The Kinetic Energy of Rolling — Question 7

Question 7

Small wheels can matter through effective translating mass

A cart body of mass $6\,\mathrm{kg}$ carries four solid-disk wheels, each of mass $0.50\,\mathrm{kg}$ and radius $R$. All translate at $V$ and roll without slip.

1. Express total kinetic energy as $\tfrac12 M_{eff}V^2$.

2. Find the speed after a lossless drop of $1\,\mathrm m$ and explain which mass appears in gravitational energy.

Question 7 - Solution

Small wheels can matter through effective translating mass

(a) Derivation and calculation

The translating mass includes the body and all wheels: $$M_{\mathrm{total}}=6+4(.50)=8\ \mathrm{kg}.$$ Each wheel contributes $I/R^2=m_w/2=.25$ kg to the energy coefficient. Therefore $$K=\frac12\left(M_{\mathrm{total}}+\sum\frac{I_i}{R^2}\right)V^2 =\frac12(8+1)V^2,$$ $$\boxed{M_{\mathrm{eff}}=9\ \mathrm{kg}}.$$

(b) Result and interpretation

Gravitational potential uses the actual mass: $$M_{\mathrm{total}}gh=\tfrac12M_{\mathrm{eff}}V^2.$$ For $h=1$ m and $g=9.8$ m/s$^2$, $$\boxed{V=\sqrt{\frac{2(8)(9.8)(1)}9}=4.174\ \mathrm{m/s}}.$$ The extra 1 kg in $M_{\mathrm{eff}}$ represents rotational energy, not additional weight.

Physical check

Ignoring wheel spin would overpredict speed by using 8 kg rather than 9 kg in the kinetic-energy coefficient.