Centre of mass

In a nutshell

Objects can be non-uniform, which means the centre of mass is not necessarily at the midpoint. Using forces and moments you can find the centre of mass.

Centre of mass

With a non-uniform rod, you may have to find the position of its centre of mass. The centre of mass can be found by considering the moment due to the weight of the rod. By taking moments about a point, you can create an equation to find the distance of the centre of mass from the point.

Example 1

A non-uniform rod $AB$ is $4m$ long and has a weight of $25N$ . It rests in a horizontal equilibrium on supports $C$ and $D$ , where $AC = 1m$ and $AD = 3m$ . The magnitude of the reaction at $C$ is four times the magnitude of the reaction at $D$ . Find the distance of the centre of mass of the rod from $A$ .

Maths; Moments; KS5 Year 13; Centre of mass

Find the magnitude of the reactions:

Take the reaction at $D$ to be $R$ , therefore the reaction at $C$ is $4R$ . Resolve vertically:

$\begin{aligned}4R+R&=25\\ 5R&=25\\ R&=5\end{aligned}$

Therefore, the reaction at $C$ is $20N$ and the reaction at $D$ is $5N$ . Take moments about point A, supposing the centre of mass of the rod acts a point $xm$ from $A$ :

$\begin{aligned}25x&=20(1)+5(3)\\ 25x&=20+15\\25x&=35\\ x&=\dfrac{35}{25}\\ x&=1.4\end{aligned}$

Therefore, the centre of the mass of the rod is $\underline{1.4m}$ from $A$ .

Example 2

Two people are sitting on a seesaw which is modelled as a non-uniform plank $AB$ with a mass of $15kg$ and a length of $4m$ . The plank rests on a pivot $P$ at the midpoint of $AB$ . The centre of mass of the plank is at a point $C$ , where $AC = 1.5m$ Person $F$ has a mass of $20kg$ and sits at point $A$ . Person $G$ has a mass of $35kg$ and is sitting on the other side of the plank. Where on the plank does $G$ have to sit for the plank to be horizontal? (G can be moved from the end of the plank B towards the pivot).

Maths; Moments; KS5 Year 13; Centre of mass

Take moments about the pivot $P$ . Take person $G$ 's distance from the midpoint to be $xm$ :

$\begin{aligned}(20g\times 2)+(15g\times 0.5)&=35g\times x\\ 40g+7.5g&=35xg\\47.5&=35x\\ 1.36&=x\end{aligned}$

Therefore for the plank to be horizontal, person $G$ must be sitting $\underline{1.36m}$ from the pivot.