I I'm calculating more energy out than I put in

[jbriggs444]

Perhaps, reanalyzing the situation where the sphere slips so the frictional force is divided into two parts one doing work as it slips resulting in the applied force providing the only torque, and one that does no work while working with the applied force to rotate the sphere. then take the limit as the friction goes to zero. Maybe?

I do not think that this is a fruitful path to follow.

If we decide that the surface is frictionless then there is nothing to divide into parts.
If we decide that the surface has friction then we need to figure out whether the coefficient of friction is sufficient for rolling without friction (in which case we already have a good approach) or insufficient (in which case we have a known and constant force that we can calculate with).

I admit that the scenario of a bar being able to exert a torque without imparting some motion perpendicular to the direction of the force may not be physically realizable.

That part does not concern me at all. It is an engineering detail, unimportant to the mathematics of the situation. The idea of a yo-yo or a reel of 9 track mag tape is a good mental image.

I'm thinking that the situation with no friction may be as shown below in which case θ=X/R sort of a Yo-Yo situation. Think of the sphere sitting on a frictionless surface with its axis of rotation perpendicular to the surface.

View attachment 348532

Sure. A spherical child's toy top being pulled from a string wound around its circumference and viewed from above.

But now there is nothing on the diagram to indicate how far the center of mass translates.

 

[A.T.]

Without friction at the bottom the mass distribution determines the resulting kinematics. The smaller the moment of inertia for a given mass, the more it will spin instead of translating.

Can you demonstrate that? The smaller the mass the greater the translational acceleration.

That's why I wrote "smaller moment of inertia for a given mass". I'm talking about changing the mass distribution, not the mass itself.

Mass concentrated closer to the center -> Greater angular acceleration for the given tangential force -> Push bar moves faster and needs less time to travel the given distance -> Less linear impulse is imparted

 

[gleem]

Consider a limiting case. We have a cylindrical pencil laying on its side on a frictionless surface and viewed end on. Give it a massless but rigid extension one light year in length (call it 1016 meters) extending vertically upward. What happens when you tap leftward with, let us say 0.01 newton-second of impulse on the top of the extension? If it helps, imagine the pencil as having a diameter of 1 cm and a mass of about 10 grams.

f it helps, imagine the pencil as having a diameter of 1 cm and a mass of about 10 grams.

What translation rate for the center of mass do you expect? Can you calculate it?
What rotation rate do you expect? Can you calculate it?

Conservation of momentum is the key here.
answers:
V_cm= 1 m/s
ω = 6x10^-12 rad/sec

A spherical child's toy top being pulled from a string wound around its circumference and viewed from above.

Not a good example as it is hard to deploy in a manner similar to the one under consideration.

I have experimented with a pool ball with a string taped to it, But the tape does not release cleanly giving the ball an extra impulse to the center of mass. but the ball rotates and moves in the direction of the pull.
It always seem to have a translation although the ball spins fast compared to the translation.

 

[jbriggs444]

Conservation of momentum is the key here.
answers:
V_cm= 1 m/s
ω = 6x10^-12 rad/sec

Your result for $V_\text{cm}$ is correct. 0.01 newton second applied to 0.01 kg results in 1 m/s.

Your calculation for $\omega$ is horridly, incredibly, absurdly wrong. Try again and show your work.

An impulse of 0.01 newton-seconds on a $10^{16}$ meter moment arm amounts to a rotational impulse of $10^{14}$ kg m^2 / sec by my reckoning. That much angular momentum applied to a pencil with a moment of inertia $I = \frac{1}{2}mr^2 = \frac{1}{2} (0.01)(0.01)^2 = 0.5 \times 10^{-6}$ kg m^2 should give something like $2 \times 10^{20}$ rad/sec.

Multiply by the one light year radius and by $3 \times 10^7$ seconds per year and we have a tangential velocity around $6 \times 10^{27}$ times the speed of light at the end of the one light year extension.

Which is a tad more than 1 m/s.

We can calculate the linear kinetic energy of the pencil: $KE_t = \frac{1}{2}mv^2 = \frac{1}{2}(0.01)(1)^2 = 0.005$ joules.

