The Forces of Rolling — Question 7

Question 7

An initially unspun disk loses speed before rolling begins

A solid disk is launched on level rough ground with center speed $6\,\mathrm{m/s}$ and zero spin. Kinetic friction coefficient is $0.20$ and $g=10\,\mathrm{m/s^2}$.

1. Find the time and center speed when slipping first ends.

2. Find the fraction of initial kinetic energy dissipated.

Question 7 - Solution

An initially unspun disk loses speed before rolling begins

(a) Derivation and calculation

While the bottom slips right, friction acts left: $$a=-\mu g=-2\ \mathrm{m/s^2},\qquad \alpha=-\frac{\mu MgR}{MR^2/2}=-\frac4R.$$ Thus $$v=6-2t,\qquad R|\omega|=4t.$$ The first rolling instant satisfies $6-2t=4t$: $$\boxed{t_r=1.00\ \mathrm s},\qquad\boxed{v_r=4.00\ \mathrm{m/s}}.$$

(b) Result and interpretation

The initial and final energies are $$K_i=\tfrac12M(6)^2=18M,\qquad K_f=\tfrac34M(4)^2=12M,$$ where the coefficients carry units J/kg. Therefore $$\boxed{\frac{K_i-K_f}{K_i}=\frac{18-12}{18}=\frac13}.$$ After rolling begins, the ideal level surface requires no horizontal force and the speed stays constant.

Physical check

Friction decreases translation while building spin; their total energy decreases during slip.