The Kinetic Energy of Rolling — Question 4

Question 4

The contact-axis energy formula is an instantaneous identity

A wheel of mass $3\,\mathrm{kg}$, radius $0.40\,\mathrm m$ and center inertia $0.24\,\mathrm{kg\,m^2}$ rolls at $2\,\mathrm{m/s}$ on fixed ground.

1. Compute its energy using center translation and rotation.

2. Recompute using the instantaneous contact axis and explain the limitation of that method.

Question 4 - Solution

The contact-axis energy formula is an instantaneous identity

(a) Derivation and calculation

The angular speed is $|\omega|=V/R=2/.40=5$ rad/s. Thus $$K_{\mathrm{trans}}=\tfrac12(3)(2)^2=6\ \mathrm J,$$ $$K_{\mathrm{rot}}=\tfrac12(.24)(5)^2=3\ \mathrm J,\qquad\boxed{K=9\ \mathrm J}.$$

(b) Result and interpretation

The instantaneous ground-contact point has zero velocity. By the parallel-axis theorem, $$I_P=I_C+MR^2=.24+3(.40)^2=.72\ \mathrm{kg\,m^2}.$$ Therefore $$\boxed{K=\tfrac12I_P\omega^2=\tfrac12(.72)(5)^2=9\ \mathrm J}.$$ This is an instantaneous kinetic-energy identity. The material contact point changes and is not a permanently fixed dynamical pivot.

Physical check

The parallel-axis term reproduces $MV^2/2$ because $R^2\omega^2=V^2$.