Centre of gravity is a concept you oft hear mentioned. Most athletes know that, with a lower centre of gravity, keeping balance is easier. Similarly with vehicles: one with a higher centre of gravity is more likely to tip over going around a hard corner. Of course, that’s the reason sports cars are built so low to the ground.<\/p>\n
For our first example moving towards finding the centre of gravity, we’re not going to find the centre of gravity itself. Rather, we’ll find the “halfway point” of the mass of a square pyramid. Since that point is not, technically, the centre of gravity, I’ve given it the name “mass midpoint”.<\/p>\n
A pyramid is not the same top to bottom; therefore, its mass midpoint is not simply at half height. We’ll make the assumption that the pyramid has uniform density. Thence, we find the point from which there is equal volume side to side, front to back, and top to bottom:<\/p>\n
Example<\/strong>: Find the mass midpoint of a square-based pyramid with base side length 10m and height 12m.<\/p>\n Solution<\/strong>: By symmetry, the mass midpoint is somewhere directly above the centre point of the square base. Let’s introduce three dimensional coordinates of (forward, across, up); aka, (x, y, z). Then the location of the pyramid’s mass midpoint will be (5, 5, z). We need to find z, as follows:<\/p>\n The formula for volume of a square-based pyramid is<\/p>\n V=w2<\/sup>h\/3, w=width, h=height<\/p>\n We seek the specific height, γ, at which the volume below is the same as the volume above. For convenience, γ will be measured from the top down, rather than from the bottom up.<\/strong> (Reason: when measured from top to bottom, the height is always 6\/5 the width; conversely, the width is 5\/6 the height.)<\/p>\n If at height γ (measured from the top) the volume above equals the volume below, then<\/p>\n Vtotal<\/sub> – (5γ\/6)2<\/sup>γ\/3 = (5γ\/6)2<\/sup>γ\/3<\/p>\n We can add (5γ\/6)2<\/sup>γ\/3 to both sides:<\/p>\n Vtotal<\/sub> = 2*(5γ\/6)2<\/sup>γ\/3<\/p>\n Therefore, <\/p>\n 102<\/sup>*12\/3 = 2*(5γ\/6)2<\/sup>γ\/3<\/p>\n which leads to<\/p>\n 400=50γ3<\/sup>\/108<\/p>\n Multiplying both sides by 108, then dividing both sides by 50, gives<\/p>\n 864=γ3<\/sup><\/p>\n so<\/p>\n γ=9.524m<\/p>\n Therefore, the height of the mass midpoint, from the bottom, is<\/p>\n 12m-9.524m=2.476m<\/p>\n So it appears that the mass midpoint of this pyramid, using the (forward, across, up) coordinate system, is at (5, 5, 2.476)m.<\/p>\n This post is meant as a first step towards exploring centre of gravity. In the case of this pyramid, its centre of gravity is not the same location as its mass midpoint. Why the two are different, and where the centre of gravity indeed is, are topics for future posts – likewise for the concept of three dimensional coordinates:)<\/p>\n