Are Angular Quantities Vectors? — Question 6

Question 6

A turning angular-velocity vector has acceleration at constant speed

A rigid body has instantaneous angular velocity $\vec\omega(t)=4[\cos(2t)\hat x+\sin(2t)\hat y]\,\mathrm{rad/s}$, with time in seconds.

1. Find angular acceleration and its magnitude.

2. At $t=0$, compare the directions of $\vec\omega$ and $\vec\alpha$. Does constant angular speed imply zero angular acceleration?

Question 6 - Solution

A turning angular-velocity vector has acceleration at constant speed

(a) Derivation and calculation

Differentiate the Cartesian components: $$\boxed{\vec\alpha=8[-\sin(2t)\hat x+\cos(2t)\hat y]\ \mathrm{rad/s^2}}.$$ $$|\vec\alpha|=8\sqrt{\sin^2(2t)+\cos^2(2t)}=\boxed{8\ \mathrm{rad/s^2}}.$$

(b) Result and interpretation

At the specified instant, $$\boxed{\vec\omega(0)=4\hat x,\qquad\vec\alpha(0)=8\hat y}.$$ At every time, $$\vec\omega\cdot\vec\alpha=32[-\cos(2t)\sin(2t)+\sin(2t)\cos(2t)]=0.$$ Thus $d|\omega|/dt=0$ while $d\vec\omega/dt\ne0$: the axis turns at constant angular speed magnitude.

Physical check

In angular-velocity space the derivative is tangent to the circle, perpendicular to its radius.