5. Centroid of an Area

by M. Bourne

Typical (straight sided) Problem:

In tilt-slab construction, we have a concrete wall (with doors and windows cut out of it) which we need to raise into position. We have only one rope and we must attach it to the centre of mass of the wall.

In this section we'll see how to find the centroid of an area with straight sides, then we'll extend this to areas with curved sides (where we'll use integration).

Moment

The moment of a mass is a measure of its tendency to rotate about a point. Clearly, the greater the mass (and the greater the distance from the point), the greater will be the tendency to rotate.

The moment is defined as:

Moment = mass × distance from a point

Example

In this case, there will be a total moment about O of:

(Clockwise is regarded as positive in this work.)

M = 2 × 1 − 10 × 3 = -28 kgm

Centre of Mass

We now aim to find the centre of mass of the system and this will lead to a more general result.

Example

We have 3 masses of 10 kg, 5 kg and 7 kg at 2 m, 2 m and 1 m distance from O as shown.

We wish to replace these masses with one single mass to give an equivalent moment. Where should we place this single mass?

Answer: Total moment = 10 × 2 + 5 × 4 + 7 × 5 = 75 kg.m

If we put the masses together, we have: 10 + 5 + 7 = 22 kg

For an equivalent moment, we need:

,

where is the distance from the centre of mass to the point of rotation.

i.e.

So our equivalent system (with one mass of 22 kg) would have:

Centre of Mass (Centroid) for a Thin Plate

1) Rectangle:

The centroid is (obviously) going to be exactly in the centre of the plate, at (2, 1).

2) Complex Shapes:

We divide the complex shape into rectangles and find (the x-coordinate of the centroid) and (the y-coordinate) by taking moments about the y and x coordinates respectively.

Because they are thin plates with a uniform density, we can just calculate moments using the area.

Example:

Find the centroid of the shape:

Answer: We divide the area into 2 rectangles and assume the mass of each rectangle is concentrated at the centre.

Left rectangle: Area = 3 × 2 = 6 sq unit. Centre (-1/2, 1)

Right rectangle: Area = 2 × 4 = 8 sq unit. Centre (2,2)

Taking moments with respect to the y-axis, we have:

Now, w.r.t the x-axis:

So the centroid is at:

This is how we would solve the tilt slab construction problem mentioned at the beginning of this section.

In general, we can say:

This idea is used more extensively in the next section.

Centroid for Curved Areas

To find the centroid, we use the same basic idea that we were using for the straight-sided case. The "typical" rectangle indicated has width Δx and height y₂ − y₁, so the total moments in the x-direction over the total area is given by:

For the y coordinate, we have to re-express the expressions y₂ and y₁ as functions of y.

Example:

Find the centroid of the area bounded by y = x³, x = 2 and the x-axis.

Answer: In this case, y₂= x³, y₁= 0, a = 0, b = 2.

Now, for the y coordinate, we need to find:

x₂=2 (this is fixed in this problem)

x₁= y^1/3 (this is variable in this problem)

c = 0, d = 8.

So the centroid is at (1.6, 2.29).