A spring, disk and pulley system

[lorenz0]

Homework Statement

A rope is wound around the axis orthogonal to a disk of mass $M$ and radius $R$ and there is a mass $m$ hanging from this rope as shown in the figure. The disk rests on a horizontal plane and connected to its center there is a spring which has its other second end fixed on a pulley.
(a) Determine the deformation of the spring at equilibrium.
(b) At a certain instant the mass $m$ is slightly lowered from the equilibrium position and then released. Determine the period of the oscillations that are caused by knowing that the disk rolls without slipping.


Assume $M, R, r, k, m, m_{pulley}, r_{pulley} = r$ all known.

Relevant Equations

$\vec{F}=m\vec{a},\ \tau=I\alpha,\ a=R\alpha$

(a) By setting up a coordinate system with the x-axis pointing to the right and the y-axis pointing downward we have $\begin{cases}-kx_{eq}+T_1+F_{s}=0\\ -RF_{s}+rT_1=0\\ r_p (T_2-T_1)=0\\ -T_2+mg=0\end{cases}\Rightarrow x_{eq}=\frac{mg}{k}\left(1+\frac{r}{R}\right)$ which coincides with the solution given.

(b) If we denote by $a_{D}$ is the the acceleration of the center of mass of the disk, I think the the goal should then be to obtain an equation of the form $a_{D}=\frac{\text{const}\times K}{\text{some combination of the masses}}(x-x_{eq})$ so that $T=\frac{2\pi}{\omega}$ where $\omega=\sqrt{\frac{\text{const}\times K}{\text{some combination of the masses}}}$. I have thus set up the following system:

$\begin{cases}-kx_{eq}+T_1+F_{s}=Ma_D\\ -RF_{s}+rT_1=I_D \frac{a_D}{R}\\ r_p (T_2-T_1)=I_{pulley}\alpha_{pulley}=I_{pulley}\frac{a_D}{r}\\ -T_2+mg=ma_D\end{cases}$

The problem is that the solution given says that in equation (3) the RHS should be $\frac{I_{pulley}}{r_{pulley}}\frac{a_D}{R}(r+R)$ and in equation (4) the RHS should be $m\frac{a_D}{R}(r+R)$ but I haven't been able to figure out why that is.
If the rope is ideal, shouldn't mass $m$ have the same acceleration as the disk?

Thanks

 

[Orodruin]

If the rope is ideal, shouldn't mass m have the same acceleration as the disk?

No, why would it? Think about what is said in the problem statement about how the disk is attached to the rope.

 

[lorenz0]

No, why would it? Think about what is said in the problem statement about how the disk is attached to the rope.

Thinking about it, since the rope is attached to axis of radius $r$ orthogonal to the disk, if the disk moves a distance $x$ then, since it moves without slipping it must be $x=R\theta$ where $\theta$ is the angle the disk (and also the axis) turn and so the length of the rope changes by an amount $\Delta l=r\theta=r\frac{x}{R}$ and by differentiating two times with respect to time we get the acceleration of a piece of the rope is $a=\frac{r}{R}a_D$. Now, since the rope is ideal I think mass $m$ should move downward with the same acceleration as a piece of the rope so I get for the RHS of equation (4) $m\frac{a_D}{R}r$ which is closer, but not equal, to $m\frac{a_D}{R}(r+R)$.

Does this make sense? How can I improve my reasoning? Thanks.

 

[TSny]

the length of the rope changes by an amount $\Delta l=r\theta$

This represents the amount of rope that unwinds when the disk rotates by $\theta$. But, this is not the amount by which the hanging mass moves downward when the disk rotates by $\theta$.

 

[erobz]

This represents the amount of rope that unwinds when the disk rotates by $\theta$. But, this is not the amount by which the hanging mass moves downward when the disk rotates by $\theta$.

Just to see if I'm following along: The speed at which the mass descends will be greater than $r \dot \theta$?

 

[Orodruin]

Just to see if I'm following along: The speed at which the mass descends will be greater than $r \dot \theta$?

It would be $r\dot \theta$ if the disk was rotating in place, which it is not.

 

[lorenz0]

I think I get it now: when the disk moves leftward for example, it brings up $R\theta$ of the rope and its axis of radius $r$ "eats" $r\theta$ of the rope (because the rope is winding around it) so the combined effect on $m$ is that it must go upward of an amount $s=(R+r)\theta$ and by differentiating twice with respect to time we get for $m$ an acceleration $a=(R+r)\alpha=(R+r)\frac{a_D}{R}$ and since for the pulley it must be $r_p\alpha_{p}=a$ we get $\alpha_p=\frac{1}{r_p}(R+r)\frac{a_D}{R}\overset{r_p=r}{=}\frac{1}{r}(R+r)\frac{a_D}{R}$.

