Angular Momentum — Question 9

Question 9

Angular momentum about an axis differs from the full vector

A $1\,\mathrm{kg}$ particle is at $(2,0,3)\,\mathrm m$ and moves with velocity $(0,4,0)\,\mathrm{m/s}$. Its instantaneous azimuthal angular speed about the $z$ axis is $2\,\mathrm{rad/s}$.

1. Find full angular momentum and its $z$ component.

2. Check $L_z=I_z\omega_z$ and explain why this does not imply $\vec L$ is parallel to $\hat z$.

Question 9 - Solution

Angular momentum about an axis differs from the full vector

(a) Derivation and calculation

The momentum is $\vec p=m\vec v=(0,4,0)$ kg m/s. Thus $$\vec L=(2,0,3)\times(0,4,0) =\boxed{(-12,0,8)\ \mathrm{kg\,m^2/s}},$$ $$\boxed{L_z=8\ \mathrm{kg\,m^2/s}}.$$

(b) Result and interpretation

The perpendicular distance to the $z$ axis is 2 m, giving $$I_z=m(x^2+y^2)=1(4)=4\ \mathrm{kg\,m^2},$$ $$\boxed{I_z\omega_z=4(2)=8=L_z}.$$ The height contributes $L_x=-zp_y=-12$, so the full vector is not parallel to $\hat z$. The scalar axis relation determines a projection, not every component.

Physical check

The same azimuthal motion at $z=0$ would remove the transverse component without changing $L_z$.