# Center of gravity: Really confused (HW problem)

by Izzybee
Tags: confused, gravity
P: 11
Hi everyone,
My question is:

Find the center of gravity for the board below: Assume that the board is uniform in thickness and that the hole in the plate is centered from top to bottom:

I only know that you have to find the area of the circle and the board and subtract the two, but I really have no clue how to solve this problem???

 P: 355 Are you sure your picture's right? Because if the radius is 20 cm, and the perpendicular distance from the left edge of the board to the right edge of the circle is 50 cm, then the length of the board comes out to be 90 cm, which is, obviously, greater than the stated length of 80 cm. Also, the diameter would be 40 cm, which is greater than the stated width of 30 cm. I'll assume you meant that the diameter is 20 cm. Anyway, I'm pretty sure, but not positive, that you can find the moment (i.e. $$\int \sigma dA$$) of the board with the hole filled in then subtract the moment of the hole. Once you have this, divide by the total mass. But don't quote me on that. Edit: oops. I just realized I forgot to define the variables I used. $$\sigma = \frac{M}{A}$$, where M = total mass, and A = total area. Anyway, I've thought it through, and the method should work. (If you haven't been given total mass, you can find it by the same method as I suggested for finding the moment.) As you know, center of gravity is defined as the average of all the points weighted by their respective masses. In other words, $$\texttf{r_{cm}} = \frac{{\sum_i \texttf{r_i} m_i}}{\sum_i m_i}.$$ Because if this, if you think about it, if you take out a portion of the board with moment $$\mu_2$$ and total mass $$M_p$$, the resulting center of mass is $$\texttf{r_{cm2}} = \frac{{\sum_i \texttf{r_i} m_i}-\mu_2}{{\sum_i m_i}-M_2}.$$