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 moment of inertia of the small element is delta m r squared}

The element has mass [maths rendering] , and is located a distance [maths rendering] from the axis through [maths rendering] . The moment of inertia of this small piece about the given axis is defined to be [maths rendering] , 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

[maths rendering]

where the sum must be taken in such a way that all parts of the lamina are included. As [maths rendering] we obtain the following integral as the definition of moment of inertia:

Key Point 5
[maths rendering]
where the limits of integration are chosen so that the entire lamina is included.

The unit of moment of inertia is [maths rendering] .

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 [maths rendering] -axis of the square lamina of mass [maths rendering] and width [maths rendering] , shown below. (The moment of inertia about the [maths rendering] -axis is a measure of the resistance to rotation around this axis.)

{A square lamina rotating about the y-axis.}

Let the mass per unit area of the lamina be [maths rendering] . Then, because its total area is [maths rendering] , its total mass [maths rendering] is [maths rendering] . 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 [maths rendering] -axis).

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

    [maths rendering]

  2. Write down the area of the strip:

    [maths rendering]

  3. With [maths rendering] as the mass per unit area write down the mass of the strip:

    [maths rendering]

  4. The distance of the strip from the [maths rendering] -axis is [maths rendering] . Write down its moment of inertia :

    [maths rendering]

  5. Adding contributions from all strips gives the expression [maths rendering] where the sum must be such that the entire lamina is included. As [maths rendering] the sum defines an integral. Write down this integral:

    [maths rendering]

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

    [maths rendering]

    Evaluate this integral:

    [maths rendering]

  7. Write down an expression for [maths rendering] in terms of [maths rendering] and [maths rendering] :

    [maths rendering]

  8. Finally, write an expression for [maths rendering] in terms of [maths rendering] and [maths rendering] :

    [maths rendering]

Task!

Find the moment of inertia of a circular disc of mass [maths rendering] and radius [maths rendering] about an axis passing through its centre and perpendicular to the disc.

{ A circular disc rotating about an axis through O.}

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 [maths rendering] . A typical ring is shown below. Note that each point on the ring is approximately the same distance from the axis of rotation.

{The lamina is divided into many circular rings.}

The ring has radius [maths rendering] and inner circumference [maths rendering] . Imagine cutting the ring and opening it up. Its area will be approximately that of a long thin rectangle of length [maths rendering] and width [maths rendering] . Given that [maths rendering] is the mass per unit area write down an expression for the mass of the ring:

[maths rendering]

The moment of inertia of the ring about [maths rendering] is its mass multiplied by the square of its distance from the axis of rotation. This is [maths rendering] .

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

[maths rendering]

Note the way that the limits have been chosen so that all rings are included in the sum. As [maths rendering] the limit of the sum defines the integral

[maths rendering]

Evaluate this integral to give the moment of inertia [maths rendering] :

[maths rendering]

Write down the radius and area of the whole disc:

[maths rendering]

With [maths rendering] as the mass per unit area, write down the mass of the disc [maths rendering] :

[maths rendering]

Finally express [maths rendering] in terms of [maths rendering] and [maths rendering] :

[maths rendering]

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 [maths rendering] . Show that this moment of inertia is [maths rendering] .
  2. Find the moment of inertia of the square lamina below about one of its sides.

    No alt text was set. Please request alt text from the person who provided you with this resource.

  3. Calculate the moment of inertia of a uniform thin rod of mass [maths rendering] and length [maths rendering] 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 [maths rendering] 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 [maths rendering] plane. It states that the moment of inertia of the lamina about the [maths rendering] -axis is equal to the sum of the moments of inertia about the [maths rendering] - and [maths rendering] -axes. Suppose that a thin circular disc of mass [maths rendering] and radius [maths rendering] lies in the [maths rendering] plane and the [maths rendering] axis passes through its centre. The moment of inertia of the disc about this axis is [maths rendering] .
    1. Use this theorem to find the moment of inertia of the disc about the [maths rendering] and [maths rendering] 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.
  1. [maths rendering] .
  2. [maths rendering] .
  3. [maths rendering] .
  4. Solution omitted
    1. The moments of inertia about the [maths rendering] and [maths rendering] axes must be the same by symmetry, and are equal to [maths rendering] .
    2. [maths rendering] .