Question

Find the center of mass of a rod of length L whose mass density changes from one end to the other quadratically. That is, if the rod is laid out along the x-axis with one end at the origin and the other end at x=L, the density is given by rho(x)=rho0+(rho1−rho0)(x/L)², where rho0 and rho1 are constant values.

a) L/4
b) L/2
c) 3L/4
d) L

Answer 1

The center of mass of a rod with a quadratic density variation is determined by integrating the mass distribution and then solving for x_cm. After performing the necessary integration, the center of mass is found to be at L/2.

Step-by-step explanation:

To find the center of mass of a rod with varying density given by ρ(x) = ρ0 + (ρ1 − ρ0)(x/L)², we need to set up an integral that will allow us to consider the mass distribution along the length of the rod. The center of mass, xcm, is found using the formula xcm = (1/M) ∫ xρ(x)dx, where M is the total mass of the rod.

First, we calculate the total mass, M, by integrating the mass density function: M = ∫ ρ(x)dx from x=0 to x=L. This integral yields M as a function of ρ0, ρ1, and L. After finding M, we then integrate x multiplied by the mass density function to find the numerator for xcm:

∫ xρ(x)dx from x=0 to x=L.

Finally, solving for xcm will give us the position of the center of mass. After performing the integration and simplifying, we find that the center of mass is L/2, which corresponds to choice (b) in the given options.