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Hi guys I have a good fundamental on basic calculations for position of center of mass, however I am having troubles creating my modelling for my problem as stated above.

To make the model easier, let's assume our non uniform object is 2Dimensional.

My objective is divided into two tasks:

  1. Finding the original position of center of mass of my non uniform object.
  2. Adding weight onto on end of the non uniform object and identifying the shift in position of the center of mass.

I understand that I should use this formula : this

However I am confused on how I should even model my non uniform object, and how to go about the calculation when we add additional weight onto it, or how to create my own weight distribution profile, and integrate the weight function from there. Example : I have a hammer and I add 10N to the end of the handle.

enter image description here

Any help or any example that you guys can create to aid me in this modelling problem will be greatly appreciated. Thank you so much!

Please feel free to add any new assumptions if I left out any details!

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2 Answers 2

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If the 10 Newton force was the weight of a mass stuck to the hammer, then yes the center of mass would move.

A good way to imagine where the new center of mass is, is to use the principle of moments (or the sum of torques is zero).

Without the 10N force the weight of the hammer $Mg$ acts through the cross. With the 10N force the center of mass moves, let's say to the position of the second vertical line, a distance $d$ to the right of the cross, and we want to find $d$.

If we were to put a thin support at that position the hammer (with the 10N) should be balanced, so from moments

$$Mg \times d = 10 (x-d)$$

where $x$ is the distance from the 10N to the cross.

rearranging gives

$$d=\frac{10x}{Mg+10}$$

that's how far the center of mass moves when a mass of weight 10N is added.

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  • $\begingroup$ Dear John, Thank you for your answer! I understood your explanation. Just to clarify, that calculation would only work if we are assuming that the hammer (our object) has a uniform density. May I ask how would you tackle this question when the object is non uniform? $\endgroup$
    – helpmepls
    Jun 8, 2021 at 7:56
  • $\begingroup$ It also works for an object of non-uniform density, such as the hammer in your diagram, as the original center of mass position includes all the information about varying density $\endgroup$ Jun 8, 2021 at 8:00
  • $\begingroup$ Okay thank you! Could you also share some knowledge on how you would find the center of mass of the hammer via mathematical calculations? I would really appreciate it if you could include a weight distribution function. $\endgroup$
    – helpmepls
    Jun 8, 2021 at 8:27
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Let the distance of $COM$ of the hammer from the Right most end be $x$ and Total length of the Hammer is $l$ .

$\rightarrow$ See , If we fix the point of $COM$ [ Or make $CO M$ the pivot ] of a Rod , the net Torque must be $0$ , i.e. $\sum\tau=0$ . So , if $COM$ is the Origin then : [in this image due to writing mistake $x$ refers to $x^*$]

enter image description here

So I can say : $$\int_{0}^{l-x^*}g\cdot k\cdot dm=10x^*+\int_{0}^{x^*}g\cdot k\cdot dm$$

Just solve these integrals .

$\rightarrow$ The initial case was : I've fixed the <span class=$COM$ of the Hammer " />

Where , $m'+m''=m$ . So , the Torque on left side of the origin will be equal to the Torque on Right side . [$\tau=\perp F\cdot r$] . Hence I can write : $$\int_{0}^{l-x}g\cdot k\cdot dm=\int_{0}^{x}g\cdot k\cdot dm$$

$\Rightarrow$ Solve for $x^*$ from here .

$$10x^*+\int_{0}^{x^*}g\cdot k\cdot dm-\int_{0}^{l-x^*}g\cdot k\cdot dm=\int_{0}^{x}g\cdot k\cdot dm-\int_{0}^{l-x}g\cdot k\cdot dm$$

$$10x^*+\frac{mgx^{*2}}{2l}-\frac{mg(l-x^*)^2}{2l}=\frac{mgx^2}{2l}-\frac{mg(l-x)^2}{2l}$$

Or ,

$$\frac{20lx^*}{mg}+2lx^*=2lx$$

And Finally ,

$$x^*=x\left(\frac{10}{mg}+1\right)^{-1}$$

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  • $\begingroup$ From here you can calculate change in position of $COM $ by calculating $x^*$ $\endgroup$
    – user303440
    Jun 8, 2021 at 12:18
  • $\begingroup$ Equate both the integrals to get to $x^*$ , i.e. new position of $COM$ $\endgroup$
    – user303440
    Jun 8, 2021 at 12:20
  • $\begingroup$ Please UpVote if you like my Solution ........... $\endgroup$
    – user303440
    Jun 8, 2021 at 12:39
  • $\begingroup$ Dear Keshav, I do not understand how you managed to reach your final answer of x*=x(1−10/mg). In your final answer, there are only 3 variables. x* (new position of COM), x (initial position of COM), the value 10(which is the external force applied). In the final answer, it doesn't include the distance of where the external force is applied. I believe we have to factor that in since the position of external force also affects the new COM position right? $\endgroup$
    – helpmepls
    Jun 9, 2021 at 8:08
  • $\begingroup$ See , I just balanced the Torque on both the sides as you can see . And the position where the Force is applied matters ....... but the position where the Force is applied is known , So there might be no problem to find $x^*$ $\endgroup$
    – user303440
    Jun 9, 2021 at 10:15

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