Find the center of mass for a disk with cut outs | Solution

[braindead101]

question:
find the location of the center of mass for the following disk of radius a. it has a three circular cut outs. one with a diameter of a and the other two have a diameter of b.
http://img479.imageshack.us/img479/2743/physics4ld.jpg

solution:
Xc = m/m(Xo - a/2)
Xc = Xo - a/2

Yc = m/2m(Yo+L+b/2) + m/2m(Yo-L-b/2)
Yc = Yo

is this right...
ive no idea.

 

[gaganpreetsingh]

Is there any figure associated with the question? I don't think this can be solved without knowing the positions of the cutouts of radius 'b'

 

[gaganpreetsingh]

oops. there was some problem. now when using firefox the images have been loaded. (was using opera before). sorry

 

[pseudovector]

Here are 2 tips for finding the centre of mass:
1. The center of mass will always be on a symmetry axis
2. Instead of thinking of a disc with holes, try imagining a full disc of radius a, and more discs (of radii a/2,b/2) with negative mass where the holes are, so the net mass density where the holes are is zero. So you multiply all the areas of all the discs by their distance from the origin point, and divide by the sum the areas. Make sure you get the signs right: the area of the biggest disc should have a plus sign, while the others should have a minus sign.
Hope this helps

 

[durt]

First of all, the picture shows up fine in Opera.
Second, I'm going to assume that the two smaller circles are tangent to both the one of radius $$\frac{a}{2}$$ and the largest one at the top and bottom.
So we first must find $$b$$ in terms of $$a$$ by constructing a right triangle connecting the centers of each the largest, medium, and one of the smallest circles. Then:
$$(\frac{a}{2})^2 + (a - \frac{b}{2})^2 = (\frac{a}{2} +\frac{b}{2})^2$$
$$b = \frac{2}{3} a$$
Now, let's say the original circle with no holes in it had a mass of $$M$$. The mass $$m_0$$ of the circle with only the two smaller holes is then:
$$m_0 = M - M \frac{2 \pi (\frac{b}{2})^2}{\pi a^2} = \frac{7}{9} M$$
The mass $$m_1$$ of the larger circle taken out is:
$$m_1 = M \frac{\pi (\frac{a}{2})^2}{\pi a^2} = \frac{1}{4} M$$
Let $$x_c$$ be the distance to the right of the center of the biggest circle at which the center of mass that you wish to find is. The center of the circle is the center of mass of this piece and the medium sized circle.
$$\frac{7}{9} M x_c - \frac{1}{4} M \frac{a}{2} = 0$$
So, $$x_c = \frac{9}{56} a$$
The y coordinate of the center of mass has obviously remained along the horizontal line from the center. Of course I have probably made a mistake, but that's how you do a problem like this.

 

[braindead101]

i don't understand why the bottom is pi a^2, it didnt say newhere in the question that the mass equals its area