I'm calculating more energy out than I put in

[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.