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

[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?

 

[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.