Q.Find the centre of mass of a uniform L-shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 kg.

The figure shows an L-shaped lamina — a flat, rigid sheet of uniform density — placed on an $xy$-coordinate grid. The shape occupies three unit squares, each of side length $1\ \text{m}$, arranged to form an L. The outline runs from the origin $O(0,0)$ to $A(2,0)$, then up to $B(2,1)$, left to $D(1,1)$, up to $E(1,2)$, left to $F(0,2)$, and back down to $O$. The axes are labelled $x$ and $y$, and the dimensions given are $2\ \text{m}$ along the longer arms and $1\ \text{m}$ along the shorter ones.

Three dots mark the centres of the individual unit squares: $C_1$ at $(\frac12,\frac12)$, $C_2$ at $(\frac32,\frac12)$, and $C_3$ at $(\frac12,\frac32)$. Each centre is the geometric centroid of its square — the point where the square’s mass would be concentrated if it were a uniform lamina. Because the whole L-shaped lamina has uniform density, the mass of each square is the same, and the total mass is simply three times the mass of one square.

The physical idea the figure teaches is this: for a system of discrete particles (or, as here, for a continuous body broken into equal-mass pieces), the centre of mass is the weighted average of the positions of the individual masses. Since each square has equal mass, the centre of mass of the whole L-shaped lamina is just the average of the three centre coordinates:

With $m_1=m_2=m_3=m$, this simplifies to

Substituting the coordinates:

So the centre of mass of the L-shaped lamina lies at $(\frac56,\frac56)$, a point inside the shape but not at any of the square centres — it is pulled toward the corner where the three squares meet.

The textbook uses this figure in Example 6.3 to demonstrate the method of dividing a continuous body into discrete, equal-mass parts and then applying the centre-of-mass formula. The key formula the example develops is the one above: for $n$ equal masses, the centre of mass is simply the arithmetic mean of their position vectors.

Concept

For a composite rigid body, the centre of mass is the mass-weighted average of the centres of mass of its parts:

Because the lamina is uniform, mass is proportional to area, so equal-area pieces carry equal mass.

Why this method …

Concept: Centre of Mass of a Composite Rigid Body

Step 1: Split the lamina into simple, symmetric pieces

The L-shape decomposes into three identical $1\text{ m}\times1\text{ m}$ squares: bottom-left ($0\le x\le1,\,0\le y\le1$), bottom-right ($1\le x\le2,\,0\le y\le1$), and top-left ($0\le x\le1,\,1\le y\le2$).

Step 2: Assign mass to each square

Total area $=3\text{ m}^2$, total mass $=3$ kg, and the lamina is uniform, so each square (area $1\text{ m}^2$) has mass $m_1=m_2=m_3=1$ kg.

Step 3: Locate each square's own centre

Step 4: Apply the composite-body centre-of-mass formula

$X=\frac{\sum m_i x_i}{\sum m_i}=\frac{1(\tfrac12)+1(\tfrac32)+1(\tfrac12)}{3}=\frac{5/2}{3}=\frac{5}{6}\text{ m}$ …