Torque Revisited — Question 10

Question 10

Torque components change with axes while rotational power does not

In a right-handed frame, $\vec\tau=(3,4,0)\,\mathrm{N\,m}$ and $\vec\omega=(2,-1,0)\,\mathrm{rad/s}$. New basis vectors are $\hat e_1=(3,4,0)/5$, $\hat e_2=(-4,3,0)/5$ and $\hat e_3=\hat z$.

1. Find torque and angular velocity components in the new frame.

2. Compute mechanical power in both frames and explain its invariance.

Question 10 - Solution

Torque components change with axes while rotational power does not

(a) Derivation and calculation

Components in the orthonormal basis are scalar projections: $$\tau'_1=(3,4,0)\cdot(3,4,0)/5=5,\quad\tau'_2=0,\quad\tau'_3=0.$$ Hence $\boxed{\vec\tau'=(5,0,0)\ \mathrm{N\,m}}$. Similarly, $$\omega'_1=\frac{2(3)-4}{5}=\frac25,\qquad \omega'_2=\frac{2(-4)-3}{5}=-\frac{11}{5},$$ so $\boxed{\vec\omega'=(2/5,-11/5,0)\ \mathrm{rad/s}}$.

(b) Result and interpretation

The original-coordinate power is $$\mathcal P=\vec\tau\cdot\vec\omega=3(2)+4(-1)=2\ \mathrm W.$$ The new components give $$\boxed{\mathcal P'=5(2/5)+0(-11/5)=2\ \mathrm W}.$$ An orthonormal change of coordinate basis preserves the dot product and therefore mechanical power.

Physical check

This is a change of basis for the same vectors, not a change to a differently rotating physical observer.