If you have been reading and thinking about this blog regularly, you may have noticed that I often ignore the dimensions of an object. For example, when we calculated the speed that a falling object hits the ground (post 17.18), we assumed that the object was a point in space. In this example, assuming that an object behaves as a point doesn’t matter. Think about the brick, in the picture below, lifted a height, h, above the flat, horizontal ground. When the brick is dropped, its lower surface (marked in red) hits the ground first. So, provided we measure h from the ground to the bottom (red) surface of the brick, it doesn’t matter that the brick is an extended object and not a point in space.
When I wrote about the earth’s gravitational field (post 16.16), I specified that the distance, r, was the distance from the centre of the earth to a point in space. In this example, I recognised that the planet earth was not a point in space but an extended object. The magnitude of its gravitational field was then given by g = GM/r2, where G represents the gravitational constant and M is the mass of the earth (post 16.16).
Why the centre of the earth? As I explained in post 16.16, this assumes that the earth is a perfect sphere (see also post 17.9). I also indicated that gravitational fields were a property of all objects with mass. So let’s think about an object whose dimensions are negligible – an object with negligible dimensions is called a particle. The gravitational field at a distance r from this particle is given by g = Gm/r2, where m now represents the mass of the particle.
The picture below shows a section though a sphere, whose centre is at O, that consists of a series of concentric shells; each shell is assumed to have a uniform density. This picture of a series of concentric spherical shells with different densities, is a reasonable model for the structure of the earth (http://www.bbc.co.uk/schools/gcsebitesize/geography/natural_hazards/tectonic_plates_rev1.shtml).
Now let’s think about the gravitational field of this sphere at point P. We define an X-axis pointing from O to P and Y and Z-axes to form a right-handed orthogonal set (post 17.3). We can think of each spherical shell as a system of particles. Each particle in a shell has another, exactly opposite it, on the opposite side of O. In the picture, a particle Q has a displacement q (post 17.4) from O where
q = xi + yj + zk.
Here, x, y and z are the Cartesian coordinates of Q and i, j and k are unit vectors defining the directions of the axes of the Cartesian coordinate system (post 17.3). In the picture, i points in the X-axis direction and j points in the Y-axis direction; in three dimensions, k would be perpendicular to the plane of the picture. Remember that i, j and k are lines whose lengths are the number 1; they are not distances and so are not measured in metres or any other units (post 17.3). Diametrically opposite Q, there is another particle whose position is given by
q’ = – xi – yj – zk.
Because of its symmetry, the whole series of concentric shells can be considered as a set of matched particles of this kind.
The displacement of P from O is given by
r = ri
since the X-axis direction was defined to be in this direction.
The displacement of P from Q is given by
p = r – q = (r – x)i – yj – zk
and the displacement of P from Q’ is
p’ = r + q = (r + x)i + yj + zk.
The average vector separation of P from Q and Q’ is
(p + p’)/2 = ri = r.
So the gravitational field at P due to Q and Q’ is the same as the gravitational field of two particles, each with the same mass, at O. If we sum up the effects of all the pairs of particles in the system, the gravitational field is the same as if the mass of all these particles were at O.
Therefore, when we are outside the system of concentric spherical shells, we can consider its mass to be acting at its centre. This point is called the centre of mass of the object.
The same arguments apply to a solid sphere with a uniform density. The centre of mass is then at the geometric centre of the sphere; the geometric centre is sometimes called the centroid. For a solid cylinder, with a uniform density, the centroid is half-way along its axis. We could obtain this results by applying the same arguments that were used for a sphere, since the cylinder is symmetric about this point. Symmetry also shows that the centroid of a cone is on its axis but not at the centre – because the mass of the cone is not evenly distributed along its length. To find the position of the centroid of the cone, we need to use the mathematical technique of integration (post 17.19); this is the general method for finding the position of a centroid and, therefore, a centre of gravity, for any object and will be the subject of a future post.