Kinetic Energy of Rotation — Question 4

Question 4

A kinetic-energy calibration can contain a zero offset

A readout obeys $E_{\mathrm{read}}=E_0+\tfrac12I\omega^2$. It displays $7\,\mathrm J$ at $\omega=2\,\mathrm{rad/s}$ and $19\,\mathrm J$ at $\omega=4\,\mathrm{rad/s}$.

1. Infer inertia and offset.

2. Predict the readout at zero speed and at $\omega=-3\,\mathrm{rad/s}$.

Question 4 - Solution

A kinetic-energy calibration can contain a zero offset

(a) Derivation and calculation

The two calibration equations are $$7=E_0+\tfrac12I(4),\qquad19=E_0+\tfrac12I(16).$$ $$12=6I\Rightarrow\boxed{I=2\ \mathrm{kg\,m^2}},\qquad \boxed{E_0=7-\tfrac12(2)(4)=3\ \mathrm J}.$$

(b) Result and interpretation

The inferred readouts are $$\boxed{E_{\mathrm{read}}(0)=3\ \mathrm J},\qquad \boxed{E_{\mathrm{read}}(-3)=3+\tfrac12(2)(-3)^2=12\ \mathrm J}.$$ The actual kinetic energies are 0 and 9 J; the offset is not stored rotational energy.

Physical check

The readout is even in angular velocity, as required by its squared-speed dependence.