We can calculate the rotational kinetic energy of the pencil: $KE_r = \frac{1}{2}Iw^2 = \frac{1}{2}(0.5 \times 10^{-6}) \times (2 \times 10^{20})^2 = 10^{34}$ joules.

Edit: Corrected $\frac{1}{2}Ir^2$ to $\frac{1}{2}I \omega^2$. Thank you, @gleem for that catch.

 

[gleem]

We can calculate the rotational kinetic energy of the pencil: KEr=12Ir2=12(0.5×10−6)×(2×1020)2=1034 joules.

In your KE_r you meant to us ω.

An impulse of 0.01 newton-seconds on a 1016 meter moment arm amounts to a rotational impulse of 1014 kg m^2 / sec by my reckoning. That much angular momentum applied to a pencil with a moment of inertia 𝐼=12𝑚𝑟2=12(0.01)(0.01)2=0.5×10−6 kg m^2 should give something like 2×1020 rad/sec.

What's going on in the red text?

 

[jbriggs444]

In your KE_r you meant to us ω.

Yes. I've corrected that now. Thank you.

What's going on in the red text?

The red text in question is the $(0.01)(0.01)^2$ in the following:

$I = \frac{1}{2}mr^2 = \frac{1}{2} (0.01)(0.01)^2 = 0.5 \times 10^{-6}$ kg m^2

The first $(0.01)$ is the mass $m$. 10 grams = 0.01 kg.

The second $(0.01)$ is intended to be the radius of the pencil. Looking back, $0.01$ meters is actually the diameter and this should have been $(0.005)^2$. So the calculated moment of inertia should have been lower by a factor of four and the calculated angular rotation rate should have been higher by a factor of four.

Allow me to digress...

Mathematics is often about being able to express complex thoughts compactly. We prefer notations that are brief. More compact is better than less compact.

One way we make things more compact is by eliminating the multiplication symbol, $\times$. If we want to express the product of two variables, we simply put them next to each other. So $a \times b$ becomes $ab$. I expect that you are quite familiar with this practice.

As long as all the quantities are represented by single character variable names ("identifiers" in computer-speak), there is no ambiguity. We know that $ab$ means $a$ times $b$ and is not a reference to some new variable named $ab$.

When you have two numeric literals ("numeral" is short for numeric literal) and want to express multiplication, life is more difficult. If, for instance, you wanted to express $0.01 \times 0.01^2$ using juxtaposition and wrote "$0.010.01^2$", your reader would have a hard time making sense of that.

So you parenthesize and get $(0.01)(0.01)^2$ or the correct $(0.01)(0.005)^2$ in this case.

Does that make sense?

I come from a background in programming and the theory of computing. So I often think of mathematical language in terms of parsing and grammars.

 

[gleem]

We are talking about a long thing rod, correct?
The moment of inertia of a thin rod is mass× length²/12.
Radius can be neglected if radius <<Length.

 

[A.T.]

It always seem to have a translation although the ball spins fast compared to the translation.

You could cut a circle from cardboard of foamboard, with two attachable weights on opposite positions of the center, which can be placed near the center (small moment of inertia), or far from the center (large moment of inertia). Then add a toothpick that sticks out a bit over the edge radially, and can be used to give it a tangential push (or two on opposite sides to preserve the center of mass).

You can then compare what happens for small and large moment of inertia, when you give it a tangential push.

But note that the conditions specified in the OP (constant force over a fixed distance) are not easy to control when pushing something with your hand. In reality the force depends on how much resistance the hand meets, and a human cannot change the hand movement quick enough to ensure a constant force during a short push.

 

[gleem]

@A.T. I've given up on any experiments that show anything more than gross effects. The pool ball is good because it has minimal resistance on a suitable hard surface but wrapping a string so that it remains on the diameter is difficult.

 

[jbriggs444]

We are talking about a long thing rod, correct?
The moment of inertia of a thin rod is mass× length²/12.
Radius can be neglected if radius <<Length.

The moment of inertia of a massless rod is zero. The pencil is the only thing here that has a non-zero moment of inertia. Its radius cannot be neglected.

