Find center of mass of L-shaped uniform lamina

Question

An L-shaped uniform lamina is formed by removing a 2 cm × 2 cm square from the top-right corner of a 4 cm × 4 cm square. Find the coordinates of the center of mass of the L-shaped lamina. (Take the bottom-left corner as origin.)

Solution — Step by Step

The original square has corners at (0,0), (4,0), (4,4), and (0,4).

The 2×2 square removed has corners at (2,2), (4,2), (4,4), and (2,4).

The L-shaped lamina = Full 4×4 square MINUS the 2×2 square.

Use the subtraction method: treat the L-shape as a full square minus a missing piece.

Full 4×4 square:

Area: $A_1 = 4 \times 4 = 16$ cm²
Centre of mass: $(x_1, y_1) = (2, 2)$ (centre of the square)

Removed 2×2 square (top-right piece):

Area: $A_2 = 2 \times 2 = 4$ cm²
Centre of mass: $(x_2, y_2) = (3, 3)$ (centre of the removed square, which spans x: 2 to 4, y: 2 to 4)

For the L-shape = Full − Removed:

x_{CM} = \frac{A_1 x_1 - A_2 x_2}{A_1 - A_2}

y_{CM} = \frac{A_1 y_1 - A_2 y_2}{A_1 - A_2}

x_{CM} = \frac{16 \times 2 - 4 \times 3}{16 - 4} = \frac{32 - 12}{12} = \frac{20}{12} = \frac{5}{3} \approx 1.67 \text{ cm}

y_{CM} = \frac{16 \times 2 - 4 \times 3}{16 - 4} = \frac{32 - 12}{12} = \frac{20}{12} = \frac{5}{3} \approx 1.67 \text{ cm}

The centre of mass of the L-shaped lamina is at $\left(\dfrac{5}{3}, \dfrac{5}{3}\right)$ cm $\approx (1.67, 1.67)$ cm.

Check: The CM of the full square was at (2, 2). The removed piece was in the top-right corner. The L-shape has more material in the bottom-left region, so the CM should shift toward the bottom-left compared to (2, 2). Our answer (1.67, 1.67) is indeed shifted toward bottom-left. ✓

Also note: the CM lies at equal x and y coordinates — this makes sense by symmetry, since the L-shape is symmetric about the diagonal $y = x$ .

Why This Works

The centre of mass is a weighted average of position, where weights are masses (or areas for uniform laminas). For a uniform lamina, mass is proportional to area.

The subtraction principle is elegant: if we know the CM of the whole and the CM of the piece removed, we can find the CM of the remainder:

A_{whole} \cdot \vec{r}_{whole} = A_{remaining} \cdot \vec{r}_{remaining} + A_{removed} \cdot \vec{r}_{removed}

Rearranging to solve for $\vec{r}_{remaining}$ gives the formula used above.

Alternative Method — Direct Integration (Dividing into Rectangles)

We can also divide the L-shape into two separate rectangles and use the addition formula.

Rectangle 1: Bottom strip: 4 cm wide, 2 cm tall (from y=0 to y=2)

Area: $A_1 = 8$ cm², CM: $(2, 1)$

Rectangle 2: Left strip: 2 cm wide, 2 cm tall (from y=2 to y=4)

Area: $A_2 = 4$ cm², CM: $(1, 3)$

x_{CM} = \frac{8 \times 2 + 4 \times 1}{8 + 4} = \frac{16 + 4}{12} = \frac{20}{12} = \frac{5}{3}

y_{CM} = \frac{8 \times 1 + 4 \times 3}{8 + 4} = \frac{8 + 12}{12} = \frac{20}{12} = \frac{5}{3}

Same answer: $\left(\frac{5}{3}, \frac{5}{3}\right)$ ✓

JEE Main CM problems on irregular laminas almost always reduce to either the subtraction method (remove a piece from a regular shape) or the addition method (divide into regular pieces). Practice both methods — sometimes one is much faster than the other depending on the shape given.

Common Mistake

Students often find the CM of the removed piece at (2,2) instead of (3,3). The removed square spans from x=2 to x=4 and y=2 to y=4 — its centre is at $\left(\frac{2+4}{2}, \frac{2+4}{2}\right) = (3, 3)$ , not (2, 2). Always compute the CM of each component piece carefully based on its actual boundaries, not by guessing.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Integration (Dividing into Rectangles)

Common Mistake

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