Find the center of mass of a plate

[Davidllerenav]

Homework Statement

Find the center of mass of the next plate if:
A) Is homogeneous
B) Its density per unit mass is $\sigma=Axy$, where $A$ is a constant.

Homework Equations

$X_{cm}=\frac{\int\sigma x dA}{\int\sigma dA}$
$Y_{cm}=\frac{\int\sigma y dA}{\int\sigma dA}$

The Attempt at a Solution

I did part A) as is shown in the picture below:

Am I correct?
My main problem is with part B), I tried to do it as I did part A), I found ##X_{cm} without any problems, like this:

But when I tried to find $Y_{cm}$ I got stucked with this:

$Y_{cm}=\frac{\int_{0}^7 \sigma ydA_1+ \int_{1}^7 \sigma ydA_2}{\int_{0}^7 \sigma dA_1 + \int_{1}^7 \sigma dA_2}=\frac{A\left[ \int_{0}^7 y^2\sqrt{y-1}dy + \int_{1}^7 y^2(y-1)dy\right]}{A\left[ \int_{0}^7 y\sqrt{y-1}dy + \int_{1}^7 y(y-1)dy\right]}$, but when I tried to integrate $\int_{0}^7 y^2\sqrt{y-1}dy$ and $\int_{0}^7 y\sqrt{y-1}dy$, I ended up with imaginary numbers, what am I doing wrong? Hope you can help me.

 

[haruspex]

Think more carefully about what your ∫(x²+1).dx and ∫(x²+1)x.dx are actually measuring.

 

[Davidllerenav]

Think more carefully about what your ∫(x²+1).dx and ∫(x²+1)x.dx are actually measuring.

$\int (x^2+1)dx$ is measuring the total area of the plate, while $\int (x^2+1)xdx$ is measuring the area of a small piece of the total area.

 

[haruspex]

$\int (x^2+1)dx$ is measuring the total area of the plate,

No.
Look at your diagram for the x>0 section. What is the length in the y direction of the strip width dx?

 

[Davidllerenav]

No.
Look at your diagram for the x>0 section. What is the length in the y direction of the strip width dx?

I think it would be $7-(x^2+1)$, right?

 

[haruspex]

I think it would be $7-(x^2+1)$, right?

Right, so what should your integrals be?

 

[Davidllerenav]

Right, so what should your integrals be?

Oh, I see know, I was calculating the area underneath the curve, so I need to write $7-x^2-1$ instead to calculate the are between the line $y=7$ and the curve.
I think the integrals would be like this, am I correct?

How would it be with $Y_{cM}$?

 

[haruspex]

the integrals would be like this, am I correct?

Yes, that answer looks better.

How would it be with $Y_{cM}$?

Please post a revised attempt. As a check, should it be more or less than 3.5?

 

[Davidllerenav]

Please post a revised attempt.

I'm trying to solve it. The length in the x direction of the strip width dy would be $x$, right? And $x=\sqrt{y-1}$.

As a check, should it be more or less than 3.5?

I think that it should be less than 3.5, but not too much.

 

[haruspex]

I think that it should be less than 3.5, but not too much.

Consider the two parts separately. How would the two y coordinates of the mass centres compare with 3.5 individually?

 

[Davidllerenav]

Consider the two parts separately. How would the two y coordinates of the mass centres compare with 3.5 individually?

Well, the coordinates of the mass center of the rectangle would be $(2.5,3.5)$ and for the parabola, I guess the would be less.

 

[Davidllerenav]

The length in the x direction of the strip width dy would be $x$, right? And $x=\sqrt{y-1}$.

@haruspex Am I correct with this?

 

[haruspex]

I'm trying to solve it. The length in the x direction of the strip width dy would be $x$, right? And $x=\sqrt{y-1}$.

Yes, but only for y in a certain range.

for the parabola, I guess the would be less.

The y coordinate would be less? Doesn't look that way to me.

 

[Davidllerenav]

Yes, but only for y in a certain range.

Yes, y would be between 1 and 7. So is my calculation of $Y_{cm}$ correct?

The y coordinate would be less? Doesn't look that way to me.

Well, then I think I didn't understand the question.

 

[haruspex]

Well, then I think I didn't understand the question.

Consider the complete rectangle, 2x7, with lower left corner at the origin. The y coordinate of its mass centre would be 3.5, yes? To get the actual shape you have to trim off the piece below the parabola. Wouid that raise or lower the mass centre?

 

[Davidllerenav]

Consider the complete rectangle, 2x7, with lower left corner at the origin. The y coordinate of its mass centre would be 3.5, yes? To get the actual shape you have to trim off the piece below the parabola. Wouid that raise or lower the mass centre?

That would raise the center of mass, so the y component would be greater than 3.5.

 

[haruspex]

That would raise the center of mass, so the y component would be greater than 3.5.

Right.

 

[Davidllerenav]

Right.

