Question

The variation of the density of a solid cylindrical rod of cross-sectional area \( \alpha \) and length \( L \) is given by: \[ \rho(x) = \rho_0 \frac{x^2}{L^2} \] Where \( x \) is the distance from one end of the rod. The position of its center of mass from one end is:

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

Hint:

For a rod with varying density, the center of mass can be calculated by integrating the mass elements and dividing the first moment by the total mass.

Answer 1

The Correct Option is D

Step 1: Problem Setup The mass of a small rod element at distance \( x \) is: \[\ndm = \rho(x) \, dx = \rho_0 \frac{x^2}{L^2} \, dx\n\] The center of mass position is: \[\nx_{\text{cm}} = \frac{\int_0^L x \, dm}{\int_0^L dm}\n\] Where: - Numerator: first moment of mass - Denominator: total mass Step 2: Total Mass Calculation The total mass \( M \) is found by integrating \( dm \): \[\nM = \int_0^L \rho(x) \, dx = \int_0^L \rho_0 \frac{x^2}{L^2} \, dx = \rho_0 \frac{1}{L^2} \int_0^L x^2 \, dx\n\] The integral of \( x^2 \) from 0 to \( L \) is: \[\n\int_0^L x^2 \, dx = \frac{L^3}{3}\n\] Therefore: \[\nM = \rho_0 \frac{1}{L^2} \cdot \frac{L^3}{3} = \frac{\rho_0 L}{3}\n\] Step 3: First Moment of Mass Calculation The first moment of mass is: \[\n\int_0^L x \, dm = \int_0^L x \rho_0 \frac{x^2}{L^2} \, dx = \rho_0 \frac{1}{L^2} \int_0^L x^3 \, dx\n\] The integral of \( x^3 \) from 0 to \( L \) is: \[\n\int_0^L x^3 \, dx = \frac{L^4}{4}\n\] Therefore: \[\n\int_0^L x \, dm = \rho_0 \frac{1}{L^2} \cdot \frac{L^4}{4} = \frac{\rho_0 L^2}{4}\n\] Step 4: Center of Mass Position Calculate the center of mass: \[\nx_{\text{cm}} = \frac{\frac{\rho_0 L^2}{4}}{\frac{\rho_0 L}{3}} = \frac{3L}{4}\n\] Step 5: Conclusion The center of mass position is \( \frac{3L}{4} \). Thus, the answer is: \[\n\boxed{(D)} \, \frac{3L}{4}\n\]