Newton's Second Law in Angular Form — Question 8

Question 8

A moving reference point contributes a transport term

A free particle has mass $1\,\mathrm{kg}$ and ground-frame position $(2,3t,0)\,\mathrm m$. A reference point $A$ has position $(t,0,0)\,\mathrm m$. Define $\vec L_A=(\vec r-\vec r_A)\times\vec p$ using ground-frame momentum.

1. Find $\vec L_A(t)$ and its time derivative.

2. Explain why that derivative is nonzero even though force and torque vanish.

Question 8 - Solution

A moving reference point contributes a transport term

(a) Derivation and calculation

The ground-frame momentum is $\vec p=(0,3,0)$ kg m/s. Relative position is $$\vec r-\vec r_A=(2-t,3t,0)\ \mathrm m.$$ Hence $$\boxed{\vec L_A=(6-3t)\hat z\ \mathrm{kg\,m^2/s}},\qquad \boxed{\frac{d\vec L_A}{dt}=-3\hat z\ \mathrm{N\,m}}.$$

(b) Result and interpretation

The force vanishes, so $\vec\tau_A=0$. Because inertial momentum is used, $$\frac{d\vec L_A}{dt}=\vec\tau_A-\vec V_A\times\vec p =0-\hat x\times3\hat y=-3\hat z.$$ The moving lever arm accounts for the change. Using momentum relative to $A$ would define a different angular momentum.

Physical check

The particle is force-free, yet its distance from the moving reference line changes with time.