Density and Pressure — Question 3

Question 3

A local density is not generally a whole-column density

A vertical sample of height $H$ has constant horizontal area $A$ and density $\rho(z)=\rho_0(1+z/H)$, where $z$ runs upward from its base. This profile is prescribed at the instant of measurement; no claim of stable equilibrium is made.

1. Find the total mass and mean density. Is the density at midheight equal to the mean?

2. Find the height of its center of mass and explain its displacement from midheight.

Question 3 - Solution

A local density is not generally a whole-column density

(a) Derivation and calculation

$$dm=A\rho(z)\,dz,\quad M=\rho_0A\int_0^H(1+z/H)\,dz=\boxed{\tfrac32\rho_0AH}.$$$$\boxed{\bar\rho=M/(AH)=\tfrac32\rho_0=\rho(H/2)}.$$The equality with the midpoint value follows specifically from the linear profile and constant area.

(b) Result and interpretation

$$Mz_{\mathrm{cm}}=\rho_0A\int_0^H(z+z^2/H)\,dz=\frac56\rho_0AH^2.$$$$\boxed{z_{\mathrm{cm}}=\frac{(5/6)\rho_0AH^2}{(3/2)\rho_0AH}=\frac{5H}{9}}.$$$$z_{\mathrm{cm}}-H/2=H/18>0.$$

Physical check

The heavier upper slices shift the center upward; the prescribed profile is not asserted to be stable.