A spring, disk and pulley system

[Orodruin]

Image a wheel rolling to the right with its CoM moving with some constant velocity without slipping. There must be no net torque on the wheel for this to be the case. However, from the perspective of the axle the static friction force ( no matter what direction its pointing ) is providing a torque? What is countering that torque?

Nothing. The frictional force is zero in the situation you describe. The wheel is already spinning at exactly the correct angular velocity to not slip.

 

[erobz]

Nothing. The frictional force is zero in the situation you describe. The wheel is already spinning at exactly the correct angular velocity to not slip.

How am I so horribly confused! So static friction is just applied while the wheel is accelerating AND when the wheel is completely stationary?EDIT:
Just to help get you guys in the head of someone about 100 IQ points lower than yourselves . I realize the static friction force adjusts itself "up to a limit" as a reaction to the acceleration, and in the stationary case it adjusts itself "up to a limit" as a reaction to the applied torque on the wheel. So the fact that it disappears in the last scenario I described is ok with me. But, if it is the static friction that is responsible for the acceleration of the CoM (as it would be in the case of walking according to my interpretation of what @kuruman was suggesting), then in the case of the wheel it seems to necessarily change direction between completely static and accelerating CoM? Or is it that it’s direction doesn’t change, and it’s magnitude drops once acceleration begins?

 

[erobz]

I’m getting myself all mixed up( hopefully not others).

In the case where the wheel is static, and not slipping there must be some other external force stopping the wheel from accelerating.

In the case were the wheel is accelerating that external force is no longer present. The static friction is in the same direction, but its magnitude is between 0 and $\mu_s N$ and there is a net torque about to axle, and the wheel is not slipping. $\mu_s N$ will act as a limit for the acceleration of the wheel.

I hope I’m dialing it back in?

 

[kuruman]

I’m getting myself all mixed up( hopefully not others).

In the case where the wheel is static, and not slipping there must be some other external force stopping the wheel from accelerating.

In the case were the wheel is accelerating that external force is no longer present. The static friction is in the same direction, but its magnitude is between 0 and $\mu_s N$ and there is a net torque about to axle, and the wheel is not slipping. $\mu_s N$ will act as a limit for the acceleration of the wheel.

I hope I’m dialing it back in?

What does "static and not slipping" mean? Can you provide an example of a wheel that does that? To me it means a wheel at rest on a horizontal surface. Setting vertical forces aside, this means that (a) the sum of all horizontal forces is zero and (b) the sum of all torques about any point in space is zero. If you postulate that static friction $f_s$ acts in some direction at the point of contact, then another force of equal magnitude but in opposite direction must be acting on the wheel to satisfy condition (a). Furthermore, to satisfy condition (b), that opposing force must applied at the same point as the $f_s$ that you postulated. The two cancel out and the horizontal force at the point of contact is zero.

 

[erobz]

What does "static and not slipping" mean? Can you provide an example of a wheel that does that? To me it means a wheel at rest on a horizontal surface. Setting vertical forces aside, this means that (a) the sum of all horizontal forces is zero and (b) the sum of all torques about any point in space is zero. If you postulate that static friction $f_s$ acts in some direction at the point of contact, then another force of equal magnitude but in opposite direction must be acting on the wheel to satisfy condition (a). Furthermore, to satisfy condition (b), that opposing force must applied at the same point as the $f_s$ that you postulated. The two cancel out and the horizontal force at the point of contact is zero.

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.

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 $R \omega \neq v_{CM}$

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

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

 

[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.