defined as mass per unit length.

The moments of inertia of rigid bodies with simple geometry (high symmetry) are relatively easy to calculate provided the rotation axis coincides with the axis of symmetry.

Examples:

Uniform Hoop

Find the moment of inertia of a uniform hoop of mass $M$ and radius $R$ about an axis perpendicular to the plane of the hoop, though its center.

Solution: All mass elements are the same distance $r=R$ from the axis, and so applying equation (2), we get for the moment of inertia about the $z$-axis through $O$:

$\displaystyle I_z=\int r^2\,dm=R^2\int\,dm=MR^2$

Uniform Rigid Rod

Calculate the moment of inertia of a uniform rigid rod of length $L$ and mass $M$ about an axis perpendicular to the rod (the $y$-axis) and passing through its center of mass.

Solution: The length element $dx$ has a mass $dm$ equal to the mass per unit length multiplied by $dx$:

$\displaystyle dm=\frac{M}{L}dx$

Substituting this expression for $dm$ into equation (2), with $r=x$, we get

$\displaystyle I_y=\int r^2\,dm=\int^{\frac{L}{2}}_{-\frac{L}{2}} x^2\frac{M}{L}\,dx=\frac{M}{L}\int^{\frac{L}{2}}_{-\frac{L}{2}} x^2\,dx=\frac{M}{L}\left[\frac{x^3}{3}\right]^{\frac{L}{2}}_{-\frac{L}{2}}=\frac{1}{12}ML^2$

Now, calculate the moment of inertia of a uniform rigid rod about an axis perpendicular to the rod and passing through one end (the $y'$-axis). Note that the calculation requires that the limits of integration be from $x=0$ to $x=L$.

$\displaystyle I_{y'}=\int r^2\,dm=\int^{L}_{0} x^2\frac{M}{L}\,dx=\frac{M}{L}\int^{L}_{0} x^2\,dx=\frac{M}{L}\left[\frac{x^3}{3}\right]^{L}_{0}=\frac{1}{3}ML^2$

What axis passing through the rod at point a, $0\le a\le L$, must we choose to minimize the moment of inertia for the uniform rigid rod?

$\displaystyle \frac{d}{dx}\frac{M}{L}\int^{L-a}_{-a}x^2\,dx=\frac{M}{L}\left((L-a)^2-(-a)^2\right)=\frac{M}{L}\left(L^2-2aL\right)=M(L-2a)=0\:\therefore\:a=\frac{L}{2}$

Uniform Solid Cylinder

A uniform solid cylinder has a radius $R$, mass $M$, and length $L$. Calculate its moment of inertia about its central axis (the $z$-axis).

Reasoning: It is convenient to divide the cylinder into many cylindrical shells each of radius $r$, thickness $dr$, and length $L$. Cylindrical shells are chosen because we want all mass elements $dm$ to have a single value for $r$, which makes the calculation more straightforward. The volume of each shell is its cross-sectional area multiplied by the length, or $dV=dA\cdot L=(2\pi dr)L$. If the mass per unit volume is $\rho$, then the mass of this differential volume element is $\displaystyle dm=\rho\cdot dV=\rho2\pi L\cdot dr$.

Solution: Substituting this expression for $dm$ into equation (2), we get

$\displaystyle I_z=\int r^2\,dm=2\pi\rho L\int^R_0 r^3\,dr=\frac{\pi\rho LR^4}{2}$

Because the total volume of the cylinder is $\pi R^2L$, we see that $\displaystyle \rho=\frac{M}{V}=\frac{M}{\pi R^2L}$. Substituting this value into the above result gives

$\displaystyle I_z=\frac{1}{2}MR^2$

Moments of Inertia of Some Rigid Objects:

Hoop or cylindrical shell: $I=MR^2$

Hollow cylinder: $\displaystyle I=\frac{1}{2}M\left(R_1^2+R_2^2 \right)$

Solid cylinder or disk: $\displaystyle I=\frac{1}{2}MR^2$

Rectangular plate of length $l$ and width $w$: $\displaystyle I=\frac{1}{12}M\left(l^2+w^2 \right)$

Long thin rod with rotation axis through its center: $\displaystyle I=\frac{1}{12}ML^2$

Long thin rod with rotation axis through end: $\displaystyle I=\frac{1}{3}ML^2$

Solid sphere: $\displaystyle I=\frac{2}{5}MR^2$

Thin spherical shell: $\displaystyle I=\frac{2}{3}MR^2$

The calculation of moments of inertia about an arbitrary axis can be somewhat cumbersome, even for a highly symmetric object. There is an important theorem, however, called the parallel-axis theorem, that often simplifies the calculation. Suppose the moment of inertia about any axis through the center of mass is $I_{CM}$. The parallel-axis theorem states that the moment of inertia about any axis that is parallel to and a distance $D$ away from the axis through the center of mass is

$I=I_{CM}+MD^2$

Proof of the Parallel-Axis Theorem. Suppose an object rotates in the $xy$ plane about the $z$-axis and the coordinates of the center of mass are $\displaystyle \left(x_{CM},y_{CM} \right)$. Let the mass element $\Delta m$ have coordinates $(x,y)$. Since this element is at a distance $r=\sqrt{x^2+y^2}$ from the $z$-axis, the moment of inertia about the $z$-axis is

$\displaystyle I=\int r^2\,dm=\int \left(x^2+y^2\right)\,dm$

However, we can relate the coordinates $(x,y)$ of the mass element $\Delta m$ to the coordinates of the mass element relative to the center of mass, $\left(x',y' \right)$. If the coordinates of the center of mass are $\left(x_{CM},y_{CM} \right)$, then the relationship between the unprimed and primed coordinates are $x=x'+x_{CM}$ and $y=y'+y_{CM}$. Therefore,

$\displaystyle I=\int \left[\left(x'+x_{CM} \right)^2+\left(y'+y_{CM} \right)^2 \right]\,dm=$

$\displaystyle \int \left[\left(x' \right)^2+\left(y' \right)^2 \right]\,dm+2x_{CM}\int x'\,dm+2y_{CM}\int y'\,dm+\left(x_{CM}^2+y_{CM}^2 \right)\int\,dm$

The first term on the right, is by definition, the moment of inertia about an axis that is parallel to the $z$-axis and passes through the center of mass. The second two terms on the right are zero because by definition of the center of mass, $\displaystyle \int x'\,dm=\int y'\,dm=0$. Finally, the last term on the right is simply $MD^2$, since $\displaystyle \int \,dm=M$ and $D^2=x_{CM}^2+y_{CM}^2$. Therefore, we conclude that

$I=I_{CM}+MD^2$