Conservation of Angular Momentum — Question 9

Question 9

A small bearing torque must be included during contraction

A rotating system starts with $L_z=12\,\mathrm{kg\,m^2/s}$. Over $2.0\,\mathrm s$ its inertia changes to $2.0\,\mathrm{kg\,m^2}$ while a constant external bearing torque $-0.50\,\mathrm{N\,m}$ acts. No other external axial torque acts.

1. Find final angular speed.

2. Quantify the error made by assuming exact angular momentum conservation.

Question 9 - Solution

A small bearing torque must be included during contraction

(a) Derivation and calculation

Include the external angular impulse: $$L_f=L_i+\int_0^2\tau_z\,dt=12+(-.50)(2)=11\ \mathrm{kg\,m^2/s}.$$ With the final inertia, $$\boxed{\omega_f=L_f/I_f=11/2=5.50\ \mathrm{rad/s}}.$$

(b) Result and interpretation

Ignoring the torque would predict $\omega_{\mathrm{ideal}}=12/2=6.00$ rad/s. The overestimate is $$\boxed{\omega_{\mathrm{ideal}}-\omega_f=.500\ \mathrm{rad/s}},$$ $$\boxed{\frac{\omega_{\mathrm{ideal}}-\omega_f}{\omega_f}=\frac{.5}{5.5}=\frac1{11}=9.09\%}.$$ The relevant measure of torque is its integrated angular impulse over the event.

Physical check

The omitted impulse is $-1$ kg m$^2$/s; dividing by final inertia gives the .50 rad/s error.