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Centre of mass

Centre of mass

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Summary

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 ABAB is 4m4m long and has a weight of 25N25N. It rests in a horizontal equilibrium on supports CC and DD, where AC=1mAC = 1m and AD=3mAD = 3m. The magnitude of the reaction at CC is four times the magnitude of the reaction at DD. Find the distance of the centre of mass of the rod from AA.

Maths; Moments; KS5 Year 13; Centre of mass


Find the magnitude of the reactions:

Take the reaction at DD to be RR, therefore the reaction at CC is 4R4R. Resolve vertically:


4R+R=255R=25R=5\begin{aligned}4R+R&=25\\ 5R&=25\\ R&=5\end{aligned}​​


Therefore, the reaction at CC is 20N20N and the reaction at DD is 5N5N. Take moments about point A, supposing the centre of mass of the rod acts a point xmxm from AA:


25x=20(1)+5(3)25x=20+1525x=35x=3525x=1.4\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 1.4m\underline{1.4m} from AA.


Example 2

Two people are sitting on a seesaw which is modelled as a non-uniform plank ABAB with a mass of 15kg15kg and a length of 4m4m. The plank rests on a pivot PP​ at the midpoint of ABAB. The centre of mass of the plank is at a point CC, where AC=1.5mAC = 1.5m​ Person FF has a mass of 20kg20kg and sits at point AA. Person GG has a mass of 35kg35kg and is sitting on the other side of the plank. Where on the plank does GG 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 PP. Take person GG's distance from the midpoint to be xmxm:


(20g×2)+(15g×0.5)=35g×x40g+7.5g=35xg47.5=35x1.36=x\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 GG must be sitting 1.36m\underline{1.36m} from the pivot.


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FAQs - Frequently Asked Questions

What does a non-uniform rod mean?

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