 

[A.T.]

@A.T. I've given up on any experiments that show anything more than gross effects. The pool ball is good because it has minimal resistance on a suitable hard surface but wrapping a string so that it remains on the diameter is difficult.

The pool ball cannot show you the effect of different mass distributions on the kinematics. For this you could also use a plastic pipe with some heavy weights inside, that can be fixed at different locations.

 

[gleem]

Consider a limiting case. We have a cylindrical pencil laying on its side on a frictionless surface and viewed end on. Give it a massless but rigid extension one light year in length (call it 1016 meters) extending vertically upward. What happens when you tap leftward with, let us say 0.01 newton-second of impulse on the top of the extension? If it helps, imagine the pencil as having a diameter of 1 cm and a mass of about 10 grams

consider the following diagram

The impulse applied to the end of the massless extension transfers its motion to the end of the rod, but the massless extension contributes nothing to the change in momentum.

The only mass moving is the pencil, and it is moving much slower than the end of the extension.

BTW the moment of Inertia of a thick rod is
$$ I_{thick-rod} = \frac{ml^{2}}{12} + \frac{mr^{2}}{4}$$⋅
see http://hyperphysics.phy-astr.gsu.edu/hbase/icyl.html

⋅

 

[jbriggs444]

consider the following diagram

[snip incorrect diagram]

The impulse applied to the end of the massless extension transfers its motion to the end of the rod, but the massless extension contributes nothing to the change in momentum.

The only mass moving is the pencil, and it is moving much slower than the end of the extension.

The pencil is laying on its side and is viewed end on. The extension is perpendicular to the pencil. It is attached to the long side of the pencil and extends vertically upward, away from the frictionless surface on which the horizontal pencil rests.

The massless extension moves rigidly with the pencil. You are correct that any impulse applied to the extension has the effect of applying that much momentum change to the pencil.

You are correct that the only mass moving is the pencil and that its center of mass is moving much slower than the end of the extension.

However, you have still not realized that the rotation rate of the pencil would be extreme. Nor have you showed any effort at performing the requisite calculation.

 

[A.T.]

but the massless extension contributes nothing to the change in momentum

It makes the torque for the given tangential force, and thus the angular acceleration very large, which makes the time of force application over the given distance, and thus the transfered linear momentum very small.

 

[gleem]

[snip incorrect diagram]

Who cares what the shape of the object near the axis of rotation is.

However, you have still not realized that the rotation rate of the pencil would be extreme. Nor have you showed any effort at performing the requisite calculation

Why do You think the rod is rotating at an extreme rate?

 

[jbriggs444]

Why do You think the rod is rotating at an extreme rate?

Because it was subject to an extreme torque. Or, to be technically correct, was subject to a large transfer of angular momentum.

You understand the idea of "impulse", right?

"Force" is the rate of transfer of momentum to the target object. But in many situations, we do not care about the rate. Instead, we just care about how much momentum was transferred. A large force over a brief interval transfers as much momentum as a smaller force over a longer interval.

In the scenario that I proposed, there is a linear "impulse" of 0.01 kg m/sec² applied. Perhaps this was 10 N applied over an interval of one millisecond. Perhaps it was 10000 N applied over an interval of one microsecond. Perhaps it was .01 N applied over an interval of one second.

How much angular momentum was transferred into the system by this linear impulse?

I claim that the answer is $\vec{p} \times \vec{R}$ where $p$ is the impulse (0.01 kg m/sec² and $R$ is the moment arm (1 light year or approximately 10¹⁶ meters). That comes to approximately 10¹⁴ kg m²/sec.

You might object to that calculation. Few physics textbooks bother to write this formula down for you. So let me justify it.

Suppose that the impulse was delivered as 1 N over a duration of 0.01 seconds...

Then the torque was 10¹⁶ kg m²/sec². Multiply that by the duration of 0.01 seconds and we have 10¹⁴ kg m²/sec of angular momentum transferred.

Suppose instead that the impulse was delivered as 0.01 N over a duration of 1 second...

Then the torque was 10¹⁴ kg m²/sec². Multiply that by the duration of 1 second and we have 10¹⁴ kg m²/sec of angular momentum transferred.

