Rotating spool on table with friction

[erfz]

Homework Statement

I am referring to this thread and question: https://www.physicsforums.com/threads/rotation-of-a-spool-about-rough-ground.295666/
Here is the problem, restated:

A cylindrically symmetric spool of mass $M$ and radius $R$ sits at rest on a horizontal table with friction. With your hand on a light string wrapped around the axle of radius $r$, you pull on the spool with a constant horizontal force of magnitude $T$ to the right. As a result, the spool rolls without slipping a distance $L$ along the table with no rolling friction.

Find the friction force $f$ acting on the spool.

Homework Equations

$\tau_{net} = I\alpha$
$\tau = Frsin(\theta)$
$F_{net} = Ma_{cm}$
$\alpha = a_{cm} / R$, when not slipping

The Attempt at a Solution

I am fine with the general method of this problem.
I would just like to make sure I understand what Doc Al means by "the rotational and translational accelerations must have the same sign" when applying the rotation without slipping condition.
This is what I make of his statement:

Consider a cylinder rolling without slipping on a table to the right. This also means, by intuition, that it must have an angular acceleration clockwise. So, if we take the convention that right is positive, then we should also take any clockwise torque contributions to be positive in $\tau_{net} = I\alpha$.
We can take the opposite conventions if desired (left is positive, counterclockwise is positive).

Is my interpretation correct?

Also, this problem (at least in my notes) does not come with a diagram. Doesn't the derivation from the link assume the T is applied at the bottom of the axle, as opposed to the top? If it were applied to the top, I get that $Ma_{cm} = T + f$ and $\tau_{net} = I\alpha = Tr - fR$. Applying the no slip condition and rearranging, the friction force is $$f = T\frac{-I + MrR}{I + MR^2}.$$
Is that correct? Doesn't this mean that if you apply $T$ to the top of the axle, $MrR > I$? For the uniform cylinder stated, $I = 1/2 ~MR^2$ so $MrR > 1/2 ~MR^2$ and $r > 1/2 ~R$. Is this true?

 

[haruspex]

Homework Statement

I am referring to this thread and question: https://www.physicsforums.com/threads/rotation-of-a-spool-about-rough-ground.295666/
Here is the problem, restated:

Homework Equations

$\tau_{net} = I\alpha$
$\tau = Frsin(\theta)$
$F_{net} = Ma_{cm}$
$\alpha = a_{cm} / R$, when not slipping

The Attempt at a Solution

I am fine with the general method of this problem.
I would just like to make sure I understand what Doc Al means by "the rotational and translational accelerations must have the same sign" when applying the rotation without slipping condition.
This is what I make of his statement:

Consider a cylinder rolling without slipping on a table to the right. This also means, by intuition, that it must have an angular acceleration clockwise. So, if we take the convention that right is positive, then we should also take any clockwise torque contributions to be positive in $\tau_{net} = I\alpha$.
We can take the opposite conventions if desired (left is positive, counterclockwise is positive).

Is my interpretation correct?

Also, this problem (at least in my notes) does not come with a diagram. Doesn't the derivation from the link assume the T is applied at the bottom of the axle, as opposed to the top? If it were applied to the top, I get that $Ma_{cm} = T + f$ and $\tau_{net} = I\alpha = Tr - fR$. Applying the no slip condition and rearranging, the friction force is $$f = T\frac{-I + MrR}{I + MR^2}.$$
Is that correct? Doesn't this mean that if you apply $T$ to the top of the axle, $MrR > I$? For the uniform cylinder stated, $I = 1/2 ~MR^2$ so $MrR > 1/2 ~MR^2$ and $r > 1/2 ~R$. Is this true?

It is important that in the original question there are two different radii. In the simplified form you have presented here, were the horizontal part of the string at the bottom it would be flat to the table and the spool could not move without slipping.
In the original thread it is never made clear whether the string is above or below the axis. Either way, @Doc Al is right that the rotational and linear acceleration directions did not match, so the torque equation in post #1 did have a wrong sign. Your interpretation is correct, but I would word it a little differently. There is nothing to prevent taking linear acceleration as positive to the right but rotational as positive anticlockwise.
If we take, arbitrarily, linear as positive right and rotational as positive clockwise, but friction as positive left, then the linear F=ma equation is correct but the frictional torque has the wrong sign. We can flip any of those three conventions and adjust the equations accordingly without changing the result.
However, there is some confusion because Doc Al assumed the string should be on top, making only the sign on the frictional torque wrong, but from the "correct" answer quoted in post #1 it appears that it should be at the bottom, so both torques had the wrong sign.

 

[erfz]

It is important that in the original question there are two different radii. In the simplified form you have presented here, were the horizontal part of the string at the bottom it would be flat to the table and the spool could not move without slipping.
In the original thread it is never made clear whether the string is above or below the axis. Either way, @Doc Al is right that the rotational and linear acceleration directions must match, so the torque equation in post #1 did have a wrong sign. Your interpretation is correct.
However, there is some confusion because Doc Al assumed the string should be on top, making only the sign on the frictional torque wrong, but from the "correct" answer quoted in post #1 it appears that it should be at the bottom, so both torques had the wrong sign.

Thank you for the clarification!
So when the $T$ is applied to the top, it seems that -- as I've shown -- $f$ is positive only for $r > 1/2 ~R$. What happens when this is not the case?

 

[haruspex]

Thank you for the clarification!
So when the $T$ is applied to the top, it seems that -- as I've shown -- $f$ is positive only for $r > 1/2 ~R$. What happens when this is not the case?

Sorry, I kept thinking of edits to what I wrote, so you might not have seen my final version.

