Center of Mass of Non-Uniform Rod Using Piecewise-Defined Function?

In summary, finding the center of mass of a non-uniform rod with a piecewise-defined function for its cross-section requires splitting the integral into sections that cover the individual domains of the function's pieces. Each volume element is calculated by squaring the cross-sectional area at a given location and multiplying it by the differential width. The linear density and center of mass can then be found using routine methods. The function does not necessarily have to be continuous, as long as the integral is split accordingly.
  • #1
Chrono G. Xay
92
3
How would one go about finding the center of mass of a non-uniform rod when its cross-section has been modeling using a piecewise-defined function?

For a specific case let's use a drumstick, noting its volume as a sum of half of a sphere (the 'butt' end), a cylinder (the 'shaft'), the majority of half of an ellipsoid (the 'shoulder'), and the majority of either a smaller sphere or smaller ellipsoid.

With an ellipsoidal tip, the piecewise-defined function for its cross-section is given by:

f(x) = sqrt( (3)^2 - ( x + 97 )^2 ), x<-97;
f(x) = 3, -97 < x < 0;
f(x) = (3)*sqrt( 1 - ( x / 36 )^2 ), 0<x<32.8456;
f(x) = (2)*sqrt( 1 - ( ( x - 36 ) / 4 )^2 ), 32.8456<x=<40

(The value "32.8456" was obtained using equations related to the 'vesica pisces')

ImageUploadedByPhysics Forums1448392716.525443.jpg


Now... If I were to take the integral of each piece... could I not easily compute the volume of the drumstick by squaring my sum and multiplying the result by π?

That might get me the volume, but I need to construct an equation for the object's linear density... I have read about one function called 'unit step' and another called 'heaviside', but am not sure which one would be appropriate, let alone how to input my pieces into the process...

The goal I am shooting for, assuming I can acquire an equation for the 'linear density' and 'center of mass' is to find the 'center of percussion'...
 
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  • #2
Integrating over x is the right idea. But you want to sum volume elements. Don't sum first and square later! Slicing the stick into thin disks of differential width dx, each having a cross sectional area related to the radius at the x location. So ##dv = \pi f(x)^2 dx##. Integrate to sum up all the volume elements. So, sum the squares rather than squaring the sums!

Fig1.png


If the material is of uniform density (such as a uniform piece of wood), multiply the volume element by that density ##\rho## to yield the mass of that volume element. So then each mass element is ##dm = \rho \pi f(x)^2 dx##.

You should be able to work out how to find the linear mass density and center of gravity with the above machinery :smile:
 
  • #3
But in order to obtain the equation for a non-uniform rod's linear density, do I not have to take the piecewise function modeling its cross-section, write the equation in unit step notation, take the Laplace transform, and so on?
 
  • #4
Chrono G. Xay said:
But in order to obtain the equation for a non-uniform rod's linear density, do I not have to take the piecewise function modeling its cross-section, write the equation in unit step notation, take the Laplace transform, and so on?
I don't see why you would need such a lot of mathematical machinery to do so. If the material that the object is made from is of a uniform density and you have its profile, then the rest seems to be accessible by routine methods. Given the cross-sectional area of the "rod" at some location along its length, the then the linear density there should be a simple calculation.
 
  • #5
I have not done this kind of thing before, so this is all new to me.

Can you explain some of the steps in the routine method you mentioned so that I can look them up? Someone I know had suggested going through a rather drawn-out process of plotting--inaccurate--points using the shadow of the object and having a computer approximate an equation connecting them all, which honestly sounds ridiculous. For the 'real world' volume it would be more accurate to measure how much water the rod pushes out of a graduated cylinder that had been filled to the very top.

I have watched a *several* videos online focused on finding the 'center of mass' of a non-uniform rod, and none of them had to model an object from the ground up, which might have involved the use of piecewise functions. They all had nice, neat, continuous functions.



...Come to think of it, if I just had an equation modeling the cross-section of the 'volume of revolution', wouldn't I just need to find the 'center of mass' along the cross-section's length? Nevertheless, it seems as though the multiple pieces would have to be written as a single, continuous function before anything else could be done...
 
