Newton's Second Law in Angular Form — Question 7

Question 7

A finite transverse angular impulse changes the magnitude too

Initial angular momentum is $\vec L_i=3\hat z\,\mathrm{kg\,m^2/s}$. A torque pulse supplies total angular impulse $\int\vec\tau\,dt=4\hat x\,\mathrm{N\,m\,s}$.

1. Find final magnitude and tilt from $+z$.

2. Why does initial perpendicularity of the impulse direction not guarantee unchanged magnitude?

Question 7 - Solution

A finite transverse angular impulse changes the magnitude too

(a) Derivation and calculation

Integrate the vector torque law: $$\vec L_f=\vec L_i+\int\vec\tau\,dt =\boxed{4\hat x+3\hat z\ \mathrm{kg\,m^2/s}}.$$ Consequently, $$\boxed{|\vec L_f|=\sqrt{4^2+3^2}=5.00\ \mathrm{kg\,m^2/s}},$$ $$\boxed{\theta=\tan^{-1}(4/3)=53.13^\circ\quad\text{from }+z}.$$

(b) Result and interpretation

For this finite impulse, $$L_f^2=L_i^2+2\vec L_i\cdot\Delta\vec L+|\Delta\vec L|^2=9+0+16=25.$$ Perpendicularity to the initial vector does not remove the positive squared-impulse term. Constant magnitude requires $\vec L(t)\cdot\vec\tau(t)=0$ at every instant, not merely at the start.

Physical check

A torque maintained along $+x$ develops a component parallel to $\vec L$ as $L_x$ grows.