Newton's Second Law in Angular Form — Question 6

Question 6

A torque perpendicular to angular momentum can turn it continuously

An angular momentum vector follows $\vec L=L_0[\cos(\Omega t)\hat x+\sin(\Omega t)\hat y]$, where $L_0$ and $\Omega$ are positive constants.

1. Find the required torque vector and magnitude.

2. Prove that this torque changes direction but not magnitude of angular momentum.

Question 6 - Solution

A torque perpendicular to angular momentum can turn it continuously

(a) Derivation and calculation

Differentiate the specified vector: $$\boxed{\vec\tau=\Omega L_0[-\sin(\Omega t)\hat x+\cos(\Omega t)\hat y]}.$$ Its magnitude is $$\boxed{|\vec\tau|=\Omega L_0\sqrt{\sin^2(\Omega t)+\cos^2(\Omega t)}=\Omega L_0}.$$

(b) Result and interpretation

The torque is perpendicular to angular momentum at every instant: $$\vec L\cdot\vec\tau=\Omega L_0^2[-\cos(\Omega t)\sin(\Omega t)+\sin(\Omega t)\cos(\Omega t)]=0.$$ Thus $$\frac{d}{dt}|\vec L|^2=2\vec L\cdot\vec\tau=0\Rightarrow\boxed{|\vec L|=L_0}.$$ The nonzero torque rotates the vector at angular rate $\Omega$ without changing its length.

Physical check

The same result can be written $\vec\tau=\Omega\hat z\times\vec L$.