Last edited:
  • #6
Chrono G. Xay said:
I have not done this kind of thing before, so this is all new to me.

Can you explain some of the steps in the routine method you mentioned so that I can look them up? Someone I know had suggested going through a rather drawn-out process of plotting--inaccurate--points using the shadow of the object and having a computer approximate an equation connecting them all, which honestly sounds ridiculous. For the 'real world' volume it would be more accurate to measure how much water the rod pushes out of a graduated cylinder that had been filled to the very top.

I have watched a *several* videos online focused on finding the 'center of mass' of a non-uniform rod, and none of them had to model an object from the ground up, which might have involved the use of piecewise functions. They all had nice, neat, continuous functions.
How the function is defined is not so important. A piecewise function can be treated in the same manner as a single continuous function except that you split the integral into sections that cover the individual domains of the function's pieces.

For example: Suppose your piecewise continuous function was defined as:

##f(x) = \left\{ \begin{array}{lr}
2 & : 0 \le x \le 3 \\
2 + x/3 & : 3 < x \le 6 \\
4 - (x - 6)^2 & : 6 \le x \le 8 \\
0 & Otherwise
\end{array}
\right.
##

Which would give a profile like this:
Fig1.png


Then for the volume an integration can be set up as:
$$V = \pi \int_0^3 2^2~dx + \pi \int_3^6 \left(2 + \frac{x}{3} \right)^2~dx + \pi \int_6^8 (4 - (x - 6)^2)^2~dx $$
Don't worry overly much about mathematical details like limits at the domain intersections -- this is supposedly a model of a real-world object, so it's an approximation to begin with. Do the integrations and you've got your volume.

Set up a similar integration to find the centroid, etc.
...Come to think of it, if I just had an equation modeling the cross-section of the 'volume of revolution', wouldn't I just need to find the 'center of mass' along the cross-section's length? Nevertheless, it seems as though the multiple pieces would have to be written as a single, continuous function before anything else could be done...
Yes, you just need to locate the center of mass along your rod's length since it is is rotationally symmetric about its lengthwise axis. No, you don't need to worry about re-writing the function. Carry out the integrations over the domains of the pieces.
 

1. What is the center of mass of a non-uniform rod using piecewise-defined function?

The center of mass of a non-uniform rod using piecewise-defined function is the point at which the entire mass of the rod can be considered to be concentrated, and where the rod will balance horizontally. It is calculated by dividing the sum of the products of each small mass element and its corresponding distance from a chosen reference point by the total mass of the rod.

2. How is the center of mass of a non-uniform rod calculated using piecewise-defined function?

The center of mass of a non-uniform rod using piecewise-defined function is calculated by dividing the sum of the products of each small mass element and its corresponding distance from a chosen reference point by the total mass of the rod. This can be represented mathematically as:

xcm = (1/M) * ∫x * dm
ycm = (1/M) * ∫y * dm
where x and y are the coordinates of each small mass element, M is the total mass of the rod, and dm is the mass element.

3. Why is the center of mass important in studying the motion of a non-uniform rod?

The center of mass is important in studying the motion of a non-uniform rod because it is the point where the entire mass of the rod can be considered to be concentrated. This means that the motion of the rod can be treated as if it were a point mass located at the center of mass. This simplifies the calculations and makes it easier to analyze the motion of the rod.

4. Can the center of mass of a non-uniform rod using piecewise-defined function be outside of the physical dimensions of the rod?

Yes, the center of mass of a non-uniform rod using piecewise-defined function can be outside of the physical dimensions of the rod. This can happen if the rod has a non-uniform distribution of mass, such as being thicker at one end. In this case, the center of mass will be closer to the heavier end, and may even be located outside of the physical dimensions of the rod.

5. How does the center of mass change if the rod is cut into smaller pieces?

If the rod is cut into smaller pieces, the center of mass will change. This is because the center of mass is affected by the distribution of mass along the length of the rod. Cutting the rod into smaller pieces will change this distribution, and therefore, change the location of the center of mass. However, if the smaller pieces are still connected to each other, the overall center of mass of the entire rod will remain the same.

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