Question:medium
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:
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:
Show 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.
Updated On: Nov 28, 2025
- \( \frac{2L}{3} \)
- \( \frac{L}{2} \)
- \( \frac{L}{3} \)
- \( \frac{3L}{4} \)