Newton's Second Law for Rotation — Question 10

Question 10

Speed-dependent resistance produces an asymptotic terminal rate

A rotor with $I=1.0\,\mathrm{kg\,m^2}$ starts at rest under constant drive torque $2.0\,\mathrm{N\,m}$. Resistance is $-b\omega$ with $b=0.50\,\mathrm{N\,m\,s}$.

1. Solve for angular velocity and find when it reaches half its terminal value.

2. Find initial acceleration and explain whether the terminal rate is reached at finite time.

Question 10 - Solution

Speed-dependent resistance produces an asymptotic terminal rate

(a) Derivation and calculation

The fixed-axis equation with linear resistance is $$I\dot\omega=\tau_0-b\omega,\qquad \dot\omega+\frac bI\omega=\frac{\tau_0}I.$$ Solving with $\omega(0)=0$ yields $$\boxed{\omega(t)=\frac{\tau_0}{b}(1-e^{-bt/I}) =4(1-e^{-t/2})\ \mathrm{rad/s}}.$$ $$\omega=\tfrac12\omega_\infty\Rightarrow e^{-t/2}=\tfrac12 \Rightarrow\boxed{t_{1/2}=2\ln2=1.386\ \mathrm s}.$$

(b) Result and interpretation

Differentiate the solution: $$\alpha(t)=2e^{-t/2}\ \mathrm{rad/s^2},\qquad\boxed{\alpha(0)=2.0\ \mathrm{rad/s^2}}.$$ For every finite $t$, $e^{-t/2}>0$, so $\omega(t)<4$ rad/s. The terminal rate is the limiting balance $$\tau_0-b\omega_\infty=0\Rightarrow\boxed{\omega_\infty=4\ \mathrm{rad/s}}.$$

Physical check

The time constant $I/b=2$ s has time units, and the initial rate of acceleration is $\tau_0/I$.