Ok, so am I correct with how I calculated $Y_{cm}$ initially?
Also, how do I calculate the center of mass when the density isn't constant, i.e. part B)? I tried to solve it as I did with part A, but $Y_{cm}$ gives me imaginary numbers.

 

[haruspex]

so am I correct with how I calculated $Yc_{cm}$initially?

Since you got a value <3.5, no. The method looks ok so there must be an arithmetical error.
By the way, for both x and y coordinate of CoM, the denominator is the area of the plate, so you don’t need to calculate that twice.

 

[Davidllerenav]

Since you got a value <3.5, no.

I think I know the problem. Do you see that I got a different value in the total mass integral? The problem is in the parabola, since I got 35 for the rectangle for both $X_{cm}$ and $Y_{cm}$. The problem is that the equatio of the parabola $y=x^2+1$ isn't complete with $x\in[0,2]$. So when I integrate $\int_{0}^2 6-x^2 dx$ I'm not getting the whole area, while when I integrate $\int_{1}^7 \sqrt{y-1}dy$ I'm indeed getting the whole area. This can be seen here:

How can I fix this? The only way I can think of would be to change the upper limit on $X_cm$ with 2.449 instead of 2.

 

[haruspex]

I think I know the problem. Do you see that I got a different value in the total mass integral? The problem is in the parabola, since I got 35 for the rectangle for both $X_{cm}$ and $Y_{cm}$. The problem is that the equatio of the parabola $y=x^2+1$ isn't complete with $x\in[0,2]$.

Ha! The problem is that the given diagram is wrong! At x=2, x²+1 is only 5, not 7.
My guess is that the distances 5 and 7 are swapped over. The plate should be 5cm high and 7+2cm wide. That fits better with the way it is drawn.

 

[Davidllerenav]

Ha! The problem is that the given diagram is wrong! At x=2, x²+1 is only 5, not 7.
My guess is that the distances 5 and 7 are swapped over. The plate should be 5cm high and 7+2cm wide. That fits better with the way it is drawn.

Yes you're right. That would be a big change though, I'll try to do it. Also, how should I solve the part where the plate isn't himoghomoge?

 

[haruspex]

Yes you're right. That would be a big change though, I'll try to do it. Also, how should I solve the part where the plate isn't himoghomoge?

It's just the same but replacing constant σ with Axy. Post what you get.

 

[Davidllerenav]

It's just the same but replacing constant σ with Axy. Post what you get.

I'll post as soon as I finish calculating with the correct measurements.

 

[Davidllerenav]

@haruspex Ok, this is what I got:
Part A):

For part B) I had the same problem, I ended up with a integral $7\int_{0}^5 \sqrt{y-1}dy$ and $7\int_{0}^5 \sqrt{y-1}ydy$ and on both I will get imaginary numbers, what am I doing wrong?

 

[haruspex]

what am I doing wrong?

The bound x²<y-1 applies to what range of y?

 

[Davidllerenav]

The bound x²<y-1 applies to what range of y?

It only works for $y\geq 1$, that's why I'm getting imaginary numbers.

 

[haruspex]

It only works for $y\geq 1$, that's why I'm getting imaginary numbers.

That's right, the constraint only applies for y>1, and that part of the plate only exists for y>1. So use the correct bounds on the integral.

 

[Davidllerenav]

That's right, the constraint only applies for y>1, and that part of the plate only exists for y>1. So use the correct bounds on the integral.

But isn't that integral supposed to go from 0 to 5? Since it is integrating the rectangle.

 

[haruspex]

But isn't that integral supposed to go from 0 to 5? Since it is integrating the rectangle.

You can run the integral from 0 if you wish, but the constraint 0<x²<y-1 only applies for y>1. So you need to break the integral into separate ranges 0 to 1 and 1 to 5. The 0 to 1 range is trivially null.

Wait - you said integrating the rectangle. You mean the x<0 part? That's not the bit that gives imaginary numbers.

 

[Davidllerenav]

You can run the integral from 0 if you wish, but the constraint 0<x²<y-1 only applies for y>1. So you need to break the integral into separate ranges 0 to 1 and 1 to 5. The 0 to 1 range is trivially null.

So I just need to integrate from 1 to 5 on both integrals?

 

[haruspex]

So I just need to integrate from 1 to 5 on both integrals?

The numerator and denominator integrals for x>0, yes.

 

[Davidllerenav]

The numerator and denominator integrals for x>0, yes.

Ok, so just to check, the two integrals on the left must be from 1 to 5? So I will have four integrals from 1 to 5?

 

[haruspex]

Ok, so just to check, the two integrals on the left must be from 1 to 5? So I will have four integrals from 1 to 5?
View attachment 240464

No, some confusion here.
You have integrals that relate to the 7x5 rectangle on the left (x<0) and integrals that relate to the curved flange on the right (x from 0 to 2).
Those for the right involve the constraint x²>y-1 and therefore only apply for y=1 to 5.
Those for the left apply for y=0 to 5, but since they do not involve the constraint x²<y-1 they should not contain any square roots so should not give imaginary nunbers.

