#1

I need to know how to do this for an exam tomorrow, but I'm pretty clueless. Any help would be great, thanks.

Find the centroid of the y-axis.

http://img22.imageshack.us/i/46683569.jpg/

Find the centroid of the y-axis.

http://img22.imageshack.us/i/46683569.jpg/

*Last edited by calnix at Mar 30, 2011,*

#2

I don't even see where a y variable is in that picture...

#3

Calculate the area of the rectangular shape and the square shape, add those two figures together, then subtract the area of the circle.

I'm assuming they want the area of the figure.

Edit: So (8x4 + 2x2)-pi(.5)^2

I'm assuming they want the area of the figure.

Edit: So (8x4 + 2x2)-pi(.5)^2

*Last edited by RU Experienced? at Mar 30, 2011,*

#4

what the.... no y variable?

#5

r=2dx42x

#6

Still don't see it.

Whatthe****amicalculating.jpg

Whatthe****amicalculating.jpg

#7

Sorry I wasn't clear, I'm supposed to find the centroid of the figure for the y-axis.

#8

Total area:

8*4+2*2=32+4=36

Subtract area of hole:

A = diameter * pi = 1*(pi) = pi

Y = 36 - pi

EDIT: Centroid is a different story.

http://www.intmath.com/applications-integration/5-centroid-area.php

Since there are no curves, it becomes a question of mass and position with linear edges, which you just plug into the calculator given the equation.

8*4+2*2=32+4=36

Subtract area of hole:

A = diameter * pi = 1*(pi) = pi

Y = 36 - pi

EDIT: Centroid is a different story.

http://www.intmath.com/applications-integration/5-centroid-area.php

Since there are no curves, it becomes a question of mass and position with linear edges, which you just plug into the calculator given the equation.

*Last edited by FaisalTMusic at Mar 30, 2011,*

#9

To do the centroid for that body i would break it up into 3 shapes: the larger 4"x8" rectangle, the 2" square on top, and the 1" circle (since it is a hole we will end up subtracting its contributions later).

First lets assign a coordinate system to the problem. Use Cartesian coord. with the origin at the bottom left corner of the body.

Next we will find the contributions of the larger rectangle on the bottom. To do this we first find its total area then find where its center of mass is in relation to the origin. The total area of that larger rectangle is 32sq in. The rectangle's center of mass is at its center. With relation to our coordinate system, its center of mass would be at point (2,4). Now multiply the area by the position of the center of mass which gives you xbar1=64 and ybar1=128.

We will now do the same thing for the square. its area is 4 sq in. and its center of mass is at its center; expressed in our coordinate system this is at (2,9). again multiplying these together we get xbar2 = 8 and ybar2=36

Finally we do the same thing for the circle the area is pi sq in. its center of mass is at the same location as the square (2,9). Multiplying this out we get xbar3= 2*pi and ybar3=9*pi.

then we need to add up the square and rec tangle and subtract the circle x bar and y bar to get the total centroid of the body. which is at (72 - 2pi, 169 - 9pi).

hope that made sense

First lets assign a coordinate system to the problem. Use Cartesian coord. with the origin at the bottom left corner of the body.

Next we will find the contributions of the larger rectangle on the bottom. To do this we first find its total area then find where its center of mass is in relation to the origin. The total area of that larger rectangle is 32sq in. The rectangle's center of mass is at its center. With relation to our coordinate system, its center of mass would be at point (2,4). Now multiply the area by the position of the center of mass which gives you xbar1=64 and ybar1=128.

We will now do the same thing for the square. its area is 4 sq in. and its center of mass is at its center; expressed in our coordinate system this is at (2,9). again multiplying these together we get xbar2 = 8 and ybar2=36

Finally we do the same thing for the circle the area is pi sq in. its center of mass is at the same location as the square (2,9). Multiplying this out we get xbar3= 2*pi and ybar3=9*pi.

then we need to add up the square and rec tangle and subtract the circle x bar and y bar to get the total centroid of the body. which is at (72 - 2pi, 169 - 9pi).

hope that made sense