Newton's Second Law for Rotation — Question 7

Question 7

A time-varying torque must be integrated twice for orientation

A rotor has fixed inertia $2\,\mathrm{kg\,m^2}$, initial angular velocity $1\,\mathrm{rad/s}$ and angle zero. Its net torque is $\tau(t)=4t\,\mathrm{N\,m}$ for numerical SI time $0\le t\le2\,\mathrm s$.

1. Find angular velocity and angle at $t=2\,\mathrm s$.

2. Compare the true angle with one obtained by assuming constant acceleration equal to its time average.

Question 7 - Solution

A time-varying torque must be integrated twice for orientation

(a) Derivation and calculation

Integrate the time-dependent rotational equation with the initial conditions: $$\alpha(t)=\frac{\tau(t)}I=2t,\qquad \omega(t)=1+\int_0^t2u\,du=1+t^2.$$ $$\theta(t)=\int_0^t(1+u^2)\,du=t+\frac{t^3}{3}.$$ $$\boxed{\omega(2)=5\ \mathrm{rad/s}},\qquad\boxed{\theta(2)=\frac{14}{3}=4.667\ \mathrm{rad}}.$$

(b) Result and interpretation

The time-average acceleration is $$\bar\alpha=\frac12\int_0^2 2t\,dt=2\ \mathrm{rad/s^2}.$$ Treating it as constant would give $$\theta_{\mathrm{const}}=1(2)+\tfrac12(2)(2)^2=6\ \mathrm{rad},\qquad \theta_{\mathrm{const}}-\theta(2)=\boxed{\frac43\ \mathrm{rad}}.$$ The average fixes final velocity, but loses the timing needed for orientation.

Physical check

Acceleration applied later has less time to contribute to angular displacement.