What sort of rotation rate do you expect when you have this much angular momentum on a pencil?

 

[gleem]

Because it was subject to an extreme torque.

I suspected that you were going to say that.

But impulse is about change in momentum and velocity. Clearly, the magnified torque at the rod is of no consequence. The tangential velocity of the rod is the velocity at the end of the extension times the radius of the rod divided by the distance of the tip of the extension to the center of the rod The angle of rotation is the same for the extension and the rod and so the angular rotation rate of the rod and extension are the same.

 

[jbriggs444]

Clearly, the magnified torque at the rod is of no consequence.

Nope. Torque is the rate of transfer of angular momentum. It has consequences.

 

[A.T.]

Clearly, the magnified torque at the rod is of no consequence.

Seems you missed my post where I explained the consequences to you:

It makes the torque for the given tangential force very large, and thus the angular acceleration very large, which makes the time of force application over the given distance very small, and thus the transferred linear momentum very small.

You have to keep the boundary conditions in mind: the applied force and the distance over which it is applied are fixed. Changing the lever arm leads to the above, given those constraints.

The angle of rotation is the same

No. That is not the boundary condition to keep fixed when comparing the cases with and without extension. The distance over which the force is applied is to be fixed.

 

[jbriggs444]

No. That is not the boundary condition to keep fixed when comparing the cases with and without extension. The distance over which the force is applied is to be fixed.

I'd fixed the applied linear impulse instead. Which amounts to pretty much the same thing, give or take a square root and a factor of two.

 

[gleem]

For a given impulse, the torque is applied for a given time determining the distance over which the force is applied and therefore defining the angle of rotation at the instant of the impulse and the angular velocity.

The rod (replace the length view with an end view) cannot rotate faster than the extension.

That is not the boundary condition to keep fixed when comparing the cases with and without extension. The distance over which the force is applied is to be fixed.

What do you mean "comparing the cases with and without extension" I'm only talking about a rod with an extension and what happens when the extension receives an impulse.

 

[A.T.]

the massless extension contributes nothing to the change in momentum.

What do you mean "comparing the cases with and without extension"

To claim that the extension contributes nothing to the change in momentum, you have to compare the cases with and without extension, under the same boundary conditions as stated in the OP (fixed force and distance it is applied over).

 

[jbriggs444]

For a given impulse, the torque is applied for a given time determining the distance over which the force is applied and therefore defining the angle of rotation at the instant of the impulse and the angular velocity.

You have to decide what you are holding fixed and what you are allowing to vary.

If impulse is held fixed while moment arm is allowed to vary then final angular velocity is proportional to moment arm.

If rotational work ($FX$) is held fixed while moment arm is allowed to vary then final angular velocity will be independent of the length of the moment arm.

Which would you like to discuss further?

 

[gleem]

If rotational work (FX) is held fixed while moment arm is allowed to vary then final angular velocity will be independent of the length of the moment arm.

This one.

 

[jbriggs444]

This one.

OK, great. So we are holding the applied rotational work ($FX$) fixed. We are, for the moment, holding the central mass and shape fixed as well. The precise mass distribution of the central object is not important. It may be $\frac{1}{3}mr^2$ or $\frac{1}{12}ml^2$. Let us simply call it $I$. Its mass is $m$.

[Note that because we are holding the applied rotational work fixed, we will not be holding the applied linear impulse fixed]

There are many things that we could calculate. We could calculate the resulting rotation rate, $\omega$, the tangential velocity at the end of the extension, $V$, the tangential velocity at the edge of the central object, $v$, the linear impulse applied ($\Delta p$), the rotational impulse applied ($L \Delta p$), the duration of the impulse ($t$), the velocity of the center of mass ($v_\text{com}$), the linear acceleration of the center of mass ($a$), the angular acceleration of the assembly $\alpha$ or the displacement of the center of mass during the interaction.

If I recall correctly, your claim is that the displacement of the center of mass is equal to the distance rotated by the tip of the extension arm.

We can begin by computing the final rotation rate $\omega$. We can do that either with kinematics (hard) or with an energy argument (easy). Both will deliver the same result.

