The Kinetic Energy of Rolling — Question 1

Question 1

The same rolling speed does not imply the same energy

A hoop and a solid disk each have mass $2\,\mathrm{kg}$, radius $0.30\,\mathrm m$ and center speed $3\,\mathrm{m/s}$. Both roll without slip. Use $I_h=MR^2$ and $I_d=MR^2/2$.

1. Compare total kinetic energies.

2. If their total energies were equal instead, which would move faster?

Question 1 - Solution

The same rolling speed does not imply the same energy

(a) Derivation and calculation

For $I=\beta MR^2$ and no slip, $|\omega|=V/R$, so $$K=\tfrac12MV^2+\tfrac12I(V/R)^2=\tfrac12M(1+\beta)V^2.$$ For the hoop and disk, $$\boxed{K_h=\tfrac12(2)(2)(3)^2=18.0\ \mathrm J},$$ $$\boxed{K_d=\tfrac12(2)(1.5)(3)^2=13.5\ \mathrm J}.$$

(b) Result and interpretation

At fixed total $K$ and mass, $$V=\sqrt{\frac{2K}{M(1+\beta)}}.$$ Consequently, $$\boxed{\frac{V_d}{V_h}=\sqrt{\frac{1+1}{1+1/2}}=\sqrt{\frac43}=1.155}.$$ The disk devotes a smaller fraction of its energy to rotation and therefore translates faster.

Physical check

At equal speed the translational energies agree; the hoop has twice the disk rotational energy.