Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 60

I’m sure you can do the integration to find the centroid of the semicircle, but I’m just going to look it up. It’s \frac{4r}{3\pi} down from the flat edge. Combining this with the triangle, we get

\bar y = \frac{(hr) \cdot \frac{h}{3} - (\frac{1}{2}\pi r^2) \cdot \frac{4r}{3\pi}}{hr + \frac{1}{2}\pi r^2} = \frac{2(h^2 - 2r^2)}{3(2h + \pi r)}

We’ve taken our origin at the center of the semicircle with the y axis pointing up. It should go without saying that \bar x = 0.

Problem 61Problem 59

Last modified: January 22, 2009 at 8:32 PM.