Question

Figure shows a composite system of two uniform rods of length as shown. The co-ordinates of the centre of mass of the system of rods are

• A. \(\left(\frac{L}{2},\frac{2L}{3}\right)\)
• B. \(\left(\frac{L}{4},\frac{2L}{3}\right)\)
• C. \(\left(\frac{L}{6},\frac{2L}{3}\right)\)
• D. \(\left(\frac{L}{6},\frac{L}{3}\right)\)

Hint:

Treat each rod as a point mass at its own centre and take the mass-weighted average of the coordinates.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The rods are uniform, so each acts like a point mass at its midpoint. The horizontal rod has length \(L\) along the x-axis and the vertical rod has length \(2L\) along the y-axis.

Step 2: Masses and Positions
With linear density \(\lambda\), the masses are \(m_1=\lambda L\) at \(\left(\dfrac{L}{2},0\right)\) and \(m_2=2\lambda L\) at \((0,L)\).

Step 3: X Coordinate
\(x_{cm}=\dfrac{\lambda L\cdot\frac{L}{2}+2\lambda L\cdot0}{3\lambda L}=\dfrac{L}{6}\).

Step 4: Y Coordinate
\(y_{cm}=\dfrac{\lambda L\cdot0+2\lambda L\cdot L}{3\lambda L}=\dfrac{2L}{3}\).

Final Answer:
The centre of mass is at (L/6, 2L/3). \[ \boxed{\text{(C) }\left(\dfrac{L}{6},\dfrac{2L}{3}\right)} \]