Angular Momentum — Question 7

Question 7

Changing translational frame changes particle angular momentum

At an instant the origins of two nonrotating frames coincide. A particle has $m=2\,\mathrm{kg}$, $\vec r=(0,3,0)\,\mathrm m$, and ground-frame velocity $(4,0,0)\,\mathrm{m/s}$. The second frame moves at $(1,0,0)\,\mathrm{m/s}$.

1. Find angular momentum in each frame at that instant.

2. Derive the relation between them.

Question 7 - Solution

Changing translational frame changes particle angular momentum

(a) Derivation and calculation

At the coincident-origin instant, $$\vec p=m\vec v=8\hat x,\qquad\vec p'=m(\vec v-\vec U)=6\hat x.$$ The same position vector then gives $$\boxed{\vec L=(3\hat y)\times(8\hat x)=-24\hat z\ \mathrm{kg\,m^2/s}},$$ $$\boxed{\vec L'=(3\hat y)\times(6\hat x)=-18\hat z\ \mathrm{kg\,m^2/s}}.$$

(b) Result and interpretation

At that instant, $\vec r'=\vec r$, so $$\vec L'=\vec r\times m(\vec v-\vec U) =\boxed{\vec L-\vec r\times m\vec U}.$$ Here the correction is $-(3\hat y)\times(2\hat x)=+6\hat z$, exactly the difference found above. Angular momentum requires both the reference origin and the velocity frame to be specified.

Physical check

The two origins separate after this instant, so subsequent comparisons must also transform the position vector.