 

[erobz]

I think that's correct.

EDIT: Just to be clear my $x$ was not the CoM coordinate $x$ you have defined. So I've changed the notation to account for that.

I figured for the rope of length $l$ you have:

$$ y - x_p = l \implies \ddot y = \ddot x_P $$

The point $P$ where the rope attaches to the wheel has a velocity w.r.t. to coordinate system:

$$ \dot x_P = r \dot \theta + R \dot \theta \implies \ddot x_P = \left( R+r \right) \ddot \theta $$

It follows that the acceleration of the hanging mass $\ddot y$ is given by:

$$ \ddot y = \left( R+r \right) \ddot \theta \implies \ddot y = \left( R+r \right) \frac{ \ddot x}{R} $$

 

[kuruman]

I think I get it now: when the disk moves leftward for example, it brings up $R\theta$ of the rope and its axis of radius $r$ "eats" $r\theta$ of the rope (because the rope is winding around it) so the combined effect on $m$ is that it must go upward of an amount $s=(R+r)\theta$ and by differentiating twice with respect to time we get for $m$ an acceleration $a=(R+r)\alpha=(R+r)\frac{a_D}{R}$ and since for the pulley it must be $r_p\alpha_{p}=a$ we get $\alpha_p=\frac{1}{r_p}(R+r)\frac{a_D}{R}\overset{r_p=r}{=}\frac{1}{r}(R+r)\frac{a_D}{R}$.

Another way to look at it is this. When the disk rolls without slipping, the point on the disk that is (instantaneously) in contact with the surface is at rest relative to the surface and acts as a pivot. An arbitrary point on the disk rotates about the contact point with angular velocity $\vec{\omega}$ and has linear velocity $\vec{v}=\vec{\omega}\times \vec{r}$ where $\vec{r}$ is its position vector relative to the point of contact.

It follows that the linear speed of the point where the string is attached is $v=\omega(R+r)$. We can also find that the speed of the disk's center is $V_{cm}=\omega~R~\implies\omega=V_{cm}/R$. Hence, we can eliminate $\omega$ and rewrite the speed of the point of interest as $v=V_{cm}\left(1+\frac{r}{R}\right).$ Of course, if the string isn't slipping at that point, its speed will be the same at all its points and so will the mass attached to the other end. If the string were wrapped around the shaft the other way and came out the bottom, the speed of the string would be $v=V_{cm}\left(1-\frac{r}{R}\right).$

 

[lorenz0]

Another way to look at it is this. When the disk rolls without slipping, the point on the disk that is (instantaneously) in contact with the surface is at rest relative to the surface and acts as a pivot. An arbitrary point on the disk rotates about the contact point with angular velocity $\vec{\omega}$ and has linear velocity $\vec{v}=\vec{\omega}\times \vec{r}$ where $\vec{r}$ is its position vector relative to the point of contact.

It follows that the linear speed of the point where the string is attached is $v=\omega(R+r)$. We can also find that the speed of the disk's center is $V_{cm}=\omega~R~\implies\omega=V_{cm}/R$. Hence, we can eliminate $\omega$ and rewrite the speed of the point of interest as $v=V_{cm}\left(1+\frac{r}{R}\right).$ Of course, if the string isn't slipping at that point, its speed will be the same at all its points and so will the mass attached to the other end. If the string were wrapped around the shaft the other way and came out the bottom, the speed of the string would be $v=V_{cm}\left(1-\frac{r}{R}\right).$

This is very helpful, thanks.

 

[erobz]

How is $x_{eq}$ measured though?

I'm imagining the free length of the spring is $l_o$ , but I see no mention of it?

Your first eq in the system would be?

$$ -k \left( x_{eq} - l_o \right) + T_1 + F_s = 0 \implies x_{eq} = l_o + \frac{mg}{k} \left( 1 + \frac{r}{R} \right)$$

In this coordinate system $x_{eq}$ and $l_o$ are both taken as negative I believe.EDIT:
I see that it's just the addition of an arbitrary constant, and I suppose that is not going to have an effect the solution to part (b).

 

[erobz]

While I'm at it:

$$ - RF_s + r T_1 = I \frac{a_D}{R} $$

$F_s$ is the static friction force. It just occurred to me that you can only find $F_s$ as:

$$ 0 < F_s = \frac{1}{R} \left( r T_1 - \frac{a_D}{R} \right) \leq \mu_s N $$

Is that accurate? You just have to check to see if the displacement doesn't cause $F_s$ escape that range, otherwise the wheel slips?

