Are Angular Quantities Vectors? — Question 2

Question 2

A negative angular acceleration can increase the speed

A fixed-axis rotor has $\vec\omega=-4\hat z\,\mathrm{rad/s}$ and $\vec\alpha=-3\hat z\,\mathrm{rad/s^2}$ at an instant.

1. Is its angular speed increasing or decreasing? Give the signed rate of change of the speed.

2. If the acceleration stays constant, find the vector after $2\,\mathrm s$ and state whether a reversal occurs.

Question 2 - Solution

A negative angular acceleration can increase the speed

(a) Derivation and calculation

Differentiate the magnitude, not the signed component: $$\frac{d|\omega|}{dt}=\frac{\vec\omega\cdot\vec\alpha}{|\omega|} =\frac{(-4)(-3)}4=\boxed{3\ \mathrm{rad/s^2}}.$$ The positive result means the rotor speeds up.

(b) Result and interpretation

With constant acceleration, $$\vec\omega(t)=\vec\omega_0+\vec\alpha t=(-4-3t)\hat z.$$ $$\boxed{\vec\omega(2)=-10\hat z\ \mathrm{rad/s}},\qquad -4-3t<0\quad(t\ge0).$$ There is no future zero crossing or reversal.

Physical check

The magnitude grows from 4 to 10 rad/s in 2 s, consistent with a rate of 3 rad/s$^2$.