The maximum force acting on a wheel that rolls without slipping as a function of the coefficient of friction

Question

If we assume that our tire is a homogenous disk and that it is rolling without slipping, with a force $F$ exerted on its center of mass pulling it forward, how would I determine the maximum force as a function of $\mu$, the coefficient of (static) friction.

I know that the friction is $\mu mg$, where $mg$ is the normal force, and that for the wheel to roll without slipping $v = R\omega$, where $v$ is the velocity of the center of mass, $R$ is the radius of the tire and $\omega$ the angular speed. We also have that for a uniform disk, $I = 1/2mR^2$

How do I now find what the maximum force is such that the tire rolls without slipping?

Answer 1

I know that the friction is μmg

Not always true. $\mu mg$ corresponds to the maximum friction, $f_{\rm max}$. The static friction force $f$ can be anywhere between $0$ and $f_{\rm max}$:

$$0\leq f\leq f_{\rm max} = \mu N$$

To solve the problem, first, write down the net force and torque equations

$$ma=F-f \qquad \text{and}\qquad I\alpha = fR$$

Using $v=\omega R\implies a=\alpha R$, eliminate $a$ and $\alpha$ to get an equation relating $f$ and $F$. You can also rewrite the inertia as $I=CmR^2$ to cancel the $R$'s.

Then, suppose that $f=f_\mathrm{max}=\mu mg$. This implies that the force of friction is maxed out, such that if $F$ gets any larger, $\mu$ must also increase. This simply means that the friction force acting on the disk cannot get any larger without increasing $\mu$.

From there, solve for $F$ and you're done.