The Angular Momentum of a Rigid Body Rotating About a Fixed Axis — Question 3

Question 3

Constant axial speed can still require a bearing torque

The tilted dumbbell has $m=1\,\mathrm{kg}$ for each mass, $a=0.50\,\mathrm m$, $b=0.20\,\mathrm m$ and constant $\omega=6\,\mathrm{rad/s}$. At $t=0$ its mass positions are $\pm(a,0,b)$. Ignore gravity.

1. Find the net external torque at $t=0$ required to maintain this rotation.

2. Find its power and explain why zero $\alpha_z$ does not imply zero vector torque.

Question 3 - Solution

Constant axial speed can still require a bearing torque

(a) Derivation and calculation

The transverse momentum rotates with the assembly: $$\vec L_\perp(t)=-2mab\omega[\cos(\omega t)\hat x+\sin(\omega t)\hat y].$$ Its axial component is constant. Therefore at $t=0$, $$\vec\tau=\left.\frac{d\vec L}{dt}\right|_0=-2mab\omega^2\hat y,$$ $$\boxed{\vec\tau=-2(1)(.50)(.20)(6)^2\hat y=-7.20\hat y\ \mathrm{N\,m}}.$$

(b) Result and interpretation

The required torque is transverse to the angular velocity: $$\boxed{\mathcal P=\vec\tau\cdot\vec\omega=(-7.2\hat y)\cdot(6\hat z)=0}.$$ The bearings turn $\vec L_\perp$ while keeping speed and $L_z$ fixed. The axial law remains valid: $$\tau_z=I_z\alpha_z=0.$$ Zero axial angular acceleration therefore does not require zero vector torque.

Physical check

The spin kinetic energy is constant because the constraint torque does no work.