The Rotational Variables — Question 8

Question 8

A velocity graph counts signed turns

A rotor's angular velocity decreases linearly from $+4\,\mathrm{rad/s}$ at $t=0$ to $-2\,\mathrm{rad/s}$ at $t=3\,\mathrm s$.

1. Find net displacement and total angular travel over the interval.

2. Find the reversal time and the fraction of total travel made clockwise.

Question 8 - Solution

A velocity graph counts signed turns

(a) Derivation and calculation

The straight-line velocity is $$\omega(t)=4+\frac{-2-4}{3}t=4-2t,\qquad\omega=0\Rightarrow t_r=2\ \mathrm s.$$ Its signed areas are $A_+=\tfrac12(2)(4)=4$ and $A_-=-\tfrac12(1)(2)=-1$ rad. Hence $$\boxed{\Delta\theta=A_++A_-=3\ \mathrm{rad}},\qquad \boxed{\mathcal S=|A_+|+|A_-|=5\ \mathrm{rad}}.$$

(b) Result and interpretation

The reversal time and clockwise fraction are $$\boxed{t_r=2\ \mathrm s},\qquad \boxed{f_{\mathrm{CW}}=\frac{|A_-|}{\mathcal S}=\frac15=0.20}.$$ Equivalently, $\Delta\theta=\int_0^3\omega\,dt$ but $\mathcal S=\int_0^3|\omega|\,dt$.

Physical check

The clockwise segment cancels one radian of displacement while adding one radian of travel.