Lesson 3 of 13 · 7 min
Motion of the centre of mass
NCERT §6.3, §6.4
At night the mela's fireworks begin. A shell leaves the launcher at 20 m/s, 60° above the ground, and bursts at the top of its flight into two equal pieces. Where do the pieces land? One simple point answers it.
The lesson in notes
In short
Differentiating MR = Σmᵢrᵢ gives MV = Σmᵢvᵢ and MA = Σmᵢaᵢ, for masses that do not change.
Internal forces come in equal and opposite pairs (third law) and cancel in the sum, so MA = F_ext: the centre of mass moves as if the whole mass were there and all the external forces acted there.
No knowledge of the internal forces is needed to find the motion of the centre of mass, whatever the body is doing inside, rotating or breaking up.
A shell that explodes in mid-flight: the explosion is internal, gravity is unchanged, so the centre of mass of the fragments keeps to the original parabola.
The total linear momentum of a system is P = Σmᵢvᵢ = MV, and dP/dt = F_ext: Newton's second law for a system.
If F_ext = 0, P is constant (conservation of linear momentum) and the centre of mass moves with constant velocity, even though the particles may follow complicated paths. The vector law is three scalar laws: Pₓ, P_y and P_z are each constant.
A radium nucleus decaying into radon and an alpha particle: the products move so that their centre of mass continues along the path of the radium nucleus. In the frame where the centre of mass is at rest, they fly apart back to back.
A binary star with no external force: the centre of mass moves in a straight line at constant speed, and in the centre-of-mass frame the two stars circle it, always on opposite sides.