Where is the centre of mass for a flat sheet of metal bounded by y=x^4 and y=5?

[TobyDarkeness]

Homework Statement

determine the position of the centre of mass of a flat sheet of metal of homogeneous density rho. The sheet lies within the area bounded by the equations y=x^4 and y=5

Homework Equations

integral centre mass

The Attempt at a Solution

a slightly trivial question, i know how to set up the correct integral when finding the centre of mass below the function curve bound by an x=? but with the area above the curve in this manner I am finding problems. Should i treat the graph on its side or change the graph form to fit in the other direction. or find the integrals below both functions and do a subtraction to find the positions?

 

[gabbagabbahey]

Homework Statement

determine the position of the centre of mass of a flat sheet of metal of homogeneous density rho. The sheet lies within the area bounded by the equations y=x^4 and y=5

Homework Equations

integral centre mass

The Attempt at a Solution

a slightly trivial question, i know how to set up the correct integral when finding the centre of mass below the function curve bound by an x=? but with the area above the curve in this manner I am finding problems. Should i treat the graph on its side or change the graph form to fit in the other direction. or find the integrals below both functions and do a subtraction to find the positions?

First, draw a picture!

Then either divide the area into horizontal slices of thickness dy, express the length of a slice at a give height y as a function of y and integrate, or divide it into veritical slices of width dx, express the height of a slice at a given x as a function of x and integrate.

 

[TobyDarkeness]

thats the method I am used to, on which I've found simple enough with other examples but i can't seem to build the correct integral with this set up.

 

[gabbagabbahey]

thats the method I am used to, on which I've found simple enough with other examples but i can't seem to build the correct integral with this set up.

Show your attempt.

 

[TobyDarkeness]

ok so I am not sure how to integrate for the area above the line, so here's my attempt.
i don't have a scanner so i don't have a diagram to show you. I turned the graph on its side and reversed the axis (tried to cheat a bit here) so my x co ords would be my y.

y=x^4, y=5

switching it, x=4th root (y) and x=5 but where i have "y" its my new x axis.

so using this i did the integration for the horizontal position as normal,
[integral xydx]/[integralydx]
doing this i get 5 as my horizontal co-ord which is actually my vertical since i switched it, here i can already see I've gone wrong. but i followed through with the vertical anyway and obtained 0.66 as my other position.

i made a second attempt with the limits +/- 4th root 5, integrating first y=5 then y=x^4 in both x and y directions then subtracting one from the other to find the positions in the area bounded. but i can see already from the first integration that this is also wrong.

how do i obtain the position of the centre of mass in this bounded area above the function line??

thanks in advance.

 

[gabbagabbahey]

so using this i did the integration for the horizontal position as normal,
[integral xydx]/[integralydx]
doing this i get 5 as my horizontal co-ord which is actually my vertical since i switched it, here i can already see I've gone wrong. but i followed through with the vertical anyway and obtained 0.66 as my other position.

What were your limits of integration?

 

[TobyDarkeness]

for this attempt i used 5, 0 as my limits treating the y=5 as x=5 bound by the 4th root curve and the origin.

 

[gabbagabbahey]

for this attempt i used 5, 0 as my limits treating the y=5 as x=5 bound by the 4th root curve and the origin.

Your problem is that the length of each ribbon with thickness dx isn't just y, you need to use a double integral. For the area of the region, you should have

A=\int_0^5 \int_{-x^{1/4}}^{x^{1/4}}dydx

Do you see why?

For the horizontal and vertical positions of the CoM, you just insert either x or y (and rho) into that double integral and then divide by the area.

 

[TobyDarkeness]

sorry i think I am missing something can you explain why i need to use the double integral, I've not been working for a long time now my maths is rusty. so i need to use the double integral but in the same way with the surface area density process, for the area should i sub in for y, y=x^4 as normal you should i do an integral subtraction to get the area of the enclosed space.

 

[Mindscrape]

rho=m/A, so then dm=rho*dA, which means that
m=\int \int \rho dA

then

x_{cm}=\frac{\int \int \rho x dA}{m}

and

y_{cm}=\frac{\int \int \rho y dA}{m}

Oh yeah, and hint: before you even start, what should x_cm be?

 

[TobyDarkeness]

should there be a integral on the bottom as well? \intdm? by xcm do you mean the limits of integration for the x position or should i be able to intuitively understand where the x position should be? at a guess i would say the origin on the axis, and therefore do i need to only calculate the y position?

 

[Mindscrape]

Yes, exactly, intuitively you should be able to guess that the x_cm will be on the axis, and mathematically you see it very soon because you are integrating an odd function over symmetric bounds.

There is an integral on the bottom, yes, it's the integral for m I have at the beginning of my post.

Oh yeah, and also you don't have to do the exact integral that gabba posted, you could integrate with respect to x first if you really wanted. You'd have to do slightly more work to get the right x bounds.