The Center of Gravity — Question 1

Question 1

A geometric midpoint can fail as a balancing point

A thin horizontal rod extends from $x=0$ to $x=L$ and has linear density $\lambda(x)=\lambda_0(1+x/L)$. Gravity is uniform and downward.

1. Find its center of gravity.

2. Find the gravity torque about a support at the geometric midpoint, in terms of total mass $M$.

Question 1 - Solution

A geometric midpoint can fail as a balancing point

(a) Derivation and calculation

Integrating the stated line density gives $$M=\int_0^L\lambda_0(1+x/L)\,dx=\frac32\lambda_0L.$$ $$Mx_G=\int_0^Lx\lambda_0(1+x/L)\,dx=\frac56\lambda_0L^2.$$ Hence $$\boxed{x_G=\frac{5L}{9}}.$$

(b) Result and interpretation

The lever arm relative to the geometric midpoint is $$x_G-L/2=\frac{L}{18}.$$ With counterclockwise positive, the downward weight produces $$\boxed{\tau_{L/2}=-Mg\frac{L}{18}}.$$ Its clockwise sense follows from the heavier right half.

Physical check

Uniform gravity makes the center of gravity coincide with the center of mass.