The rotational work supplied is $FX$. The resulting rotational kinetic energy is $\frac{1}{2}I \omega^2$. Equating the two and solving for $\omega$, we get $\omega = \sqrt{\frac{2FX}{I}}$.

The angular acceleration rate is easy. $\alpha = \frac{\text{torque}}{\text{moment of inertia}} = \frac{FL}{I}$

The duration for the interaction will then be $t = \frac{\omega}{\alpha}$ = $\frac{\sqrt{\frac{2FX}{I}}}{\frac{FL}{I}} = \sqrt{\frac{2XI}{FL}}$

The linear acceleration rate is easy: $a = \frac{F}{m}$

The linear displacement is then: $x_\text{com} = \frac{1}{2}at^2 = \frac{F}{2m}\frac{2XI}{FL} = \frac{XI}{mL}$

 

[A.T.]

We can begin by computing the final rotation rate $\omega$. We can do that either with kinematics (hard) or with an energy argument (easy). Both will deliver the same result.

The rotational work supplied is $FX$. The resulting rotational kinetic energy is $\frac{1}{2}I \omega^2$. Equating the two ...

Not sure about that. $FX$ is the total energy input, which goes into linear KE (motion of CoM) and rotational KE (rotation around CoM).

I think you can only equate $FX$ and $\frac{1}{2}I \omega^2$, if $I$ is the moment of inertia around the instantaneous center of rotation (not around the CoM). Then you basically express the entire motion as rotation, which accounts for all KE.

 

[jbriggs444]

Not sure about that. $FX$ is the total energy input, which goes into linear KE (motion of CoM) and rotational KE (rotation around CoM).

I agree. $FX$ only quantifies the rotational portion, not the translational portion. I tried to use wording to reflect that.

I think you can only equate $FX$ and $\frac{1}{2}I \omega^2$, if $I$ is the moment of inertia around the instantaneous center of rotation (not around the CoM). Then you basically express the entire motion as rotation, which accounts for all KE.

You can partition the total energy into a rotational and a linear component differently by picking and choosing different axes of rotation. In each case, the total energy is given by $\frac{1}{2}I \omega^2 + \frac{1}{2}mv_\text{axis}^2$

Yes, one choice is to use the instantaneous center of rotation. In this case, $v_\text{axis}$ is zero by definition so that the total system energy is purely rotational.

For the purposes of my analysis, I was choosing an axis of rotation at the center of mass. This seemed to match the choice that @gleem was implicitly making. I did not want to bring in an unnecessary point of contention.

To be clear, I considered $X$ to be defined as the length of the circular arc traced out by the end of the tip as described in the center of mass frame, not the length of a cycloidal arc traced out in some other frame.

 

[A.T.]

I agree. $FX$ only quantifies the rotational portion, not the translational portion.

That is the opposite of what I wrote. $FX$ is total energy input and equals the final sum of rotational and translational KE.

 

[jbriggs444]

That is the opposite of what I wrote. $FX$ is total energy input and equals the final sum of rotational and translational KE.

I consider $X$ to be the length of the circular arc traced out in the COM frame, not the length of a cycloidal arc traced out in some other frame.

I apologize for not understanding your intent.

 

[A.T.]

I consider $X$ to be the length of the circular arc traced out in the COM frame, not the length of a cycloidal arc traced out in some other frame.

Oh, I see. I assumed $X$ is the translation of the bar applying $F$, so total energy input is $FX$.

 

[jbriggs444]

Oh, I see. I assumed $X$ is the translation of the bar applying $F$, so total energy input is $FX$.

A reasonable interpretation.

 

[gleem]

As I began reading your post in response to my request I found a few things that I do not agree with so I stopped there to see if my observation will affect your following explanation.
First:

the rotational impulse applied (LΔp)

Typo or not?

second:

If I recall correctly, your claim is that the displacement of the center of mass is equal to the distance rotated by the tip of the extension arm.

No, I don't think so. Our discussion of late has only been about rotation.

 

[jbriggs444]

the rotational impulse applied (L \Delta p)

Typo or not?

