The Rotational Variables — Question 4

Question 4

Average acceleration does not require constant acceleration

A rotor has $\omega(t)=2+t^2\,\mathrm{rad/s}$ for SI numerical $t$ between $0$ and $3\,\mathrm s$.

1. Find average angular acceleration and the time when instantaneous acceleration equals that average.

2. Find angular displacement and compare with the trapezoidal average of endpoint angular velocities.

Question 4 - Solution

Average acceleration does not require constant acceleration

(a) Derivation and calculation

Use the endpoint difference for the average and the derivative for the instantaneous value: $$\bar\alpha=\frac{\omega(3)-\omega(0)}{3} =\frac{11-2}{3}=\boxed{3\ \mathrm{rad/s^2}}.$$ $$\alpha(t)=2t,\qquad 2t=3\quad\Longrightarrow\quad\boxed{t=1.5\ \mathrm s}.$$

(b) Result and interpretation

The exact angular displacement is the area under the curve: $$\Delta\theta=\int_0^3(2+t^2)\,dt =\left[2t+\frac{t^3}{3}\right]_0^3=\boxed{15\ \mathrm{rad}}.$$ $$\Delta\theta_{\mathrm{trap}}=\frac{2+11}{2}(3)=19.5\ \mathrm{rad},\qquad \Delta\theta_{\mathrm{trap}}-\Delta\theta=4.5\ \mathrm{rad}.$$ Endpoint averaging requires a linear velocity history; this curve is quadratic.

Physical check

The convex velocity curve lies below its endpoint chord, so the trapezoidal estimate is too large.