Question

A rod has length \( L \). The linear mass density is \( 2x \, \text{kg/m} \), where \( x \) is the distance from the left end. The center of mass of the rod from the left end lies at a distance of

• A. \( \frac{L}{2} \)
• B. \( \frac{3L}{4} \)
• C. \( \frac{L}{3} \)
• D. \( \frac{L}{5} \)

Hint:

When dealing with a varying mass density, the center of mass is found by taking the weighted average of position \( x \) using the mass distribution \( \lambda(x) \) as the weight.

Answer 1

The Correct Option is C

We are given a rod of length \( L \), with a linear mass density that varies along the length of the rod. The linear mass density \( \lambda(x) \) at a point at a distance \( x \) from the left end of the rod is given by: \[ \lambda(x) = 2x \, \text{kg/m} \] The center of mass \( x_{\text{cm}} \) of the rod can be found using the following formula: \[ x_{\text{cm}} = \frac{\int_0^L x \, \lambda(x) \, dx}{\int_0^L \lambda(x) \, dx} \] First, we calculate the mass of the rod using the linear mass density. The total mass \( M \) of the rod is given by the integral of \( \lambda(x) \) over the length of the rod: \[ M = \int_0^L \lambda(x) \, dx = \int_0^L 2x \, dx \] Performing the integration: \[ M = \left[ x^2 \right]_0^L = L^2 \] Thus, the total mass of the rod is \( M = L^2 \, \text{kg} \). Now, we calculate the moment of mass distribution about the left end of the rod: \[ \int_0^L x \, \lambda(x) \, dx = \int_0^L x \cdot 2x \, dx = 2 \int_0^L x^2 \, dx \] Performing the integration: \[ \int_0^L x^2 \, dx = \frac{L^3}{3} \] Thus, the integral becomes: \[ \int_0^L x \, \lambda(x) \, dx = 2 \times \frac{L^3}{3} = \frac{2L^3}{3} \] Finally, the center of mass \( x_{\text{cm}} \) is: \[ x_{\text{cm}} = \frac{\frac{2L^3}{3}}{L^2} = \frac{2L}{3} \] Thus, the center of mass of the rod is at a distance \( \frac{L}{3} \) from the left end. So the correct answer is (C) \( \frac{L}{3} \).