Relating the Linear and Angular Variables — Question 6

Question 6

At reversal the velocity vanishes but acceleration need not

A disk rotates about a fixed axis with $\omega(t)=4-2t\,\mathrm{rad/s}$ for SI time. A rim point is at radius $0.30\,\mathrm m$.

1. Find its speed and the radial and tangential acceleration magnitudes at $t=2\,\mathrm s$.

2. Compare radial acceleration just before and after reversal at $t=1.5$ and $2.5\,\mathrm s$.

Question 6 - Solution

At reversal the velocity vanishes but acceleration need not

(a) Derivation and calculation

The angular acceleration remains constant even at the zero of velocity: $$\omega(2)=4-2(2)=0,\qquad\alpha=\frac{d\omega}{dt}=-2\ \mathrm{rad/s^2}.$$ $$\boxed{v=r|\omega|=0},\qquad\boxed{a_{\mathrm{in}}=r\omega^2=0},\qquad \boxed{|a_\theta|=r|\alpha|=0.60\ \mathrm{m/s^2}}.$$

(b) Result and interpretation

At the two neighboring instants, $$\omega(1.5)=+1,\qquad\omega(2.5)=-1\quad\mathrm{rad/s}.$$ $$\boxed{a_{\mathrm{in}}(1.5)=a_{\mathrm{in}}(2.5)=0.30(1)^2=0.30\ \mathrm{m/s^2}}.$$ The signed tangential component remains negative, slowing the rotor before reversal and speeding it afterward.

Physical check

Zero instantaneous velocity does not imply zero acceleration.