I'm calculating more energy out than I put in

[jbriggs444]

I think I'll leave it the way it is.

OK. I will try to follow your analysis and see where it goes wrong.

The force being constant produces a constant translational acceleration a = F/M and a constant rotational acceleration α. = L / I where L (torque) =F⋅R and I is the moment of inertia. The work required to do this has two parts one rotating it and one moving the center of mass horizontally. The force is applied for a distance of its length x causing the sphere to rotate θ radians where X=R⋅θ.

You appear to be assuming that although there is rolling without slipping that the frictional force of the sphere on the platform imparts no momentum, provides no torque and does no work. That is, that the frictional force turns out to be zero.

I disagree with this, but will accept it for the purposes of argument.

Let us go ahead and apply the force $F$. The claim is that the resulting acceleration is $a = F/M$.

We can go ahead and apply the torque as well. The claim is that the resulting angular acceleration $\alpha = \frac{L}{I} = \frac{F \cdot R}{I}$. The moment of inertia of a sphere is $\frac{2}{5}MR^2$. So we end up with $\alpha = \frac{5F}{2MR}$. If we multiply by $R$ to convert this angular acceleration rate to a rate of horizontal acceleration, we get $a = \frac{5}{2}F/M$.

The angular acceleration is 2.5 times too fast. We would not actually maintain rolling without slipping under these conditions.

Non-zero friction with the platform has to provide enough retarding torque and advancing horizontal force so that the rotational acceleration and the horizontal acceleration match.

 

[A.T.]

The sphere moves without slipping horizontally on a flat surface as shown below.

Note that the OP doesn't specify this. And as @jbriggs444 points out, you cannot assume zero friction for accelerated rolling without slipping in general.

 

[gleem]

You appear to be assuming that although there is rolling without slipping that the frictional force of the sphere on the platform imparts no momentum, provides no torque and does no work. That is, that the frictional force turns out to be zero.

I disagree with this, but will accept it for the purposes of argument.

The OP did not specify a coefficient of friction so I think my assumption of not slipping is justified. If the frictional force is zero, there would be no rotation, so there must be friction. The frictional force is not moving so it does no work (neglecting rolling friction) and does not affect the final kinetic energy.

My post was to demonstrate that the work done was equal to the final kinetic energy. that is

$$KE_{tot}= \frac{1}{2}mv^2 +\frac{1}{2}I\omega ^2 = FX+ L\theta=maX+I\alpha \theta $$
$$v^{2}=2aX $$ and $$ \omega^{2} =2\alpha \theta$$
$$KE_{final}=maX+I\alpha \theta =mXv^{2}/2X +I\theta \omega ^{2}/2\theta $$




Now I did find this analysis surprisingly simple since when I considered using V² = 2Xa, I could almost see where it was going to take me.

 

[PeterDonis]

@gleem please use the LaTeX feature for equations. Your equations are not very readable.

 

[jbriggs444]

The OP did not specify a coefficient of friction so I think my assumption of not slipping is justified. If the frictional force is zero, there would be no rotation, so there must be friction. The frictional force is not moving so it does no work (neglecting rolling friction) and does not affect the final kinetic energy.

My post was to demonstrate that the work done was equal to the final kinetic energy. that is

½ mV² +½ Iω² = FX + Lθ = maX + Iαθ

What is $F$ here? The applied force? If so, that is incorrect.
What is $L$ here? The torque from the applied force? If so, that is incorrect.

 

[gleem]

What is F here? The applied force? If so, that is incorrect.
What is L here? The torque from the applied force? If so, that is incorrect.

Yes to both. Perhaps I was sucked into an error by my approach. Can you see it?

 

[jbriggs444]

Yes to both. Perhaps I was sucked into an error by my approach. Can you see it?

$F \ne ma$
$\sum F = ma$

One can use an energy approach. But then should treat $a$ and $\alpha$ as unknowns to be determined.

You have $F$ (the applied force) and $X$ (the displacement of the center of mass) to work with. And you have the assumption of rolling without slipping.

