A spring, disk and pulley system

[haruspex]

1. Wheel with radius $R$ on a horizontal surface: No friction
If the wheel sees an external torque it rotates, but does not translate. It slips.

2. Wheel on a horizontal surface: With Friction
If the wheel sees an external torque, it rotates and translates. If it sees a torque greater than $R \mu_s N$ it slips. slips means $R \omega \neq v_{CM}$

3. Wheel on a horizontal surface with friction being held by an external force that stops translation. For some range of torque less than $\mu_s N$ the wheel remains static and does not slip. If the torque exceeds $\mu_s N$ the wheel slips, but does not translate.

Are all these scenarios consistent with some understanding, or do I still not get something?

1. Yes
2. Not quite. The radius of inertia matters. If you consider moments and forces you get $\tau\leq R\mu_sN(1+k^2)$, where the radius of inertia is kR.
3. This one is $R \mu_s N$. $\mu_s N$ is not a torque.

 

[haruspex]

So static friction is just applied while the wheel is accelerating AND when the wheel is completely stationary?

There being no other horizontal forces, there is a nonzero frictional force if and only if there is a nonzero acceleration.
When there is no friction (on a frictional surface) the wheel could be stationary or rolling at constant velocity with a matching angular velocity.

 

[erobz]

1. Yes
2. Not quite. The radius of inertia matters. If you consider moments and forces you get $\tau \leq R \mu_s N (1+k^2)$, where the radius of inertia is kR.
3. This one is $R \mu_s N$. $\mu_s N$ is not a torque.

3 was a typo

Can you elaborate on 2. Is that otherwise known as the radius of gyration?

 

[haruspex]

3 was a typo

Can you elaborate on 2. Is that otherwise known as the radius of gyration?

Yes, same as radius of gyration. For a uniform disk, $\tau\leq R\mu_sN\frac 32$.

 

[erobz]

Yes, same as radius of gyration. For a uniform disk, $\tau\leq R\mu_sN\frac 32$.

So that is making a torque from the IC of zero velocity. Why does that arise as the criterion for slip? I notice that for the case of a flat disk it just differs from the frictional torque by a constant.

 

[haruspex]

So that is making a torque from the IC of zero velocity. Why does that arise as the criterion for slip? I notice that for the case of a flat disk it just differs from the frictional torque by a constant.

If no slip, $mr\alpha=ma=F_{sf}\leq N\mu_s$.
Moments about point of contact: $\tau=\alpha (mr^2+I_{cm})=\alpha mr^2(1+k^2)$
The result follows.