The Forces of Rolling — Question 8

Question 8

Excess spin can accelerate a wheel forward

A hoop is set on rough level ground with zero center speed and clockwise angular speed $\Omega_0$. Its radius is $R$, kinetic coefficient is $\mu$, and $I=MR^2$.

1. Find the time to rolling and final center speed.

2. Find the dissipated fraction of its initial kinetic energy.

Question 8 - Solution

Excess spin can accelerate a wheel forward

(a) Derivation and calculation

Initially the clockwise spin makes the bottom slip left, so friction is rightward: $$a=\mu g,\qquad\alpha=+\mu g/R.$$ Hence $$v=\mu gt,\qquad R|\omega|=R\Omega_0-\mu gt.$$ Their first equality gives $$\boxed{t_r=\frac{R\Omega_0}{2\mu g}},\qquad\boxed{v_r=\frac{R\Omega_0}{2}}.$$

(b) Result and interpretation

Initially only rotation contributes: $$K_i=\tfrac12MR^2\Omega_0^2.$$ At rolling, hoop translation and rotation are equal: $$K_f=Mv_r^2=\tfrac14MR^2\Omega_0^2.$$ Thus $\boxed{(K_i-K_f)/K_i=1/2}$. Friction increases center kinetic energy while removing an even larger amount of rotational energy.

Physical check

The final clockwise rim speed is half its initial value and equals the acquired center speed.