# Find centre of mass

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1. Dec 15, 2015

### Elena14

• Moved from a technical forum, so homework template missing
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.

2. Dec 15, 2015

### Staff: Mentor

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?

3. Dec 15, 2015

### 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?

4. Dec 15, 2015

### Staff: Mentor

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?

5. Dec 15, 2015

### Elena14

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

6. Dec 15, 2015

### Staff: Mentor

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]

Last edited: Dec 15, 2015
7. Dec 15, 2015

### Elena14

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

8. Dec 15, 2015

### Staff: Mentor

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

9. Dec 15, 2015

### Elena14

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

10. Dec 15, 2015

### Staff: Mentor

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?

11. Dec 15, 2015

### Elena14

That's what I am struggling with.

12. Dec 15, 2015

### Elena14

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

13. Dec 15, 2015

### Staff: Mentor

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.

14. Dec 15, 2015

### Elena14

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

15. Dec 15, 2015

### Staff: Mentor

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

16. Dec 15, 2015

### Staff: Mentor

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.

17. Dec 15, 2015

### 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.

18. Dec 15, 2015

### Staff: Mentor

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.

19. Dec 15, 2015

### Elena14

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

20. Dec 15, 2015