Centre of mass

Up to now we treated every body as a point. A real body has size. A rigid body keeps its shape, and it can translate (every point moves the same way), rotate about a fixed axis, or do both at once, like a rolling wheel.

For a system of particles there is one special point, the centre of mass. It moves as if all the mass were packed there and all the external forces acted there. For two masses it lies on the line joining them, nearer the heavier one, with distances in the inverse ratio of the masses.

Example: 40 kg and 60 kg are 6 m apart. From the 40 kg, X = (40 × 0 + 60 × 6)/(40 + 60) = 3.6 m, so it is 2.4 m from the heavier one.

For many particles, take a weighted average of their positions, the weights being the masses:

• X = Σmᵢxᵢ/M, Y = Σmᵢyᵢ/M, Z = Σmᵢzᵢ/M, where M is the total mass.
• For a continuous body, the sum becomes an integral, X = ∫x dm/M. A uniform rod gives X = L/2, its middle, as you would expect.

Symmetry saves work. A uniform body's centre of mass lies on every line of symmetry, so a uniform rod, disc, sphere or rectangle has it at the geometric centre, and a triangle has it where the medians meet. It need not be inside the material: for a ring it is at the empty centre.

For a body with a piece cut out, treat the whole as (remainder) + (piece). Each piece's mass is in proportion to its area, so the cut-out piece is put in with a negative mass.

MA_cm = F_ext: only external forces can change how the centre of mass moves. Internal forces, however large, cannot. A shell that bursts in mid-air sends pieces everywhere, but their centre of mass carries on along the same parabola.

Example: an 80 kg man walks 6 m from one end of a 40 kg plank to the other, on smooth ice. No external force acts sideways, so the centre of mass must stay where it was. As the man goes forward 2 m over the ice, the plank slides 4 m back (80 × 2 = 40 × 4).