Centre of mass

The idea

Every equilibrium and dynamics problem puts the weight $mg$ at the centre of mass. For a uniform rod that’s obviously the middle, but for a bracket, a robot chassis or a gripper finger you have to work it out.

The idea comes straight from the parallel force system in Forces and moments: the moment of the whole object’s weight about a point equals the sum of the moments of all its parts about that point.

$\bar{x}\sum W_i=\sum W_i{x}_i\Rightarrow \boxed{\bar{x}=\frac{\sum A_i{x}_i}{\sum A_i}}$

For a flat plate of uniform thickness and density, the weight of each part is proportional to its area, so you can use areas directly. The same works for $\bar{y}$ with the $y_i$ values.

Composite shapes

Split the shape into simple pieces whose centroids you know:

Shape	Centroid
Rectangle	The middle
Triangle	$\frac{1}{3}$ of the height up from the base
Circle or semicircle	Centre; a semicircle’s is $4r/3\pi$ from the flat edge

Then tabulate area × position and divide. Holes count as negative area.

Worked example: an L-shaped bracket

The bracket is a 20 × 100 mm upright with an 80 × 20 mm foot attached at the bottom right. Find its centre of mass from the bottom-left corner.

Answer

Part	A (mm²)	$x_i$	$y_i$	$A{x}_i$	$A{y}_i$
1 upright	2000	10	50	20 000	100 000
2 foot	1600	60	10	96 000	16 000
Total	3600			116 000	116 000

$\bar{x}=116000/3600=$ 32.2 mm · $\bar{y}=116000/3600=$ 32.2 mm G sits in the empty corner of the L, outside the material. That’s normal for hollow or bent shapes.

Curved shapes: integrate with strips

When the edge is a curve $y(x)$, slice the shape into thin vertical strips of width $\delta x$:

1. Whole shape: area $A=\int ydx$, and its moment is $A\bar{x}$.
2. One strip at distance x: area $y\delta x$, moment $xy\delta x$.
3. Equate the whole shape’s moment with the total of all the strips:

$\boxed{\bar{x}=\frac{\int xydx}{A}}\boxed{\bar{y}=\frac{\int \frac{1}{2}{y}^{2}dx}{A}}$

For $\bar{y}$, each vertical strip’s own centroid is halfway up it, at $y/2$, which is where the $\frac{1}{2}{y}^2$ comes from. (Equivalently, use horizontal strips of width $x(y)$: $\bar{y}=\int yx(y)dy/A$.)

Worked example: under a parabola

Find the centroid of the region under $y=x^2$ from $x=0$ to $x=2$.

Answer

$A={\int }_0^2{x}^{2}dx=[x^3/3{]}_0^2=8/3$ ${\int }_0^{2}x\cdot x^{2}dx=[x^4/4{]}_0^2=4$ → $\bar{x}=4/(8/3)=$ 1.5 ${\int }_0^2\frac{1}{2}{x}^{4}dx=[x^5/10{]}_0^2=3.2$ → $\bar{y}=3.2/(8/3)=$ 1.2 Sense check: the shape is fatter on the right, so $\bar{x}$ is past the halfway point (1), and it’s flat along the bottom, so $\bar{y}$ is low.

Why it matters for robots

• Tipping: a mobile robot tips over when its centre of mass moves outside the wheelbase, for example when an arm reaches out or it climbs a slope. Keeping G low and central keeps it stable.
• Motor torque: an arm joint has to hold the load’s weight at the distance to its centre of mass, $T=mgd$. Moving the motor close to the base reduces that d for the joints above it.
• Inertia: a uniform slender rod’s moment of inertia about its centre, $I=\frac{1}{12}m{L}^2$, assumes the mass is spread evenly with G at the middle. Real links with holes, bosses and fittings have their G elsewhere.

Common mistakes

• Adding a hole instead of subtracting it.
• Measuring every $x_i$ from a different origin. Pick one corner and stick to it.
• Forgetting the ½ in the $\bar{y}$ integral.
• Assuming G must be on the material. It doesn’t have to be.

Check yourself

1. A 100 × 60 mm plate has a 20 mm diameter hole centred at (70, 30) mm. Find $\bar{x}$.

Plate: 6000 mm² at x = 50. Hole: $\pi \times 10^2=314$ mm² at x = 70. $\bar{x}=(6000\times 50-314\times 70)/(6000-314)=278009/5686=$ 48.9 mm. The hole pulls G slightly left, away from it. $\bar{y}=30$ mm by symmetry.

2. A right-angled triangle has a 90 mm base along the x-axis and a 60 mm height up the y-axis. Where's its centroid?

One third along each leg from the right angle: (30, 20) mm.

Back to WM1AC, or practise in WM1AC practice questions.