Ranking the CM distance in dumbbells

Problem

Ranking. Four dumbbells (two masses joined by a rigid rod of negligible mass) with masses $m_1, m_2$ and length $L$:

• (A) $m_1 = 1$, $m_2 = 3$, $L = 4$;
• (B) $m_1 = 2$, $m_2 = 2$, $L = 2$;
• (C) $m_1 = 1$, $m_2 = 5$, $L = 6$;
• (D) $m_1 = 2$, $m_2 = 4$, $L = 3$.

Rank the distances of the centre of mass from mass $m_1$, from smallest to largest.

Solution

The distance of the CM from $m_1$ is $d = \dfrac{m_2}{m_1 + m_2}\,L$. Calculating for each dumbbell: $d_A = \frac{3}{4}\cdot 4 = 3 \qquad d_B = \frac{2}{4}\cdot 2 = 1$ $d_C = \frac{5}{6}\cdot 6 = 5 \qquad d_D = \frac{4}{6}\cdot 3 = 2$ Ordering $1 < 2 < 3 < 5$: $\ev{\text{B} < \text{D} < \text{A} < \text{C}}$

Links

Topics: Centre of mass Concepts: Centre of mass Skills: Calculating the centre of mass · Ranking reasoning