It is an interesting feature of "horizontal force applied to rolling body" questions that there is a critical height of application where the rolling does not require any friction. Above that height, friction acts in the opposite direction from what you might assume. This is because, in the absence of friction, the rotational acceleration would "outpace" the linear, so friction acts to impede the rotation and assist the linear acceleration.

 

[erfz]

Above that height, friction acts in the opposite direction from what you might assume.

Very interesting.
When you say that friction acts opposite to what I would assume above a critical height, is it when $r > 1/2 ~R$ in particular (applied to top of axle). And, in that case, friction will assist the linear acceleration s.t. $F_{net} = T + f$, as you say? This is what I actually expected though, except that I thought that this would hold for any $r$ for $T$ applied to the top of the axle.
Then, what happens when $r < 1/2 ~R$? Does friction impede the linear acceleration and aid the rotation?

 

[haruspex]

is it when r>1/2 R in particular (applied to top of axle)

For a uniform disc or cylinder, yes. Different for a hollow sphere, solid sphere, etc.

what happens when r<1/2 R? Does friction impede the linear acceleration and aid the rotation?

Yes. Note which way the bobbin moves when -R<r<-½R.
What if r<-R? (This could be arranged with, say, two disks on rails with a larger, massless bobbin between the two.)

 

[erfz]

Note which way the bobbin moves when -R<r<-½R.

So, what I gather is that there are three scenarios for $-R < r < R$:

1. $-R < r < 0$. We get $F_{net} = T - f$ and $\tau_{net} = fR - Tr$
2. $0 < r < 1/2 ~R$. We get $F_{net} = T - f$ and $\tau_{net} = fR + Tr$ (both $T$ and $f$ act to aid rotation)
3. $1/2 ~R < r < R$. We get $F_{net} = T + f$ and $\tau_{net} = Tr - fR$

For $r < -R$, I assume we get situation 1. I have no intuition of this, but I see no mathematical issue with this from $$f = T\frac{I + MrR}{I + MR^2}.$$ Here $r$ is the positive distance from the center of mass to a point of application below it. $R$ is also positive. All we require, as far as I can tell, is that $I + MrR > 0$ so, for a uniform cylinder, $r > -1/2 ~R$. This is always true.

 

[haruspex]

So, what I gather is that there are three scenarios for $-R < r < R$:

1. $-R < r < 0$. We get $F_{net} = T - f$ and $\tau_{net} = fR - Tr$

Ah.. this could get confusing. I used negative r to indicate the string is below the centre. That's ok but we need to be consistent, i.e. the r on the right is also negative, so with clockwise positive it is $\tau_{net} = fR + Tr$.
Likewise, we can simplify matters by taking f as always being positive to the right. We end up with one pair of equations for all cases:
$F_{net} = T + f$ and $\tau_{net} = -fR + Tr$ (because the friction acts at -R).

1. $r < \frac 12R$: f<0
2. $\frac 12R < r$: f>0

For $r < -R$, I assume we get situation 1.

Yes, but which way does it move?

 

[erfz]

Ah.. this could get confusing. I used negative r to indicate the string is below the centre. That's ok but we need to be consistent, i.e. the r on the right is also negative, so with clockwise positive it is $\tau_{net} = fR + Tr$.
Likewise, we can simplify matters by taking f as always being positive to the right. We end up with one pair of equations for all cases:
$F_{net} = T + f$ and $\tau_{net} = -fR + Tr$ (because the friction acts at -R).

1. $r < \frac 12R$: f<0
2. $\frac 12R < r$: f>0

That's my mistake. All my expression should have $|r|$ instead of just $r$, and indeed we can replace that absolute value with $\pm r$ where appropriate. And then we can combine expressions and the intervals on which they are valid to reconstruct your more elegant presentation. Fundamentally though, the expressions I've written are fine though right?

Yes, but which way does it move?

Hmm.. So $$f = T\frac{I + MrR}{I + MR^2}$$ still applies here. However, for a uniform solid sphere, we get that $$f = 1/3 ~T(MR^2 + 2MrR)/(MR^2) = T(1/3 + 2r ~/ ~3R) > T(1/3 + 2/3) = T$$ using positive $r$. This means that $F_{net} = T - f < 0$ so it's moving to the left but it is still rotating to the right (I think) since $$0 < \tau_{net} = fR - Tr < fR - TR = (f - T)R.$$

 

[haruspex]

All my expression should have |r| instead of just r

Ok, but my preference would be to leave it as r so that a single pair of equations covers all cases.

Fundamentally though, the expressions I've written are fine though right?

If you prefer to write |r| everywhere, yes.

Hmm.. So $$f = T\frac{I + MrR}{I + MR^2}$$ still applies here. However, for a uniform solid sphere,

The direction of movement does not depend on whether it is a sphere or disk.
The net force is given by $T\frac{R(R+r)}{R^2+k^2}$ where I=mk².
If |r|>R and the string is below the axis then r<-R, so the net force is to the left.

 

[erfz]

Ok, but my preference would be to leave it as r so that a single pair of equations covers all cases.

If you prefer to write |r| everywhere, yes.

The direction of movement does not depend on whether it is a sphere or disk.
The net force is given by $T\frac{R(R+r)}{R^2+k^2}$ where I=mk².
If |r|>R and the string is below the axis then r<-R, so the net force is to the left.

And it rotates clockwise?

 

[haruspex]

And it rotates clockwise?

No, it still rolls, so the rotation must be anticlockwise.

 

[erfz]

No, it still rolls, so the rotation must be anticlockwise.

Ah ok. So when it's rolling left I actually should write $\tau_{net} = Tr - fR$. I'm beginning to see how much more element the single pair of equations is compared to the case analysis I did.