If the center of mass is displaced by $X$ then the cumulative distance moved by the rod that is supplying force $F$ is $2X$. The mechanical work supplied by $F$ is then obviously: $W=2\vec{F} \cdot \vec{X}$.

If you like, you can interpret the work done as center of mass work $W_\text{com} = FX$ and mechanical work in the center of mass frame $W_\text{mech} = \tau \theta$. Torque $\tau$ is given by $FR$ while rotation angle $\theta$ is given by $\frac{X}{R}.$ It follows that $W_\text{mech} = FX$ and by no coincidence, $W_\text{tot} = W_\text{com} + W_\text{mech} = FX + FX = 2FX$

By the work-energy theorem, the final kinetic energy is equal to the initial kinetic energy plus the work done:$KE_\text{final} = 2FX$.

Final kinetic energy is also given by $\frac{1}{2}mv^2 + \frac{1}{2}I \omega^2$

But rotation rate $\omega$ is $\frac{v}{R}$ and moment of inertia $I$ is $\frac{2}{5}mR^2$. So we can substitute in and rewrite the previous equation as $\frac{1}{2}mv^2 + \frac{1}{2} \frac{2}{5}mv^2$ for a total of: $$KE = 2FX = 0.7 mv^2$$We can solve for $v$ and get:$$v = \sqrt{\frac{20FX}{7m}}$$Given the total distance covered and the final velocity we can do the boring work of determining the horizontal acceleration $a$ and the rotational acceleration $\alpha$.

Let us go ahead and do that. Elapsed time $t$ is given by distance $X$ divided by average velocity $\frac{v}{2}$. That gives us:$$t = \frac{X}{0.5\sqrt{\frac{20FX}{7m}}} = \sqrt{\frac{7mX}{5F}}$$Acceleration is then given by the change in velocity divided by the elapsed time. That gives us:$$a = \frac{v}{t} = \frac{\sqrt{\frac{20FX}{7m}}}{\sqrt{\frac{7mX}{5F}}} = \frac{10F}{7m}$$More directly, we could have simply used the fact that the "effective mass" of a rolling sphere is $\frac{7}{5} = 1.4$ times its regular mass and the fact that we have a 2:1 mechanical advantage.

Edit: Massive edit to add the equations.

 

[gleem]

This is my thought. The forces are the applied force and the frictional force I assumed. But the frictional force does no work so it was not included in the equation for work. One might think the frictional force might contribute to the rotational kinetic energy. But consider this if the sphere were in free space the applied tangential force would cause a rotation which if placed in contact with a frictional stationary surface underneath would not change the motion, would it? So no additional torques would be present.

Now if the force were applied to the center of the sphere there would be no rotation until it was in contact with a stationary surface which would produce a torque.

 

[jbriggs444]

This is my thought. The forces are the applied force and the frictional force I assumed. But the frictional force does no work so it was not included in the equation for work. One might think the frictional force might contribute to the rotational kinetic energy. But consider this if the sphere were in free space the applied tangential force would cause a rotation which if placed in contact with a frictional stationary surface underneath would not change the motion, would it? So no additional torques would be present.

Now if the force were applied to the center of the sphere there would be no rotation until it was in contact with a stationary surface which would produce a torque.

See the work editted into #77 above. Both acceleration $a$ and angular acceleration $\alpha$ have contributions from the frictional force on the bottom. Although this does not contribute to energy, it does contribute to acceleration.

Think about it this way if you like... The frictional force on the bottom takes away from horizontal acceleration and adds to rotational acceleration. There is no net change in energy since this force does zero work. But it does change the balance between the kinetic energy of linear motion and the kinetic energy of rotation.

 

[gleem]

If the center of mass is displaced by X then the cumulative distance moved by the rod that is supplying force F is 2X.

I just rolled a cylinder without slipping with a ruler on top as in the OP. The center of the cylinder moved the same distance as the ruler.

