# 6. Moments of Inertia by Integration

by M. Bourne

The **moment of inertia** is a measure of the resistance of a rotating body to a change in motion.

The moment of inertia of a particle of mass *m *rotating about a particular point is given by:

`"Moment of inertia" = md^2`

where *d* is the radius of rotation.

## Inertia for a Collection of Particles

If a group of particles with masses *m*_{1}, *m*_{2},* m*_{3}, ... , *m*_{n} is rotating around a point with distances *d*_{1}, *d*_{2}, *d*_{3}, ... *d*_{n}, (respectively) from the point, then the moment of inertia *I* is given by:

I=m_{1}d_{1}^{2 }+m_{2}d_{2}^{2 }+m_{3}d_{3}^{2}+... +m_{n}d_{n}^{2}

If we wish to place all the masses at the one point (*R *units from the point of rotation) then

d_{1}=d_{2}=d_{3}= ... =d_{n }=Rand we can write:

I= (m_{1}+m_{2}+m_{3}... +m_{n})R^{2}

*R* is called the **radius of gyration.**

### Example 1

Find the moment of inertia and the radius of gyration w.r.t. the origin (0,0) of a system which has masses at the points given:

Mass |
`6` | `5` | `9` | `2` |

Point |
`(-3, 0)` | `(-2, 0)` | `(1, 0)` | `(8, 0)` |

**Moment of Inertia for Areas**

We want to find the moment of inertia, *I*_{y} of the given area, which is rotating around the *y*-axis.

Each "typical" rectangle indicated has width *dx* and height *y*_{2} − *y*_{1}, so its area is (*y*_{2} − *y*_{1})*dx*.

If *k* is the mass per unit area, then each typical rectangle has mass *k*(*y*_{2} −* y*

_{1})

*.*

*dx*The moment of inertia for each typical rectangle is [*k*(*y*_{2} −* y*

_{1})

*]*

*dx**x*

^{2}, since each rectangle is

*x*units from the

*y*-axis.

We can add the moments of inertia for all the typical rectangles making up the area using integration:

`I_y=kint_a^bx^2(y_2-y_1)dx`

Using a similar process that we used for the collection of particles above, the radius of gyration *R*_{y} is given by:

`R_y=sqrt((I_y)/m`

where *m* is the mass of the area.

### Example 2

For the first quadrant area bounded by the curve

`y = 1 − x^2`,

find:

a) The moment of inertia w.r.t the *y* axis. (*I*_{y})

b) The mass of the area

c) Hence, find the radius of gyration

## Rotation about the *x*-axis

For rotation about the *x*-axis, the moment of inertia formulae become:

`I_x=kint_c^(d)y^2(x_2-x_1)dy`

and

`R_x=sqrt((I_x)/m`

### Example 3

Find the moment of inertia and the radius of gyration for the area `y=x^2+1` from `x=1` to `x=2`, and `y>1`, when rotated around the *x*-axis. The mass per unit area is `3` kg m^{−2}.

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