The Center of Mass — Question 4

Question 4

A rod that grows denser

A thin rod occupies $0\le x\le L$ and has linear density $\lambda(x)=\lambda_0(1+2x/L)$, with $\lambda_0>0$.

1. Find its total mass and center of mass.

2. For what value of $x$ is half its mass to the left? Is this the center of mass?

Question 4 - Solution

A rod that grows denser

(a) Derivation and calculation

Integrate the density and its first moment: $$M=\lambda_0\int_0^L\left(1+\frac{2x}{L}\right)dx=\boxed{2\lambda_0L}.$$ $$Mx_c=\lambda_0\int_0^L\left(x+\frac{2x^2}{L}\right)dx =\frac76\lambda_0L^2\Rightarrow\boxed{x_c=\frac{7L}{12}}.$$

(b) Result and interpretation

Let the half-mass coordinate be $x=uL$. Then $$M_{\mathrm{left}}=\lambda_0L(u+u^2)=\frac M2=\lambda_0L.$$ $$u^2+u-1=0\Rightarrow\boxed{x_{1/2}=\frac{\sqrt5-1}{2}L=0.618L}.$$ This differs from $x_c=0.583L$: equal mass on either side does not impose equal moments about that point.

Physical check

Both coordinates lie beyond the geometric midpoint because density increases toward the right.