 

[jbriggs444]

I just rolled a cylinder without slipping with a ruler on top as in the OP. The center of the cylinder moved the same distance as the ruler.

What, exactly, did you measure?

If you measured the forward progress of the center of the cylinder and separately measured the forward progress of one end of the ruler then you should clearly witness a 2:1 ratio, ruler:cylinder.

If, however, you measured the forward progress of the center of the cylinder and separately measured the rearward progress of the point of contact where the cylinder rolls on the ruler, you should clearly witness a 1:1 ratio. (The cylinder rolls forward 1 unit, the ruler forward by 2 units and the delta covered by the point of contact relative to the ruler is -1 unit).

 

[A.T.]

I just rolled a cylinder without slipping with a ruler on top as in the OP. The center of the cylinder moved the same distance as the ruler.

It should be obvious that the top of a rolling cylinder moves twice as fast as it's center. So if you don't have any slippage (neither at the ground, nor at the ruler), then a horizontal ruler touching the top will also move twice as fast as the center.

From: https://www.miniphysics.com/uy1-rolling-motion.html

 

[A.T.]

$$KE_{tot}= \frac{1}{2}mv^2 +\frac{1}{2}I\omega ^2 = FaX + L\alpha =FaX+I\alpha \theta $$

The dimensions in ths equation don't match.

 

[gleem]

First, @jbriggs444
I see said the blind man as he picked up his hammer and saw that the speed at the top of the sphere is moving twice as fast as the center so the bar moves twice the distance compared to the center. Thanks for your patience.

Second: @A.T.

It should be obvious that the top of a rolling cylinder moves twice as fast as it's center

Actually, I always knew that but somehow did not see its relevancy until @jbriggs444 kept my attention on the actual work that was being performed.

The dimensions in ths equation don't match.

As for this, I was requested to use LaTex for my equations and in a rush accidentally forgot to replace F with m. but thanks anyway.

I will edit my previous posts to show appropriate corrections and errors suitably struck out.

 

[jbriggs444]

$$KE_{tot}= \frac{1}{2}mv^2 +\frac{1}{2}I\omega ^2 = FaX + L\alpha =FaX+I\alpha \theta $$

The dimensions in ths equation don't match.

As I reconstruct the intent we have several typos and some idiosyncratic usage.

@gleem has been using $ma$ as a synonym for $F$. One $FaX$ should read $FX$ and the other should read $maX$.
@gleem has been using $L$ for torque. $L\alpha$ should have read $L\theta$. I will use $\tau$ for torque instead.

With those corrections in mind we arrive at:$$KE_\text{tot} = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2 = FX + \tau \theta = maX + I \alpha \theta$$That equation is, at least, dimensionally consistent.

Although it is wrongly motivated, the resulting equation is even correct.

While $F \ne ma$ and $\tau \ne I \alpha$, the sums of the two terms: $FX + \tau \theta$ and $maX + I \alpha \theta$ do indeed happen to match. The discrepancy in the one exactly cancels with the discrepancy in the other. Because the unaccounted for force does zero work.

 

[A.T.]

With those corrections in mind we arrive at:$$KE_\text{tot} = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2 = FX + \tau \theta = maX + I \alpha \theta$$That equation is, at least, dimensionally consistent.

Although it is wrongly motivated, the resulting equation is even correct.

While $F \ne ma$ and $\tau \ne I \alpha$, the sums of the two terms: $FX + \tau \theta$ and $maX + I \alpha \theta$ do indeed happen to match. The discrepancy in the one exactly cancels with the discrepancy in the other. Because the unaccounted for force does zero work.

OK, but $KE_\text{tot} = FX$, not $FX + \tau \theta$.

 

[jbriggs444]

OK, but $KE_\text{tot} = FX$, not $FX + \tau \theta$.

If $X$ is the distance that the center of mass moves then the mechanical work done is $2FX$. And since $\tau = FR$ and $\theta = \frac{X}{R}$, it turns out that $\tau \theta = FX$.

