To calculate the resisting moment of an object, you must determine its moment of inertia. The moment of inertia comprises a mathematical description of how easily an object rotates about a fixed axis due to the application of a force. It depends directly on the mass of the object, and how far the object extends from the axis of rotation.

As an example, think of a spinning merry-go-round at a playground. As the children on the merry-go-round move about, the rotation speed of the toy changes, because the children are changing the location of their mass.

Using a scale, find the mass of the object for which you need the moment of inertia. In this example, we'll suppose that you are studying a 5-kilogram steel rod.

Measure the length of the object if it is a rod, or the radius if it is a cylinder or sphere. Take these measurements in centimetres.

Here, we'll assume that the length of the steel rod is 50 centimetres.

Convert the length or radius measurement to meters by dividing it by 100 (since 100 centimetres equal 1 meter).

In the case of the steel rod, 50 centimetres divided by 100 gives us a total of 0.50 meters.

Use a table to locate the formula for the moment of inertia for the given object. The appropriate formula depends on how the object rotates around its axis. An axis is a line that passes through some portion of the object, usually an end or the centre.

The moment of inertia for a rod rotated around its centre equals

(1/12) x M x L^2

where "M" and "L" express the mass and length of the rod, respectively. The symbol "^" represents an exponent, and translates as "to the power." Thus, "^2" means "to the power of 2."

Plug the measured quantities of mass and length (or radius) into the moment of inertia formula. Using the example of the steel rod, we have

(1/12) x (5 kilograms) x (0.50 meters)^2

which gives us a resisting moment of 0.10 kilograms x meter^2.

The unit "meter^s" measures area. Therefore, 0.10 kilograms x meter^2 indicates a spreading out of mass. If the steel rod were longer, its mass would be even more spread out. This would increase its resisting moment, and make the bar more difficult to rotate.