Find Centre of Mass of a Metal Sheet

[Elena14]

A thin sheet of metal of uniform thickness is cut into the shape bounded by the line x=a, y=kx^2 and y=-kx^2 . Find coordinates of center of mass.
My attempt at the solution : To apply the formula r(c.m) = (Σm1a1)/Σa1 ; a is the area; we need to know the area but we have just been given the equation of curve. I thought of calculating the area under the graph using integration but even that didn't get me to the answer. Integrating gave the area under one curve = (2*k*a^3)/3
I am not even sure if my approach is correct.

 

[gneill]

Hi Elena14, Welcome to Physics Forums.

How does your center of mass formula change when the distribution of mass is continuous rather than discrete? In other words, what is the formula when you need to integrate over the object rather than sum contributions from individual lumps of mass?

 

[Elena14]

That formula can be derived when thickness and density is same for a system of particles, and the question talks about cutting shapes from the same piece of metal. So I thought it will work.
ps: I am not exactly sure what you mean?

 

[gneill]

That formula can be derived when thickness and density is same for a system of particles, and the question talks about cutting shapes from the same piece of metal. So I thought it will work.
ps: I am not exactly sure what you mean?

When the scenario is a system of discrete particles you find the center of mass using a summation formula, summing the contributions of each particle. When the scenario is a continuous mass distribution you need to use integration instead. How does the formula look in that form? What comprises the "mass elements" in that case?

 

[Elena14]

But then how else can that equation y=kx^2 be used in the question if not for calculating area?

 

[gneill]

But then how else can that equation y=kx^2 be used in the question if not for calculating area?

It's also used to form the differential mass area elements (dm dA) that you will be using in the center of mass formula.

[Edit: Updated the text and image to indicate area elements rather than mass elements]

 

[Elena14]

I am reattempting the question using ∫x dm/ ∫ dm but what should I substitute with x?

 

[gneill]

I am reattempting the question but I am confused as to what should be dx!

dx is the differential length element of x. You'll be integrating over x.

 

[Elena14]

I am reattempting the question using ∫x dm/ ∫ dm but what should I substitute with x?

 

[gneill]

Okay, perhaps I made a poor choice by calling "dm" a mass element since in this case we are dealing with areas (no mass density for the material was supplied). So think of the dm's as area elements. They have a width dx. What's their length?

 

[Elena14]

That's what I am struggling with.

 

[Elena14]

Can you provide me with a more comprehensible hint to this?

 

[gneill]

That's what I am struggling with.

Study the diagram I supplied. You should be able find the area of the dm element. What's its width and height? For integrating the area elements they are approximated as rectangles.

 

[Elena14]

Is the area of the dm element kx^2 * dx

 

[gneill]

Is the area of the dm element kx^2 * dx

Almost, you've only accounted for half its total height (from the x-axis to the upper curve).

 

[gneill]

By the way, I've updated the image in post #6 to reflect that we're summing area elements in this case, not mass elements. Sorry if there was any confusion there.

 

[Elena14]

Alright, so dm is 2kx^2 * dx. Integrating using the formula ∫x dm/ ∫ dm with limits 0 to a, we get 3a/4 which is the answer. But how do we find out that dm is the area element and not the length element.

 

[gneill]

dm is an area element because it was defined as the width (dx) multiplied by the height (2 kx²), which is a rectangular area. The integration limits are determined by the boundaries defined in the problem. The object is bounded on the right by the line x = a.

 

[Elena14]

Thank you so much. You have been a great help.

 

[gneill]

Thank you so much. You have been a great help.

No problem! Glad to help.

 

[Elena14]

Could not we have taken a smaller element dm which will be only defined as the length so we don't have to go into that "area element" ? What would be wrong with that? And is there a thing such as a volume element as well?

 

[gneill]

Could not we have taken a smaller element dm which will be only defined as the length so we don't have to go into that "area element" ? What would be wrong with that?
A line, with no width, would not have an area. The 2D center of mass (area) formula depends upon finding a weighted sum of the area elements and dividing it by the total area. You might also consider that the differential width "dx" is about as narrow (line-like) as you can get without it not existing at all. dx → 0 in the limit.

And is there a thing such as a volume element as well?

Yes indeed. This happens when you have a 3D object. Then you have double or triple integrals to deal with.