Are Angular Quantities Vectors? — Question 9

Section 10-2 · Physics pages 1408–1409

Question 9 Coaxial rates add after the reference sense is fixed A turntable rotates at +3ˆz rad/s relative to the lab. A small coaxial rotor turns at −8ˆz rad/s relative to the turntable. The turntable's rate increases at 1 rad/s2 while the relative rate stays fixed. x y + Coaxial rotations share one laboratory positive axis; their signed rates may oppose. (a) Find the rotor's laboratory angular velocity and acceleration. (b) Is its laboratory angular speed initially increasing or decreasing, and when does it reverse if the acceleration persists?
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Question 9 - Solution Coaxial rates add after the reference sense is fixed x y + (a) Derivation and calculation All rates are expressed about the same laboratory axis:⃗ ωlab =⃗ωtable +⃗ωrel = (3 −8)ˆz = −5ˆz rad/s .⃗ αlab = (1 + 0)ˆz = ˆz rad/s2 . (b) Result and interpretation Under the continued constant acceleration, ωz(t) = −5 + t, ωz(tr) = 0 ⇒tr = 5 s . Initially, d|ω| dt 0 = (−5)(1) 5 = −1 rad/s2. The rotor slows to rest, then accelerates in the positive sense. Physical check The relative rotor rate stays negative even when the laboratory rate changes sign.

Question 9

Coaxial rates add after the reference sense is fixed

A turntable rotates at +3ẑrad/s+3\hat z\,\mathrm{rad/s} relative to the lab. A small coaxial rotor turns at −8ẑrad/s-8\hat z\,\mathrm{rad/s} relative to the turntable. The turntable's rate increases at 1rad/s21\,\mathrm{rad/s^2} while the relative rate stays fixed.

Diagram shown in the original SVG worksheet.

  1. Find the rotor's laboratory angular velocity and acceleration.

  2. Is its laboratory angular speed initially increasing or decreasing, and when does it reverse if the acceleration persists?

Question 9 - Solution

Coaxial rates add after the reference sense is fixed

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

All rates are expressed about the same laboratory axis: ω→lab=ω→table+ω→rel=(3−8)ẑ=−5ẑrad/s.\vec\omega_{\mathrm{lab}}=\vec\omega_{\mathrm{table}}+\vec\omega_{\mathrm{rel}} =(3-8)\hat z=\boxed{-5\hat z\ \mathrm{rad/s}}. α→lab=(1+0)ẑ=ẑrad/s2.\vec\alpha_{\mathrm{lab}}=(1+0)\hat z=\boxed{\hat z\ \mathrm{rad/s^2}}.

(b) Result and interpretation

Under the continued constant acceleration, ωz(t)=−5+t,ωz(tr)=0⇒tr=5s.\omega_z(t)=-5+t,\qquad \omega_z(t_r)=0\Rightarrow\boxed{t_r=5\ \mathrm s}. Initially, d|ω|dt|0=(−5)(1)5=−1rad/s2.\left.\frac{d|\omega|}{dt}\right|_0=\frac{(-5)(1)}5=-1\ \mathrm{rad/s^2}. The rotor slows to rest, then accelerates in the positive sense.

Physical check

The relative rotor rate stays negative even when the laboratory rate changes sign.