Newton's Second Law in Angular Form — Question 4

Question 4

Only the transverse impulse changes angular momentum at impact

During a very short impact a particle remains effectively at $\vec r=(0,2,0)\,\mathrm m$. It receives impulse $\vec J=(3,-1,0)\,\mathrm{N\,s}$.

1. Find the angular impulse about the origin.

2. Which part of the impulse changes linear momentum without changing angular momentum about this origin?

Question 4 - Solution

Only the transverse impulse changes angular momentum at impact

(a) Derivation and calculation

During the short impact the position is effectively fixed, so $$\Delta\vec L=\int\vec r\times\vec F\,dt=\vec r\times\vec J.$$ Substitute the impulse: $$\Delta\vec L=(2\hat y)\times(3\hat x-\hat y) =\boxed{-6\hat z\ \mathrm{kg\,m^2/s}}.$$

(b) Result and interpretation

Resolve the impulse into radial and transverse parts: $$\vec J_r=-\hat y\ \mathrm{N\,s},\qquad\vec J_\perp=3\hat x\ \mathrm{N\,s}.$$ Their moments are $$\vec r\times\vec J_r=0,\qquad\vec r\times\vec J_\perp=-6\hat z.$$ Both change linear momentum, $\Delta\vec p=\vec J$, but only the transverse component changes angular momentum about this origin.

Physical check

The approximation requires the particle displacement during the impulse to be negligible on the stated lever-arm scale.