Ranking the CM height of plane figures

Problem

Four uniform plane figures:

• A: square of side $L$;
• B: equilateral triangle of side $L$;
• C: circle of diameter $L$;
• D: rectangle $L \times 2L$.

Rank the height of the centre of mass above the base, from lowest to highest.

Solution

For uniform figures the CM is the geometric centroid. Height above the base:

• A (square): $h_\text{CM} = L/2$;
• B (triangle): the centroid is at $1/3$ of the height from the base, $h_\text{CM} = L/3$;
• C (circle of diameter $L$): the centre is halfway up, $h_\text{CM} = L/2$;
• D (rectangle $L \times 2L$, base $L$, height $2L$): $h_\text{CM} = L$.

Ordering $L/3 < L/2 = L/2 < L$: $\ev{\text{B} < \text{A} = \text{C} < \text{D}}$

Links

Topics: Centre of mass Concepts: Centre of mass Skills: Ranking reasoning