Where is the centroid of a parabolic wire located?

[StirlingA]

I know I'm making this more difficult than it needs to be. I need to find the centroid of a wire bent into the shape of a parabola, defined to be y=x^2 with -2<X>2 and 0<y>4. Obviously due to symetry X-bar =0... but what's y-bar?? No dimensions are given for the width of the wire, so I assume it is not needed. I'm guessing it has something to do with arc length... I'm still lost though.

Thanks everyone.

 

[Sunshine]

Using Pappus second theorem is probably the easiest way to solve this... :)

 

[thepatientmental]

Finding the centroid of a parabolic wire involves determining the average location of all the points along the wire. Since the wire is bent into the shape of a parabola, we can use the formula for the centroid of a parabola, which is located at (x,y) = (0, 4/3).

To understand this, imagine the parabola as a shape made up of many small rectangles stacked on top of each other. The centroid of each rectangle is located at its center, which is (x,y) = (0, 4/3). Now, to find the centroid of the entire parabolic wire, we need to find the average location of all these centroids.

Since the wire is symmetrical, the centroid will also be located at (x,y) = (0, 4/3). This means that the centroid of the wire is simply the midpoint of the parabola's height, which is 4/3. This does not require any dimensions of the width of the wire, as it is a 2-dimensional shape and the centroid is just a point.

In summary, the centroid of a parabolic wire is located at (x,y) = (0, 4/3) and does not require any dimensions for the width of the wire. It is simply the midpoint of the parabola's height. I hope this helps clarify things for you.