 

[kuruman]

I'm imagining the free length of the spring is $l_o$, but I see no mention of it?

One shouldn't have to worry about $l_0$. If the spring is not compressed, it exerts zero force. If it is compressed, it will exert a force of magnitude $kx_{eq}$ directed to the left. That's enough to draw the FBD. A second equation is needed expressing the torque balance condition. This will provide a system of two equations and two unknowns, $x_{eq}$ and $f_s$.

 

[erobz]

Perhaps I'm missing something obvious, but $x_{eq}$ is not going to be measured from the coordinate system in that case (Which is I believe fixed at the pulley)?

 

[kuruman]

Is that accurate? You just have to check to see if the displacement doesn't cause $F_s$ escape that range, otherwise the wheel slips?

If the disk is at equilibrium, it doesn't slip by definition.

Perhaps I'm missing something obvious, but xeq is not going to be measured from the coordinate system in that case (Which is I believe fixed at the pulley)?

The way I see it, $x_{eq}$ is the amount by which the spring is compressed from its relaxed length. As such, it is a displacement and therefore independent of origin choice.. The force it exerts has magnitude $kx_{eq}$ which is also independent of the choice of origin.

 

[erobz]

If the disk is at equilibrium it doesn't slip by definition.

But there (probably) is some displacement or acceleration ( $a_D$ ) of the mass that would cause the disk to slip, and negate the solution.

For the purposes of the problem I understand that it doesn't need to be checked because the prof says its in equilibrium "by definition" as you say. But in general, if you were just solving this problem, that "no slip criterion" should be verified. That is what I was asking.

 

[kuruman]

But in general, if you were just solving this problem, that "no slip criterion" should be verified. That is what I was asking.

When a system is in static equilibrium, it is at rest and remains at rest. Therefore it doesn't slip. What is there to verify? That it doesn't slip? We know that already by looking at it. We can calculate the force of static friction and, if we know $\mu_s$, we can verify whether the force is below or at its upper limit, but to what purpose?

 

[erobz]

When a system is in static equilibrium, it is at rest and remains at rest. Therefore it doesn't slip. What is there to verify? That it doesn't slip? We know that already by looking at it. We can calculate the force of static friction and, if we know $\mu_s$, we can verify whether the force is below or at its upper limit, but to what purpose?

Sorry, I thought I was more clear than I was. I was talking about part (b), where we are to displace the hanging mass from equilibrium and determine the characteristics of the resulting oscillation. Not the static solution.

I see that in part (b) it is stated that "it doesn't slip". But I am asking in general "If we were just modeling the system", we should verify that we haven't displaced it so much so that the wheel slips. There will be a limit to the displacement where the system of equations used to generate the motion no longer hold.

 

[erobz]

Homework Statement:: A rope is wound around the axis orthogonal to a disk of mass $M$ and radius $R$ and there is a mass $m$ hanging from this rope as shown in the figure. The disk rests on a horizontal plane and connected to its center there is a spring which has its other second end fixed on a pulley.
(a) Determine the deformation of the spring at equilibrium.
(b) At a certain instant the mass $m$ is slightly lowered from the equilibrium position and then released. Determine the period of the oscillations that are caused by knowing that the disk rolls without slipping.Assume $M, R, r, k, m, m_{pulley}, r_{pulley} = r$ all known.
Relevant Equations:: $\vec{F}=m\vec{a},\ \tau=I\alpha,\ a=R\alpha$

(a) By setting up a coordinate system with the x-axis pointing to the right and the y-axis pointing downward we have $\begin{cases}-kx_{eq}+T_1+F_{s}=0\\ -RF_{s}+rT_1=0\\ r_p (T_2-T_1)=0\\ -T_2+mg=0\end{cases}\Rightarrow x_{eq}=\frac{mg}{k}\left(1+\frac{r}{R}\right)$ which coincides with the solution given.

(b) If we denote by $a_{D}$ is the the acceleration of the center of mass of the disk, I think the the goal should then be to obtain an equation of the form $a_{D}=\frac{\text{const}\times K}{\text{some combination of the masses}}(x-x_{eq})$ so that $T=\frac{2\pi}{\omega}$ where $\omega=\sqrt{\frac{\text{const}\times K}{\text{some combination of the masses}}}$. I have thus set up the following system:

$\begin{cases}-kx_{eq}+T_1+F_{s}=Ma_D\\ -RF_{s}+rT_1=I_D \frac{a_D}{R}\\ r_p (T_2-T_1)=I_{pulley}\alpha_{pulley}=I_{pulley}\frac{a_D}{r}\\ -T_2+mg=ma_D\end{cases}$

The problem is that the solution given says that in equation (3) the RHS should be $\frac{I_{pulley}}{r_{pulley}}\frac{a_D}{R}(r+R)$ and in equation (4) the RHS should be $m\frac{a_D}{R}(r+R)$ but I haven't been able to figure out why that is.
If the rope is ideal, shouldn't mass $m$ have the same acceleration as the disk?