The rotational impulse applied will indeed be given by $L \Delta p$ where $L$ is the length of the moment arm and $\Delta p$ is the impulse applied at the end of that arm.

No, I don't think so. Our discussion of late has only been about rotation.

Then I do not know what your claim is.

 

[gleem]

The angular acceleration rate is easy. α=torquemoment of inertia=FL/I

You cannot write that equation assuming I is the moment of inertia of the rod. $$I = I_{rod} + I_{extention}$$. Assume $I_{extention}=kmL^{2}$ where m is a small mass . $$\alpha = \frac {FL} {I_{extentioin} +I_{rod}}= \frac {FL} { k\cdot m\cdot L^{2} +I_{rod}} =\frac {(M+m)a} { k\cdot m\cdot L +I_{rod}/L} $$

But we know that the rotation of the rod and the extension must be the same.

Whatever acceleration is at the end of the extension is reduced at the rod so the rod does not spin faster than the end of the extension. As long as the extension has mass meaning as long as it is physically connected to the rod there is no problem. I suggest that when the extension mass is zero it cannot transfer momentum to the rod. The product m⋅L must be greater than zero.

 

[jbriggs444]

You cannot write that equation assuming I is the moment of inertia of the rod.

First of all, I can and I did. The extension was specified as massless. If you prefer that it have mass, that is fine. Make it negligibly massive.

$$I = I_{rod} + I_{extention}$$

Right. With $I_\text{extension} \approx 0$.

Assume $I_{extention}=kmL^{2}$ where m is a small mass . $$\alpha = \frac {FL} {I_{extentioin} +I_{rod}}= \frac {FL} { k\cdot m\cdot L^{2} +I_{rod}} =\frac {(M+m)a} { k\cdot m\cdot L +I_{rod}/L} $$

Note what happens when $m$ is negligible. You then have $L$ in the numerator.

But we know that the rotation of the rod and the extension must be the same.
View attachment 348825
Whatever acceleration is at the end of the extension is reduced at the rod so the rod does not spin faster than the end of the extension.

Are you talking about tangential acceleration or angular acceleration?

The angular acceleration is not reduced by the presence of the extension. On the contrary, it is increased due to the increased torque. Provided, of course, that the moment of inertia of the extension remains negligible.

As long as the extension has mass meaning as long as it is physically connected to the rod there is no problem.

"Having mass" and "being connected" are not the same thing. We can imagine a rigid framework with negligible mass.

I suggest that when the extension mass is zero it cannot transfer momentum to the rod. The product m⋅L must be greater than zero.

I do not want to be drawn into a discussion about the physical reality of massless extensions. So let us simply make the extension negligibly massive instead.

Do you agree that an increase to $L$ while holding $F$ constant results in an increase in angular acceleration, so long as the increase in moment of inertia is not commensurate.

 

[A.T.]

I suggest that when the extension mass is zero it cannot transfer momentum to the rod.

We use massless ropes that transfer momentum in idealized calculations all the time.

The whole point of that massless extension is to have an object with a very small $I$ for a given mass and size, to make the dependence on mass-distribution obvious. But since this confuses you even more, I suggest you forget about that extreme case and compare a disc of uniform density to a thin hoop, both of the same mass and radius. A bar applies a given constant force $F$ on the top (with prefect traction) while it moves the given distance $X$ ($F$ and $X$ are the same for both objects). The ground is frictionless.

 

[jbriggs444]

We use massless ropes that transfer momentum in idealized calculations all the time.

The whole point of that massless extension is to have an object with a very small $I$ for a given mass and size, to make the dependence on mass-distribution obvious. But since this confuses you even more, I suggest you forget about that extreme case and compare a disc of uniform density to a thin hoop, both of the same mass and radius. A bar applies a given constant force $F$ on the top (with prefect traction) while it moves the given distance $X$ ($F$ and $X$ are the same for both objects). The ground is frictionless.

Let me try with an arbitrary shape using $k$ as the mass distribution parameter.

