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: Understanding the Problem The mass of a small element of the rod at a distance \( x \) from one end is given by: \[ dm = \rho(x) \, dx = \rho_0 \frac{x^2}{L^2} \, dx \] The position of the center of mass is given by the formula: \[ x_{\text{cm}} = \frac{\int_0^L x \, dm}{\int_0^L dm} \] Where: - The numerator is the first moment of mass, - The denominator is the total mass of the rod. Step 2: Total Mass of the Rod The total mass \( M \) of the rod is obtained by integrating the mass element \( dm \) over the length of the rod: \[ M = \int_0^L \rho(x) \, dx = \int_0^L \rho_0 \frac{x^2}{L^2} \, dx = \rho_0 \frac{1}{L^2} \int_0^L x^2 \, dx \] The integral of \( x^2 \) from 0 to \( L \) is: \[ \int_0^L x^2 \, dx = \frac{L^3}{3} \] Thus: \[ M = \rho_0 \frac{1}{L^2} \cdot \frac{L^3}{3} = \frac{\rho_0 L}{3} \] Step 3: First Moment of Mass The first moment of mass is: \[ \int_0^L x \, dm = \int_0^L x \rho_0 \frac{x^2}{L^2} \, dx = \rho_0 \frac{1}{L^2} \int_0^L x^3 \, dx \] The integral of \( x^3 \) from 0 to \( L \) is: \[ \int_0^L x^3 \, dx = \frac{L^4}{4} \] Thus: \[ \int_0^L x \, dm = \rho_0 \frac{1}{L^2} \cdot \frac{L^4}{4} = \frac{\rho_0 L^2}{4} \] Step 4: Position of the Center of Mass Now we can calculate the position of the center of mass: \[ x_{\text{cm}} = \frac{\frac{\rho_0 L^2}{4}}{\frac{\rho_0 L}{3}} = \frac{3L}{4} \] Step 5: Conclusion The position of the center of mass from one end is \( \frac{3L}{4} \). Thus, the correct answer is: \[ \boxed{(D)} \, \frac{3L}{4} \]