Center of mass for sheet metal

In summary, the conversation is about finding the center of mass of a parabolic sheet of metal, specified by the equation y=ax^2, with a curved edge and a flat top. The participants discuss different methods for finding the area and integral needed to compute the center of mass, including the use of a density factor and the importance of properly setting up the integral. Ultimately, the correct answer is determined to be 3b/5, with the method involving integrating ydm over the surface of the parabola.
  • #1
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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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  • #2
The formula for y_cm looks good. What r u using for "dm"?
 
  • #3
dm

at first i had (3M)/(ab^3)ydx
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.
 
  • #4
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"?
 
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  • #5
Did you get it?

I find 3b/4.
 
  • #6
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
 
  • #7
That is just saying the density is constant i think. Anyway u don't 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
 
  • #8
could you show me the steps of how you arrived to this answer
maybe the answer in the book is wrong.
 
  • #9
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...
 
  • #10
What does "weighted" means?

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

[tex]Y_{CM} = \iint\frac{y}{A}dA[/tex]

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

[tex]A = \int_0^b2\sqrt{\frac{y}{a}}dy = \frac{4b}{3}\sqrt{\frac{b}{a}}[/tex]

So

[tex]Y_{CM} = \frac{3}{4b}\sqrt{\frac{a}{b}}\int_{-\sqrt{b/a}}^{\sqrt{b/a}}\int_0^b ydydx = \frac{3}{4}b[/tex]
 
  • #11
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 because 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.
 
  • #12
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?
 
  • #13
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.
 
  • #14
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

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

should be

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

add: u were integrating over a rectangle
 
  • #15
HackaB said:
this

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

should be

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

add: u were integrating over a rectangle

HAHAHA. ooook. :tongue2:

It could have been

[tex]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[/tex]

too... which is what I tought I was doing.
 
  • #16
quasar987 could you show me how you arrive at getting the stuff before the integral sign and the logic behind the two integrals, thanks.
 

1. What is the center of mass for sheet metal?

The center of mass for sheet metal is a point that represents the balance of mass distribution in a sheet metal object. It is the point where the object would balance perfectly if supported at that point.

2. How is the center of mass calculated for sheet metal?

The center of mass for sheet metal can be calculated by finding the weighted average of the individual masses of the sheet metal object. Each point on the object is assigned a weight based on its mass and distance from a reference point. The center of mass is then located at the point where the sum of the individual masses multiplied by their distances is equal on both sides of the reference point.

3. Why is the center of mass important for sheet metal?

The center of mass is important for sheet metal because it affects the stability and balance of the object. Knowing the location of the center of mass can help determine the best way to handle and transport the object, as well as ensure its structural integrity.

4. How does the shape of the sheet metal affect its center of mass?

The shape of the sheet metal can greatly affect the location of its center of mass. Objects with irregular shapes or uneven distributions of mass will have a different center of mass compared to symmetrical objects with uniform mass distributions.

5. Can the center of mass be outside of the physical boundaries of the sheet metal object?

Yes, the center of mass can be outside of the physical boundaries of the sheet metal object. This is possible when the object has an irregular shape or an uneven distribution of mass. In these cases, the center of mass may be located in empty space or outside of the physical boundaries of the object.

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