Angular Momentum — Question 4

Question 4

Angular momentum fixes tangential speed but leaves radial motion open

A $2\,\mathrm{kg}$ particle is at radius $3\,\mathrm m$ in a plane and has angular momentum $12\hat z\,\mathrm{kg\,m^2/s}$ about the plane origin.

1. Find tangential velocity and the minimum possible kinetic energy.

2. If its actual kinetic energy is $13\,\mathrm J$, determine the possible radial velocities.

Question 4 - Solution

Angular momentum fixes tangential speed but leaves radial motion open

(a) Derivation and calculation

Only tangential velocity contributes to angular momentum: $$L_z=mr v_\theta\Rightarrow\boxed{v_\theta=\frac{12}{2(3)}=2.00\ \mathrm{m/s}}.$$ The energy is $$K=\tfrac12m(v_r^2+v_\theta^2)\ge\tfrac12mv_\theta^2,$$ so $\boxed{K_{\min}=4.00\ \mathrm J}$, attained at $v_r=0$.

(b) Result and interpretation

For $K=13$ J, $$13=\tfrac12(2)(v_r^2+2^2)\Rightarrow v_r^2=9,$$ $$\boxed{v_r=+3.00\ \mathrm{m/s}\quad\text{or}\quad-3.00\ \mathrm{m/s}}.$$ Energy fixes the magnitude of radial motion; angular momentum fixes the tangential component. Neither measurement distinguishes outward from inward radial motion at this instant.

Physical check

The extra 9 J above the minimum is precisely the radial kinetic energy.