Question

The centre of mass of a thin uniform rod of length L lies at a distance (from one end) of ________.

• A. $\frac{2L}{3}$
• B. $\frac{3L}{4}$
• C. $\frac{L}{2}$
• D. $\frac{L}{3}$
• E. $\frac{L}{4}$

Hint:

Center of mass of uniform objects is always the geometric center.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The center of mass of an object is the unique point where the weighted relative position of the distributed mass sums to zero. For a rigid body with a uniform density and a symmetric shape, the center of mass is located at its geometric center.
Step 2: Detailed Explanation
1. Analyze the object.
The object is a "thin uniform rod".
- "Uniform" means that its mass is distributed evenly along its length. The linear mass density (mass per unit length) is constant.
- A "thin rod" is a one-dimensional object, and its geometric shape is a line segment.
2. Identify the geometric center.
The geometric center of a line segment of length L is its midpoint.
3. Locate the center of mass.
Because the rod is uniform and symmetric, its center of mass coincides with its geometric center.
The midpoint of a rod of length L, measured from one end, is at a distance of half its length.
Distance from one end = \(\frac{L}{2}\).
Calculation using Integration (Formal Proof):
Let the rod lie along the x-axis from \(x=0\) to \(x=L\). Let \(\lambda\) be the constant linear mass density. A small element of length \(dx\) at position \(x\) has mass \(dm = \lambda dx\). The x-coordinate of the center of mass, \(x_{CM}\), is given by: \[ x_{CM} = \frac{\int x \, dm}{\int dm} = \frac{\int_0^L x (\lambda dx)}{\int_0^L \lambda dx} \] Since \(\lambda\) is constant, it cancels out: \[ x_{CM} = \frac{\int_0^L x \, dx}{\int_0^L dx} = \frac{[\frac{1}{2}x^2]_0^L}{[x]_0^L} = \frac{\frac{1}{2}L^2 - 0}{L - 0} = \frac{\frac{1}{2}L^2}{L} = \frac{L}{2} \] Step 4: Final Answer
The centre of mass is at a distance of \(\frac{L}{2}\) from one end.