Conservation of Angular Momentum — Question 4

Question 4

Opposite rotors can stop each other without external angular impulse

Two coaxial rotors with inertias $2$ and $1\,\mathrm{kg\,m^2}$ rotate at $+3$ and $-6\,\mathrm{rad/s}$. An internal clutch locks them together; external axial torque is negligible.

1. Find final common angular speed.

2. Find energy converted into internal energy and explain why zero final speed is allowed.

Question 4 - Solution

Opposite rotors can stop each other without external angular impulse

(a) Derivation and calculation

The initial signed momenta cancel: $$L_i=I_1\omega_1+I_2\omega_2=2(3)+1(-6)=0.$$ After clutch engagement, $$(I_1+I_2)\omega_f=L_i\Rightarrow\boxed{\omega_f=0}.$$

(b) Result and interpretation

The initial energy is positive despite zero total angular momentum: $$K_i=\tfrac12(2)(3)^2+\tfrac12(1)(-6)^2=9+18=27.0\ \mathrm J.$$ Since $K_f=0$, $$\boxed{E_{\mathrm{internal}}=K_i-K_f=27.0\ \mathrm J}.$$ The clutch converts rotational kinetic energy into heat and deformation while conserving the signed angular-momentum sum.

Physical check

Opposite angular momenta cancel algebraically; positive kinetic energies cannot cancel.