Angular Momentum — Question 6

Question 6

Projectile angular momentum about the launch point changes

A projectile of mass $m$ is launched from the origin with horizontal velocity $u>0$ and vertical velocity $w$. Ignore air drag, and take gravity downward with magnitude $g$.

1. Derive $L_z(t)$ about the launch point.

2. For $m=1$, $u=4$, $g=10$ and $t=2$ in SI units, evaluate $L_z$ and explain why $w$ cancels.

Question 6 - Solution

Projectile angular momentum about the launch point changes

(a) Derivation and calculation

The projectile position and velocity are $$x=ut,\quad y=wt-\tfrac12gt^2,\qquad v_x=u,\quad v_y=w-gt.$$ Therefore $$L_z=m[xv_y-yv_x] =m[ut(w-gt)-u(wt-\tfrac12gt^2)],$$ $$\boxed{L_z=-\tfrac12mugt^2}.$$

(b) Result and interpretation

At the stated instant, $$\boxed{L_z=-\tfrac12(1)(4)(10)(2)^2=-80\ \mathrm{kg\,m^2/s}}.$$ The initial vertical-motion terms $muwt$ cancel. As an independent check, $$\tau_z=-mgx=-mgut=\frac{dL_z}{dt}.$$ Gravity changes angular momentum about the launch point even though it supplies no torque about the particle itself.

Physical check

The result starts at zero and becomes clockwise for $u>0$ and $t>0$.