2 Calculating the moment of inertia

Suppose a lamina is divided into a large number of small pieces or elements . A typical element is shown in Figure 9.

Figure 9 :

The element has mass $\delta m$ , and is located a distance $r$ from the axis through $O$ . The moment of inertia of this small piece about the given axis is defined to be $\delta m{r}^2$ , that is, the mass multiplied by the square of its distance from the axis of rotation. To find the total moment of inertia we sum the individual contributions to give

$\sum r^2\delta m$

where the sum must be taken in such a way that all parts of the lamina are included. As $\delta m\to 0$ we obtain the following integral as the definition of moment of inertia:

We shall illustrate how the moment of inertia is actually calculated in practice, in the following Tasks.

Task!

Calculate the moment of inertia about the $y$ -axis of the square lamina of mass $M$ and width $b$ , shown below. (The moment of inertia about the $y$ -axis is a measure of the resistance to rotation around this axis.)

Let the mass per unit area of the lamina be $\rho$ . Then, because its total area is $b^2$ , its total mass $M$ is $b^2\rho$ . Imagine that the lamina has been divided into a large number of thin vertical strips. A typical strip is shown in the figure above. The strips are chosen in this way because each point on a particular strip is approximately the same distance from the axis of rotation (the $y$ -axis).

1. Referring to the figure, write down the width of each strip:

   Answer

   $\delta x$

2. Write down the area of the strip:

   Answer

   $b\delta x$

3. With $\rho$ as the mass per unit area write down the mass of the strip:

   Answer

   $\rho b\delta x$

4. The distance of the strip from the $y$ -axis is $x$ . Write down its moment of inertia :

   Answer

   $mb{x}^2\delta x$

5. Adding contributions from all strips gives the expression $\sum \rho b{x}^2\delta x$ where the sum must be such that the entire lamina is included. As $\delta x\to 0$ the sum defines an integral. Write down this integral:

   Answer

   $I={\int \; <!--nolimits\; -->}_{-b∕2}^{b∕2}\rho b{x}^{2}dx$

6. Note that the limits on the integral have been chosen so that the whole lamina is included. Then

   $I=\rho b{\int \; <!--nolimits\; -->}_{-b∕2}^{b∕2}{x}^{2}dx$

   Evaluate this integral:

   Answer

   $I=\rho b{\left[\frac{x^3}{3}\right]}_{-b∕2}^{b∕2}=\frac{\rho b^4}{12}$

7. Write down an expression for $M$ in terms of $b$ and $\rho$ :

   Answer

   $M=b^2\rho$

8. Finally, write an expression for $I$ in terms of $M$ and $b$ :

   Answer

   $I=\frac{M{b}^2}{12}$

Task!

Find the moment of inertia of a circular disc of mass $M$ and radius $a$ about an axis passing through its centre and perpendicular to the disc.

The figure above shows the disc lying in the plane of the paper. Because of the circular symmetry the disc is divided into concentric rings of width $\delta r$ . A typical ring is shown below. Note that each point on the ring is approximately the same distance from the axis of rotation.

The ring has radius $r$ and inner circumference $2\pi r$ . Imagine cutting the ring and opening it up. Its area will be approximately that of a long thin rectangle of length $2\pi r$ and width $\delta r$ . Given that $\rho$ is the mass per unit area write down an expression for the mass of the ring:

Answer

The moment of inertia of the ring about $O$ is its mass multiplied by the square of its distance from the axis of rotation. This is $\left(2\pi r\rho \delta r\right)\times r^2=2\pi r^3\rho \delta r$ .

The contribution from all rings must be summed. This gives rise to the sum

${\sum }_{r=0}^{r=a}2\pi r^3\rho \delta r$

Note the way that the limits have been chosen so that all rings are included in the sum. As $\delta r\to 0$ the limit of the sum defines the integral

${\int \; <!--nolimits\; -->}_0^{a}2\rho \pi r^{3}dr$

Evaluate this integral to give the moment of inertia $I$ :

Answer

Write down the radius and area of the whole disc:

Answer

With $\rho$ as the mass per unit area, write down the mass of the disc $M$ :

Answer

Finally express $I$ in terms of $M$ and $a$ :

Answer

Exercises

1. The moment of inertia about a diameter of a sphere of radius 1 m and mass 1 kg is found by evaluating the integral $\frac{3}{8}{\int \; <!--nolimits\; -->}_{-1}^1{\left(1-x^2\right)}^{2}dx$ . Show that this moment of inertia is $0.4\text{kg}{\text{m}}^2$ .
2. Find the moment of inertia of the square lamina below about one of its sides.

   

3. Calculate the moment of inertia of a uniform thin rod of mass $M$ and length $\ell$ about a perpendicular axis of rotation at its end.
4. Calculate the moment of inertia of the rod in Exercise 3 about an axis through its centre and perpendicular to the rod.
5. The parallel axis theorem states that the moment of inertia about any axis is equal to the moment of inertia about a parallel axis through the centre of mass, plus the mass of the body $\times$ the square of the distance between the two axes. Verify this theorem for the rod in Exercise 3 and Exercise 4.
6. The perpendicular axis theorem applies to a lamina lying in the $xy$ plane. It states that the moment of inertia of the lamina about the $z$ -axis is equal to the sum of the moments of inertia about the $x$ - and $y$ -axes. Suppose that a thin circular disc of mass $M$ and radius $a$ lies in the $xy$ plane and the $z$ axis passes through its centre. The moment of inertia of the disc about this axis is $\frac{1}{2}M{a}^2$ .
   1. Use this theorem to find the moment of inertia of the disc about the $x$ and $y$ axes.
   2. Use the parallel axis theorem to find the moment of inertia of the disc about a tangential axis parallel to the plane of the disc.

Answer