Thanks

$$ -kx_{eq}+T_1+F_{s}=Ma_D $$

Is that the correct equation for part (b). Don't you have to include the displacement from equilibrium in the restoring force from the spring?

 

[kuruman]

Sorry, I did not realize you were talking about part (b). I saw the $-kx_{eq}$ which belongs in part (a) and I thought that's what you were talking about.

For part (b) $-kx_{eq}+T_1+F_{s}=Ma_D$ is not the correct equation. The force from the spring should be $-kx$ not $-kx_{eq}$. In part (b) we are told to assume that the system executes simple harmonic motion about $x_{eq}$ with the disk rolling without slipping.

 

[erobz]

Sorry, I did not realize you were talking about part (b). I saw the $-kx_{eq}$ which belongs in part (a) and I thought that's what you were talking about.

For part (b) $-kx_{eq}+T_1+F_{s}=Ma_D$ is not the correct equation. The force from the spring should be $-kx$ not $-kx_{eq}$. In part (b) we are told to assume that the system executes simple harmonic motion about $x_{eq}$ with the disk rolling without slipping.

I'm still not exactly following though. Where is $x$ to be taken from? Is it the displacement from the equilibrium position or the CoM coordinate of the disk measured from the fixed pulley? Maybe I'm missing it completely. I'm sorry for being hopelessly confused about this.

 

[erobz]

Is what I'm referring to as $D$ in the diagram, what you are referring to as $x$?

 

[kuruman]

Case I: Measure horizontal positions from the center of the pulley.
With this choice position increases to the left so to the left is positive and to the right is negative.
Part (a)
System is at rest
Let $L$ be the relaxed length of the pulley. Then the force balance equation is
$k(L-x_{eq})-T_1+f_s=0$
Using the second torque balance equation, one can find $x_{eq}$ which is constant.
Part (b)
Spring is compressed additional distance D from equilibrium
$k(L-x_{eq}-D)-T_1+f_s=M_Da$

Case II: Measure horizontal positions from the left end of the relaxed spring.
With respect to case I, the origin has moved to the left by $L$. With this choice, position increases to the right so to the right is positive and to the left is negative.
Part (a)
System is at rest
The force balance equation is
$-k(x_{eq})+T_1-f_s=0$
Using the second torque balance equation, one can find $x_{eq}$ which is constant.
Part (b)
Spring is compressed additional distance D from equilibrium
$-k(x_{eq}+D)+T_1-f_s=M_Da$

In either case the quantity that you have defined as $x$ in the figure, namely the instantaneous length of the spring does not appear. If you really insist on using $x$ as a position variable, then you must write the spring force as $F=k(L-x)$ so that it changes sign when the spring changes from being compressed (x < L) to being extended (x > L). This is equivalent to defining $x=x_{eq}+D$ in case I.

The bottom line is that the system will oscillate with the same frequency regardless of where you and I pick the origin. I favor Case II because then I don't have to worry about how long the spring is.

 

[erobz]

Case I: Measure horizontal positions from the center of the pulley.
With this choice position increases to the left so to the left is positive and to the right is negative.
Part (a)
System is at rest
Let $L$ be the relaxed length of the pulley. Then the force balance equation is
$k(L-x_{eq})-T_1+f_s=0$
Using the second torque balance equation, one can find $x_{eq}$ which is constant.
Part (b)
Spring is compressed additional distance D from equilibrium
$k(L-x_{eq}-D)-T_1+f_s=M_Da$

Case II: Measure horizontal positions from the left end of the relaxed spring.
With respect to case I, the origin has moved to the left by $L$. With this choice, position increases to the right so to the right is positive and to the left is negative.
Part (a)
System is at rest
The force balance equation is
$-k(x_{eq})+T_1-f_s=0$
Using the second torque balance equation, one can find $x_{eq}$ which is constant.
Part (b)
Spring is compressed additional distance D from equilibrium
$-k(x_{eq}+D)+T_1-f_s=M_Da$

In either case the quantity that you have defined as $x$ in the figure, namely the instantaneous length of the spring does not appear. If you really insist on using $x$ as a position variable, then you must write the spring force as $F=k(L-x)$ so that it changes sign when the spring changes from being compressed (x < L) to being extended (x > L). This is equivalent to defining $x=x_{eq}+D$ in case I.

