Newton's Second Law in Angular Form — Question 9

Question 9

A small unmeasured torque gives a rigorous conservation tolerance

Angular momentum initially has magnitude $10\,\mathrm{kg\,m^2/s}$. The net torque about a fixed origin has magnitude at most $0.030\,\mathrm{N\,m}$ for $20\,\mathrm s$, with unknown time-dependent direction.

1. Bound the change in the angular momentum vector.

2. Give a guaranteed interval for the final magnitude and explain whether the bound must be attained.

Question 9 - Solution

A small unmeasured torque gives a rigorous conservation tolerance

(a) Derivation and calculation

The angular impulse and integral triangle inequality give $$\Delta\vec L=\int_0^{20}\vec\tau(t)\,dt,$$ $$|\Delta\vec L|\le\int_0^{20}|\vec\tau(t)|\,dt \le(.030)(20).$$ Thus $$\boxed{|\Delta\vec L|\le.600\ \mathrm{kg\,m^2/s}}.$$

(b) Result and interpretation

By the reverse and ordinary triangle inequalities, $$|L_i|-|\Delta\vec L|\le|L_f|\le|L_i|+|\Delta\vec L|.$$ Therefore $$\boxed{9.40\le|L_f|\le10.60\ \mathrm{kg\,m^2/s}}.$$ This is a guaranteed bound, not a statistical uncertainty. Direction changes can cancel angular impulse; equality requires appropriately aligned torque throughout.

Physical check

Constant maximum torque parallel or antiparallel to the initial vector attains the two limiting magnitudes.