What is the angular momentum of a falling mass about the origin at any instant?

Answer 1

Depends on where you locate the origin. If the mass falls along the y-axis (x = 0) then the angular momentum with respect to the origin is zero. If you offset the mass so that it falls parallel to the y-axis, but has a position in x of x0, then you could compute:

r = sqrt(y^2+x0^2) where y = 1/2*gt^2 or

r = sqrt(1/4 g^2t^4 +x0^2)

Speed is dr/dt = 1/4*g^2*t^3/r

Angular momentum = L = r x p ---> |L| = rp*sin(q)

q = angle between vector r and vector p

If you work out the trig you get

sin(q) = x0/r

So |L| = (m/4)*(x0/r)*g^2t^3

Note that for x0 = 0, |L| = 0 as it must. Also note that |L| is not constant in time - it is not conserved. Gravity is exerting a torque on the system.

Answer 2

What do you mean by 'falling'?

What do you mean by 'origin'?

Unless there is a Torque on the body its angular momentum does not change. Gravity alone doesn't produce a torque on a falling body.