JEE Main: Centre of Mass of L-Shaped Rod — Smart Method 💡

 

❓ Question

A uniform rod of total length 5L is bent at a right angle, such that one arm has length 2L and the other arm has length 3L.

Find the position of the centre of mass of the system.

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🖼️ Question Image

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✍️ Short Solution

This is a standard JEE Centre of Mass problem.
No integration needed — treat each arm separately and use weighted average 🔥

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🔹 Step 1 — Split the system into two rods

Since the rod is uniform:

• Mass ∝ length

Let:

• Rod 1: length = 2L, along x-axis

• Rod 2: length = 3L, along y-axis

Assume:

• Total mass = M

• Mass per unit length = $\lambda$

Then:

• Mass of rod 1 = $2\lambda L$
  

• Mass of rod 2 = $3\lambda L$
  

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🔹 Step 2 — Locate centre of mass of each part

For a uniform straight rod, COM lies at its midpoint.

• COM of rod 1 (along x-axis):

  $$(x_1, y_1) = (L, 0)$$
  

• COM of rod 2 (along y-axis):

  $$(x_2, y_2) = (0, \tfrac{3L}{2})$$
  

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🔹 Step 3 — Use Centre of Mass formula

For a system of particles (or rigid bodies):

$$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}, \quad y_{cm} = \frac{\sum m_i y_i}{\sum m_i}$$

Total mass:

$$M = 2\lambda L + 3\lambda L = 5\lambda L$$

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🔹 Step 4 — Calculate x-coordinate of COM

$$x_{cm} = \frac{(2\lambda L)(L) + (3\lambda L)(0)}{5\lambda L} = \frac{2\lambda L^2}{5\lambda L} = \frac{2L}{5}$$

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🔹 Step 5 — Calculate y-coordinate of COM

$$y_{cm} = \frac{(2\lambda L)(0) + (3\lambda L)\left(\tfrac{3L}{2}\right)}{5\lambda L} = \frac{\tfrac{9}{2}\lambda L^2}{5\lambda L} = \frac{9L}{10}$$

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✅ Final Answer

The centre of mass of the bent rod is located at:

$$\boxed{\left(\frac{2L}{5},\ \frac{9L}{10}\right)}$$

(measured from the corner where the rod is bent)

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⭐ Golden JEE Insight

• Bent rod → split into straight parts

• Uniform rod → COM at midpoint

• Mass proportional to length

• Use weighted average, not intuition

🧠 One-line memory:

COM of composite body = weighted average of COMs