Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 61

There’s this joke about engineers…

A mathematician, a physicist, and an engineer were all given a red rubber ball and told to find the volume. The mathematician measured the diameter and did the triple integral. The physicist filled a beaker with water, put the ball in the water, and measured its displacement. The engineer looked up the model number in his Red Rubber Ball Handbook.

Can we integrate to find the centroid of a cone and a hemisphere? Of course. In fact, that’s what we did for the hemisphere in Problem 58. But working engineers don’t do integrals just for the sake of doing integrals. We already know the c.g. of the hemisphere is 3/8 r below its base. And if we go to our Cone Handbook (or here), we’ll find that the centroid of the cone is 1/4 h above its base. So, the c.g. of the composite body is

\bar z = \frac{(1/3 \pi r^2 h)(1/4 h) + (2/3 \pi r^3)(-3/8 r)}{1/3 \pi r^2 h + 2/3 \pi r^3} = \frac{h^2 - 3 r^2}{4(h+2r)}

where the origin of z is on the plane where the cone and hemisphere meet. Note that the c.g. will be right on that plane if h = \sqrt{3} r, below it if h is smaller, and above it if h is larger.

Problem 62Problem 60

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