The bottom line is that the system will oscillate with the same frequency regardless of where you and I pick the origin. I favor Case II because then I don't have to worry about how long the spring is.

Thank you, but they say were "one door closes another opens". Unfortunately, the same thing applies to questions I have!

For case 1:

You have $f_s$ pointing to the left. I don't understand that. Doesn't the force of static friction have to counter the torque from the tension force $T_1$ so that the wheel doesn't slip?

This FBD seems like there has to be a net torque on the wheel?

 

[kuruman]

The force of static friction is what is necessary to provide the observed acceleration. When you have to balance both forces and torques, the direction of the static friction is not always obvious. The way to proceed is to assume a direction, solve the equations for the unknowns, and see what you get for $f_s$. SInce in the original equations $f_s$ stands for the magnitude of the friction, if this number comes out negative it means that you have chosen the direction incorrectly; if it comes out positive, it means that you you have chosen the direction correctly.

This FBD seems like there has to be a net torque on the wheel?

It does, doesn't it? Have you solved this problem, part (a)? If not, do it assuming that $f_s$ is as shown in the diagram and see what you get.

Thank you, but they say were "one door closes another opens". Unfortunately, the same thing applies to questions I have!

That is not at all unfortunate. I commend you for following your curiosity and opening closed doors instead of just walking past them.

 

[haruspex]

Doesn't the force of static friction have to counter the torque from the tension force T1 so that the wheel doesn't slip?

The torques would have to balance for the system to be static, but it is not a requirement for rolling contact. How do the wheels of a car get to rotate faster as the car accelerates without skidding?

 

[erobz]

It does, doesn't it? Have you solved this problem, part (a)? If not, do it assuming that fs is as shown in the diagram and see what you get.

$$ k \left( L - x_{eq} \right) - T_1 + f_s = 0 \tag{1} $$

For the sum of the torques: $\circlearrowright^+$

$$ Rf_s + rT_1 = 0 \implies f_s = -\frac{r}{R}T_1 \tag{2}$$

Since the system is static $T_1 = mg$

$$ f_s = -\frac{r}{R}mg \tag{3} $$

Subbing back into $(3) \to (1)$

$$ k \left( L - x_{eq} \right) - mg -\frac{r}{R}mg = 0 \implies x_{eq} = L - \frac{mg}{k} \left( 1 + \frac{r}{R}\right)$$

Which is consistent with expectations.

$(3)$ is saying that the $f_s$ is opposite the assumed convention?

 

[erobz]

The torques would have to balance for the system to be static, but it is not a requirement for rolling contact. How do the wheels of a car get to rotate faster as the car accelerates without skidding?

I tricked myself into believing I understood this somehow, and I clearly don't. So I'm just going to answer your question. They must have a net torque on them to accelerate. I can't figure out what that exactly implies for the force of static friction and the fact that they don't slip relative to the road.

I'm thinking if we took torques about the point of application of the static friction force $P$ the force of static friction $f_s$ will have no moment arm, and thus apply no torque. However, letting/forcing the wheel to slip...still no moment arm for the kinetic friction $f_k$ either in that perspective...so that is probably not the proper treatment.

 

[kuruman]

I tricked myself into believing I understood this somehow, and I clearly don't.

A common misconception is that "friction always opposes motion". That is true when surfaces in contact slide past each other but not necessarily true when they are at rest with respect to each other. Take the simple act of starting to walk from rest. Clearly, you cannot do that on a perfectly frictionless surface. What external force provides the forward horizontal acceleration that changes your velocity from zero to non-zero? Hint: It cannot be gravity because it acts along the vertical. What is left?

The bottom line, consistent with Newton's laws, is that the force of static friction is whatever is needed to provide the observed acceleration.

 

[erobz]

A common misconception is that "friction always opposes motion". That is true when surfaces in contact slide past each other but not necessarily true when they are at rest with respect to each other. Take the simple act of starting to walk from rest. Clearly, you cannot do that on a perfectly frictionless surface. What external force provides the forward horizontal acceleration that changes your velocity from zero to non-zero? Hint: It cannot be gravity because it acts along the vertical. What is left?

The bottom line, consistent with Newton's laws, is that the force of static friction is whatever is needed to provide the observed acceleration.

I'm not conflicted about letting the laws decide the direction, I'm conflicted about the magnitude of $f_s$.

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? Furthermore, it would seem that there must be a force that is opposing static friction? What is countering that torque? where is the countering force?

 

[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?