The Kinetic Energy of Rolling — Question 9

Question 9

Rolling resistance can be inferred from an energy deficit

A $2\,\mathrm{kg}$ disk rolls without slip from rest down a path of length $4\,\mathrm m$ and vertical drop $1\,\mathrm m$. Its final speed is $3\,\mathrm{m/s}$. Model all losses by a constant effective resisting force $f_r$ whose work is $-f_r s$. Use $g=9.8$.

1. Infer $f_r$.

2. Find the fraction of released potential energy dissipated; distinguish this model from ideal static friction.

Question 9 - Solution

Rolling resistance can be inferred from an energy deficit

(a) Derivation and calculation

For a rolling disk, $$K_f=\tfrac34MV^2=\tfrac34(2)(3)^2=13.5\ \mathrm J.$$ The energy balance with the stipulated loss model is $$Mgh-f_rs=K_f.$$ Hence $$\boxed{f_r=\frac{2(9.8)(1)-13.5}{4}=1.525\ \mathrm N}.$$

(b) Result and interpretation

The dissipated energy and its fraction are $$E_{\mathrm{loss}}=19.6-13.5=6.10\ \mathrm J,\qquad \boxed{\frac{E_{\mathrm{loss}}}{Mgh}=\frac{6.1}{19.6}=.311}.$$ For ideal static contact on stationary ground, $$\mathcal P_f=\vec f\cdot\vec v_{\mathrm{contact}}=0.$$ The effective resistance therefore describes deformation or other dissipation beyond ideal no-slip static friction.

Physical check

The inferred work $f_rs=6.10$ J closes the energy balance exactly.