 

[Davidllerenav]

No, some confusion here.
You have integrals that relate to the 7x5 rectangle on the left (x<0) and integrals that relate to the curved flange on the right (x from 0 to 2).
Those for the right involve the constraint x²>y-1 and therefore only apply for y=1 to 5.
Those for the left apply for y=0 to 5, but since they do not involve the constraint x²<y-1 they should not contain any square roots so should not give imaginary nunbers.

But as you can see in here, I ended up with square roots on the left:

 

[haruspex]

But as you can see in here, I ended up with square roots on the left:
View attachment 240465

Consider finding the mass of some shape. The general form is ∫σ.dA. In Cartesian, that's ∫σ.dxdy.
In the present problem, σ=Axy, so ∫∫Axy.dxdy. For the rectangle (x<0) the bounds are easy: ∫_x=-7^x=0_y=0^y=5Axy.dxdy.
See if you can solve that and do the right hand portion and the numerator integrals the same way.

 

[Davidllerenav]

Consider finding the mass of some shape. The general form is ∫σ.dA. In Cartesian, that's ∫σ.dxdy.
In the present problem, σ=Axy, so ∫∫Axy.dxdy. For the rectangle (x<0) the bounds are easy: ∫_x=-7^x=0_y=0^y=5Axy.dxdy.
See if you can solve that and do the right hand portion and the numerator integrals the same way.

I'm new to the doble integral notation, but I think that it would be $\int_{0}^2\int_{1}^5 Axy\cdot dydx$.

 

[haruspex]

I'm new to the doble integral notation, but I think that it would be $\int_{0}^2\int_{1}^5 Axy\cdot dydx$.

No, that would be a rectangle. You need to relate the bounds on the inner integral to the value of the dummy variable in the outer integral.
E.g. suppose we integrate wrt y first. That means for the purpose of that integral x is a constant. What is the range of y in terms of that x?

 

[Davidllerenav]

No, that would be a rectangle. You need to relate the bounds on the inner integral to the value of the dummy variable in the outer integral.
E.g. suppose we integrate wrt y first. That means for the purpose of that integral x is a constant. What is the range of y in terms of that x?

Sorry, but I don't understand what you mean.

 

[haruspex]

Sorry, but I don't understand what you mean.

Look at the top right diagram in your post #1. You show a vertical band width dx and a horizontal band height dy in the rectangle and another pair in the x>0 portion.
Consider the vertical band in the x>0 area. What is the range of values of y in that band?

 

[Davidllerenav]

Look at the top right diagram in your post #1. You show a vertical band width dx and a horizontal band height dy in the rectangle and another pair in the x>0 portion.
Consider the vertical band in the x>0 area. What is the range of values of y in that band?

The range of the y values are from 5-y to 5.

 

[haruspex]

The range of the y values are from 5-y to 5.

No, the range of y can't depend on y. It depends on x.

 

[Davidllerenav]

No, the range of y can't depend on y. It depends on x.

Then the range would be from x=0 to X=2, right?

 

[haruspex]

Then the range would be from x=0 to X=2, right?

No, you are looking for a range of y as a function of x.
Look at your diagram. For that shaded vertical band, what is the value of y at the top? What is it at the bottom?

 

[Davidllerenav]

No, you are looking for a range of y as a function of x.
Look at your diagram. For that shaded vertical band, what is the value of y at the top? What is it at the bottom?

For y at the top the value is 7, and for y at the bottom it is 7-y. Right?

 

[haruspex]

For y at the top the value is 7, and for y at the bottom it is 7-y. Right?

If y equals 7-y then y=3.5. We don't want y as a function of itself, that's not helpful. We want y as a function of x.

The bottom of the shaded strip lies on the curve y=x²+1. If the X coordinate is x, what is its Y coordinate in terms of x?

 

[Davidllerenav]

If y equals 7-y then y=3.5. We don't want y as a function of itself, that's not helpful. We want y as a function of x.

The bottom of the shaded strip lies on the curve y=x²+1. If the X coordinate is x, what is its Y coordinate in terms of x?

Y in terms of x would be function, right? $y=x^2+1$?

 

[haruspex]

Y in terms of x would be function, right? $y=x^2+1$?

Right, so that is your lower bound for y in that strip. The y integral is $\int_{y=x^2+1}^5Axy.dy$.

 

[Davidllerenav]

Right, so that is your lower bound for y in that strip. The y integral is $\int_{y=x^2+1}^5Axy.dy$.

Ok, and should I change that on every integral? Even on part A?

 

[haruspex]

Ok, and should I change that on every integral? Even on part A?

For part A it was not necessary to do a double integral because the density was constant. You could just write down that the mass of the vertical strip on the right was σ(5-(x²+1))dx. That was instead of the integral (∫_y=x²+1^y=5σ.dy).dx