 

[A.T.]

If $X$ is the distance that the center of mass moves then the mechanical work done is $2FX$.

I went by diagram by @gleem where $X$ is the distance of the upper bar applying the force $F$, so $FX$ would be the total work done and thus the total change in $KE$.

 

[jbriggs444]

I went by diagram by @gleem where $X$ is the distance of the upper bar applying the force $F$, so $FX$ would be the total work done and thus the total change in $KE$.


View attachment 348445

OK, I agree with that.

I had not interpreted the drawing that way. But it makes perfect sense, now that you've drawn my attention to it. It does mean that evaluating the work done by the same force twice would be double dipping, just as you've said.

 

[gleem]

OK, I agree with that.

Wait. I am currently working on revising my post per your suggestions and have changed my diagram to

It would have been posted by now but overnight I lost a lot of my work for some reason and will have to redo it.

 

[gleem]

This post is a revised version of my previous post 66 due to suggestions of @jbriggs444 and other posts that changed my thinking.

A sphere or radius R and mass M is pushed tangentially with constant force $F$ for $X$ meters as with a flat bar of length $X$ moving across the top of the sphere. The sphere moves without slipping due to a frictional force $f$ horizontally on a flat surface as shown below. The frictional force contributes to the net force on the sphere as well as the rotation.

The force being constant produces a constant translational acceleration $a = F/M-f/M$ and a constant rotational acceleration $\alpha = L / I$ where$L =(F+f)⋅R$ and $I$ is the moment of inertia. I will assume for this discussion that $I=kmR^2$. The work required to do this has two parts one rotating it and one moving the center of mass horizontally. The force is applied for a distance of twice the length $X$ of the bar causing the sphere to move only a distance$X$ and rotates the sphere θ radians where $\theta =X/R$.
Using the above relationships we determine that $$a=2F/M(1+k)$$ and $$\alpha= 2kRF/I(1+k)$$

The translational work is$W_{T}=F⋅X$ since f is not applied over a distance along the surface. The rotational work is $W_{R} = (F+f)R \theta$. The total work $W_{total} = 2FX$

After the force has been applied the sphere moves with constant velocity $V$ and rotates with an angular velocity $\omega$. The two which are related by $$V = R\omega$$.

This results in the kinetic energy of translation $KE_T=\frac{1}{2}⋅mV^{2}$ and the kinetic energy of rotation $KE_R=\frac{1}{2}⋅I\omega^{2}$.

Because of the constant acceleration, the translational and rotational velocities are related to their respective distances of the applied force by similar equations.
$$V ^{2} = 2⋅ a ⋅x $$
and
$$\omega ^{2} = 2⋅ \alpha \theta$$.
Using these relationships avoids the consideration of the time the force is applied.

Using the above relationships we find
$$V^2= 4FX/(M(1+k))$$
and
$$\omega^2=4kFX/(I(1+k))$$

The total kinetic energy is
$$KE_{final}= \frac{1}{2}⋅I\omega^{2}+\frac{1}{2}⋅mV^{2}$$
So $$KE_{final}= \frac{1}{2}⋅I\omega^2=4kFX(I(1+k))+ \frac{1}{2}⋅M4FX/(M(1+k)$$ thus

$$KE_{final}= 2kFX/(1+k)+ 2FX/(1+k)$$ which is $2F$ the work done by $F$

If the frictional force is eliminated and the sphere slides, it will rotate due to $FR$ translate due to $F$. We find interestingly that $KE_{T} =FX$ and $KE_{R} =FX$. The work is shared equally between translation and rotation.

 

[A.T.]

If the frictional force is eliminated and the sphere slides, it will rotate due to $FR$ translate due to $F$. We find interestingly that $KE_{T} =FX$ and $KE_{R} =FX$. The work is shared equally between translation and rotation.

Does the ratio of $KE_{T}$ to $KE_{R}$ depend on the mass distribution here? Are they equal only for spheres of uniform density?

