# Center of mass for sheet metal

I am having the hardest time figuring out the center of mass of this problem.

A sheet of metal is cut in the shape of a parabola (imagine that you have a parabola shape with the top being flat). The curved edge of the sheet is specified by the equation y=ax^2, and y ranges from 0 to b. Find the center of mass in terms of a and b. (You will need to find the area first.)

There's the question. The answer is 3b/5. I used center of mass equation

Ycm=(1/m)(integral of y dm)

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The formula for y_cm looks good. What r u using for "dm"?

dm

then i found the area a different way and i came up with (Mx^3)ydx/(2ab)

what is the correct evaulation of this area. this dm is the most important part of the integration.

The width of the metal sheet at distance y is 2*Sqrt(y/a). Intergate this from 0 to b to get the area,

BTW, what is "M"?

Last edited:
quasar987
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Did you get it?

I find 3b/4.

M comes from this

Total mass (M) / Total Area (the integration of the function) = dm (tiny portion of mass) / da (tiny portion of area)

M/A=dm/da

i just solved for dm so i could then substitute in for dm when i find the center of mass

That is just saying the density is constant i think. Anyway u dont need the density. Were u able to get the area?

quasar987: do u get 3b/4 for the final answer or for area or what? hbombs first post says its 3b/5

could you show me the steps of how you arrived to this answer
maybe the answer in the book is wrong.

i doubt the book is wrong. First find the area, that is compute the integal

integral from 0 to b of (2*sqrt(y/a)) dy = Area

then compute the integral of y "weighted" by mass, or area in this case

integral from 0 to b of (2*sqrt(y/a)*y)dy

and divide what u get by the area....

quasar987
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What does "weighted" means?

According to the dm you found, the integral we want to compute is

$$Y_{CM} = \iint\frac{y}{A}dA$$

So I found A. HackaB's way it the shortest:

$$A = \int_0^b2\sqrt{\frac{y}{a}}dy = \frac{4b}{3}\sqrt{\frac{b}{a}}$$

So

$$Y_{CM} = \frac{3}{4b}\sqrt{\frac{a}{b}}\int_{-\sqrt{b/a}}^{\sqrt{b/a}}\int_0^b ydydx = \frac{3}{4}b$$

what i meant by weighted is this: when u compute the center of mass of an ojbect, u are finding the average position "weighted" by how much mass is at a given position. In this prob., we know the center o mass is on the y axis by symmetry so we just have to find the "weighted average" of y. The amount of mass at distance y above the x-axis is dm = 2*sqrt(y/a)*dy. Actuallly this is da...area...but we can take density to be 1 cuz it will cancel anyway. So what you need is

integral of y dm = integral of 2*y*sqrt(y/a)*dy from 0 to b. For this I get (4/5) * b^(5/2) / sqrt(a). Divide this by what quasar987 gave for A to get the answer.

quasar987
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Could you tell me what is wrong with my method? How is the integral I evaluated not the right one? Isn't it the integral of y/A over the surface of the parabola?

hbomb,

I get 3b/5

I turned the parabola on its side so the new function is:

y = (x/a)^1/2

then I integrated (Xcm - x)*y*dx from 0 to b, set the integral to zero and solved for Xcm.

quasar987 said:
Could you tell me what is wrong with my method? How is the integral I evaluated not the right one? Isn't it the integral of y/A over the surface of the parabola?
this

$$Y_{CM} = \frac{3}{4b}\sqrt{\frac{a}{b}}\int_{-\sqrt{b/a}}^{\sqrt{b/a}}\int_0^b ydydx = \frac{3}{4}b$$

should be

$$Y_{CM} = \frac{3}{4b}\sqrt{\frac{a}{b}}\int_0^b\int_{-\sqrt{y/a}}^{\sqrt{y/a}} ydxdy = \frac{3}{5}b$$

add: u were integrating over a rectangle

quasar987
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HackaB said:
this

$$Y_{CM} = \frac{3}{4b}\sqrt{\frac{a}{b}}\int_{-\sqrt{b/a}}^{\sqrt{b/a}}\int_0^b ydydx = \frac{3}{4}b$$

should be

$$Y_{CM} = \frac{3}{4b}\sqrt{\frac{a}{b}}\int_0^b\int_{-\sqrt{y/a}}^{\sqrt{y/a}} ydxdy = \frac{3}{5}b$$

add: u were integrating over a rectangle
HAHAHA. ooook. :tongue2:

It could have been

$$Y_{CM} = \frac{3}{4b}\sqrt{\frac{a}{b}}\int_{-\sqrt{b/a}}^{\sqrt{b/a}}\int_{ax^2}^b ydydx = \frac{3}{4}b$$

too... which is what I tought I was doing.

quasar987 could you show me how you arrive at getting the stuff before the integral sign and the logic behind the two integrals, thanks.