Calculating the change in position of center of mass when additional weight is added to a non uniform object

Question

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 :

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.

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!

Answer 1

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.

Answer 2

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 $x^*$]

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 : $COM$ 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 ,

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