 

[jbriggs444]

Does the ratio of $KE_{T}$ to $KE_{R}$ depend on the mass distribution here? Are they equal only for spheres of uniform density?

Ahhh, good catch. I believe that they will only be the same for a thin hoop.

In the case of rolling without slipping for a thin hoop, this obviously holds. $I = mr^2$ so $KE_{R} = \frac{1}{2}I \omega^2 = \frac{1}{2}mr^2 \frac{v}{r}^2 = \frac{1}{2}mv^2$.

In the case of rolling without slipping for mass distributions where $I \ne mr^2$ the equality will not hold.

In the case of rolling without friction for a thin hoop, it will turn out that this yields rolling without slipping. The bottom contact point rolls smoothly on the supporting platform, even without friction.

[Sadly, I lack access to a hula hoop that I can suspend from a thread while tapping horizontally on a screw afixed to the top to see whether the bottom deflects as an immediate result]

The interesting case is for rolling without friction and without a moment of inertia given by $I = mr^2.$ In this case, slipping occurs.

Let us consider the case of a sphere with moment of inertia given by $I = \frac{2}{5}mr^2$ on a frictionless surface.

The assertion is that a force applied to the top of the object will result in bulk kinetic energy of linear motion ($KE_T = \frac{1}{2}mv^2$) equal to the rotational kinetic energy that it also produces, $KE_R = \frac{1}{2}I \omega^2$.

This object rotates 2.5 times more easily than it translates. If we apply a force, that force will result in 2.5 times more distance rolled (of the top surface relative to the COM) than the COM translates. That means 2.5 times the rotational work done and 2.5 times the rotational kinetic energy gained. By that argument, the claim that $KE_{R} = KE_{T}$ fails.

Arguing a slightly different way, the rotary motion involves 2.5 times more "velocity" on an "effective mass" that is 2.5 times less resistive. But the energy formula has only one factor of mass and two factors of velocity. So again, 2.5 times more energy is rotational.

We may be misled by the picture which suggests that the object rolls through an angle of $X/R$ radians while the center of mass moves $X$. However, this only holds for rolling without slipping. In the case of a sphere being pushed on a frictionless surface, the rotation angle should turn out to be $2.5X/R$.

 

[gleem]

Does the ratio of KET to KER depend on the mass distribution here? Are they equal only for spheres of uniform density?

This statement is made with no frictional force as in the OP. The moment of inertial cancels when ω²=2F/I is plugged into KE_R =½Iω².

 

[jbriggs444]

This statement is made with no frictional force as in the OP. The moment of inertial cancels when ω²=2F/I is plugged into KE_R =½Iω².

You are going to need to flesh that out more completely. Start by justifying that $\omega^2 = 2F/I$.

Note that as it stands, this equation is not even dimensionally consistent.

 

[gleem]

ou are going to need to flesh that out more completely. Start by justifying that ω2=2F/I.

Oops! forgot X. ω²= 2FX/I
$$\omega^{2} =2\alpha\theta =2LX/RI =2FRX/RI =2FX/I$$

 

[jbriggs444]

Oops! forgot X. ω²= 2FX/I
$$\omega^{2} =2\alpha\theta =2LX/RI =2FRX/RI =2FX/I$$

What makes you think that $\theta = \frac{X}{R}$?

When you give up friction, you give up on rolling without slipping. Now the rotation angle is de-coupled from the translation distance.

I'd called this potential error out in the last paragraph in #93 above.

 

[A.T.]

This statement is made with no frictional force as in the OP.

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.

 

[gleem]

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.

When you give up friction, you give up on rolling without slipping. Now the rotation angle is de-coupled from the translation distance.

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 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.

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.

 

[jbriggs444]

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

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 $10^{16}$ 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.

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?

Let us agree not to worry about the limits to rigidity or rotation rate imposed by Special Relativity and keep this purely Newtonian. Without even starting the calculation, I predict a fiendishly high rotation rate with a truly ludicrous tangential velocity.

 

[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?