The Kinetic Energy of Rolling — Question 5

Question 5

A translating observer measures a different energy partition

A solid disk of mass $2\,\mathrm{kg}$ rolls at $4\,\mathrm{m/s}$ on stationary ground. A nonrotating observer moves right at $4\,\mathrm{m/s}$.

1. Find kinetic energy in ground and observer frames.

2. Why is substituting the new center speed into $\omega=V/R$ incorrect?

Question 5 - Solution

A translating observer measures a different energy partition

(a) Derivation and calculation

In the ground frame, $|\omega|=V/R$ and $I=MR^2/2$: $$K=\tfrac12MV^2+\tfrac14MV^2=\boxed{\tfrac34(2)(4)^2=24\ \mathrm J}.$$ In the translating observer frame, $V'=V-U=0$, while $\omega'=\omega$. Hence $$\boxed{K'=\tfrac12I\omega^2=\tfrac14(2)(4)^2=8\ \mathrm J}.$$

(b) Result and interpretation

The surface now moves with $U_g'=-4$ m/s. Its contact constraint is $$V'+\omega R=U_g',\qquad0+\omega R=-4.$$ Thus $\omega=-4/R$ remains nonzero. Setting $\omega=-V'/R=0$ would incorrectly treat the moving ground as stationary.

Physical check

The energy difference is the original 16 J of center translation; spin energy is unchanged.