$m$ = mass of disc
$R$ = radius of disc
$F$ = force applied to top rim of disc
$X_\text{f}$ = cumulative distance moved by the push bar that rides over the top of the disc at the end of the interaction. This will combine the effects of the movement of the shape and the rotation of the shape.
$I = kmR^2$ = moment of inertia of the shape
$k$ = fraction reflecting mass distribution. $k = \frac{1}{2}$ for a disc; $k = 1$ for a hoop; $k=\frac{2}{5}$ for a sphere.

The goal is to compute the final angular velocity ($\omega$), the final linear velocity ($v_\text{COM}$) and the final tangential velocity of rotation ($v_\text{rot}$) as functions of $m$, $R$, $F$, $X$ and $k$

We can begin by computing translational acceleration $a_\text{com}$ and angular acceleration $\alpha$.

The idea is that if we know the accelerations then we can calculate how the cumulative distance moved ($X(t)$) evolves over time during the interaction. We then hope to be able to solve for $t$ where $X(t) = X_\text{f}$.

Linear acceleration:$$a_\text{com} = \frac{F}{m}$$Angular acceleration:$$\alpha = \frac{FR}{kmR^2} = \frac{F}{kmR}$$Tangential acceleration (of rim relative to the center of mass):$$a_\text{rot} = \frac{FR}{kmR} = \frac{F}{km}$$

It is perhaps worth noting that $R$ has dropped out of our calculations.

The acceleration of the push bar that rides over the top of the shape will be given by the sum of the acceleration of the center of mass and the acceleration of the top rim relative to the center of mass:$$a_\text{tot} = a_\text{com} + a_\text{rot} = \frac{F}{m} + \frac{F}{km} = \frac{F}{m}\frac{k+1}{k}$$With this formula in hand, we can compute the cumulative distance moved by the push bar:$$X(t) = \frac{1}{2}a_\text{tot}t^2 = \frac{1}{2}\frac{F}{m}\frac{k+1}{k}t^2$$We solve for $t$ where$$X(t) = \frac{1}{2}\frac{F}{m}\frac{k+1}{k}t^2 = X_\text{f}$$$$t^2 = \frac{2mkX_\text{f}}{F(k+1)}$$$$t = \sqrt{\frac{2mkX_f}{F(k+1)}}$$
With $t$ in hand, we are now in a position to easily compute the final translation rate and the final angular rotation rate:$$v_\text{com} = at = \frac{F}{m}\sqrt{\frac{2mkX_\text{f}}{F(k+1)}} = \sqrt{\frac{2FkX_\text{f}}{m(k+1)}}$$
$$\omega = \alpha t = \frac{F}{kmR} \sqrt{\frac{2mkX_\text{f}}{F(k+1)}} =\frac{\sqrt{\frac{2FX_\text{f}}{k(k+1)m}}}{R}$$
$$v_\text{rot} = R \omega = \sqrt{\frac{2FX_\text{f}}{k(k+1)m}}$$

Edit: repaired algebra where I'd dropped $X_\text{f}$ halfway through.

 

[gleem]

Do you agree that an increase to L while holding F constant results in an increase in angular acceleration, so long as the increase in moment of inertia is not commensurate.

Yes of course. Increasing the torque on any rotating object increases its angular acceleration and angular velocity and the angle of rotation for a fixed time.

After considerable consideration of the meaning of the equations I now understand my error.

I wish to apologize for my stubbornness and thank you for hanging in there while I unraveled my issue. If I had just followed the math things would have been OK.

 

[A.T.]

$$v_\text{com} = \sqrt{\frac{2FkX_\text{f}}{m(k+1)}}$$
$$\omega = \frac{\sqrt{\frac{2FX_\text{f}}{k(k+1)m}}}{R}$$

If your write them like this:
$$v_\text{com} = \sqrt{\frac{2FkX_\text{f}}{m(k+1)}} = \sqrt{\frac{2FX_\text{f}}{m+ \frac{m}{k}}}$$
$$\omega = \frac{\sqrt{\frac{2FX_\text{f}}{k(k+1)m}}}{R} = \sqrt{\frac{2FX_\text{f}}{k(k+1)mR^2}} $$
then it's obvious that making $k$ larger makes $v_\text{com}$ larger and $\omega$ smaller.