Where the centre of gravity falls (rod and ball)

Problem

Where the centre of gravity falls. A metal rod $1\;\text{m}$ long and a lead ball of twice the rod’s mass are welded together: the rod at the centre, the ball at one end. Determine where the centre of gravity of the system falls and why.

Solution

Take the origin at the centre of the rod. The rod (mass $m$) has its centre of gravity at its own centre, $x = 0$. The ball (mass $2m$) is at the end $x = +L/2 = +0{,}5\;\text{m}$. Total mass $3m$: $x_\text{CM} = \frac{m\cdot 0 + 2m\cdot \tfrac{L}{2}}{3m} = \frac{2m\cdot 0{,}5}{3m} = \frac{1}{3}\;\text{m} \approx 0{,}33\;\text{m}$ $\ev{x_\text{CM} = \tfrac{1}{3}\;\text{m from the centre, toward the ball}}$ Since the rod runs from $-0{,}5$ to $+0{,}5\;\text{m}$, the value $0{,}33\;\text{m}$ falls inside the rod. The heavier ball pulls the centre of gravity toward itself, but not enough to take it outside the rod.

Links

Topics: Centre of mass Concepts: Centre of mass Skills: Calculating the centre of mass