Rolling down a slope vs to slide down a slope

AI Thread Summary
The discussion centers on the differences in motion between a rolling cylinder and a sliding cylinder on an incline. It highlights that while both can have the same translational motion, the rolling cylinder also incorporates rotational motion, resulting in different kinetic energy distributions. The conversation emphasizes that the center of mass (COM) motion is crucial for understanding translational acceleration, regardless of where forces are applied. Additionally, the role of friction in rolling versus sliding is debated, with the conclusion that the forces and resulting accelerations differ due to the effects of rotation. Ultimately, both cases demonstrate that while translational speeds may be similar, the kinetic energy outcomes differ significantly due to rotational dynamics.
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Consider the rolling cylinder in the figure

img1437.png


To describe the cylinders translational motion, I have

Ma=G sin \theta - f

a is the acceleration of the cylinder parallel to the slope.

If you have an object (cylinder or not) sliding down the same slope without friction, with the same mass as the cylinder in the figure. And if you then apply a force through the center of mass equally big as the friction force f above, with a direction parallel to the slope upwards (so that it works against the direction of the acceleration) - will this sliding body act exactly as the rolling body above (considering translational motion, not rotational)?
 
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Why not?
 
Maybe because the fact that the body is rotating will make it accelerate differently. Plus I don't find it intuitive that the friction force working on the edges of the cylinder will be as effective in changing the motion as a force acting directly in the COM. The frictional force creates a torque, while the force in the COM does not.
 
Well, speaking about COM, in my opinion, it's just a mathematical point which is useful for perceiving the motion. Since it's math, then let's talk about math. Math says, the equation of motion of COM is the equation of motion of a mass point, whose mass equal to the mass of the body, experiencing a force equal to the sum of all forces on the body. So the fact that the body rotates and frictional force acts at the edge or anywhere else (which directly affects only torque, and again, only rotation) doesn't affect the motion of COM (points don't rotate!). Trust math!
 
I really don't think that's called math.

Anyway, consider this: You have a rod laying down on a frictionless surface.

If you hit the rod with a force near the edges two things will happen: a) it will get translational motion in the direction of the force, b) it will get rotational motion due to the torque given by the force.

Hit the rod again: If you hit the rod in COM, only a) occurs. This means that it will translate faster than the above. The translational acceleration is greater here than above. In my own clumsy words: "The force acting in COM is more effective in giving translational acceleration of the object".

Thus, the rod moves very differently in the two cases. And silly me draw a parallel to the rolling cylinder case. When the cylinder is rolling, you have a (friction) force (f) acting on the edges, and this gives a rotational motion. When the cylinder is sliding, and you apply a force (f), of equal magnitude as the friction force, through the COM, parallel to the direction of the friction force. You thus have no rotational translation, and my intuition then tells me that the force acting through COM is more effective in "stopping" the object from sliding downwards.

Well, I can't make sense out of it. So I asked here so that maybe some kind soul could clarify this for me.

Thanks
 
Yes, it's more "effective" because you gain rotation besides translational motion. Only rotation makes the difference, agree? If you leave rotation aside, as you wrote in the parentheses in post #1, then the rest is translational motion, and there is no difference in translational motion, correct? Plus the fact that translational motion of the body is actually described by the COM's motion, consequently, there is no difference in translational motion between when f is applied at the edge and when f is applied right at the center.

the force acting through COM is more effective in "stopping" the object from sliding downwards.

As I have pointed out, this is not true. The downward motion is in fact the translational motion, which has nothing to do with rotation.

Now the main source of your problem is, I think, the way people usually choose to perceive the motion of a body. In a body, there are infinitely points, which have various motions. You may arbitrarily pick one point and describe the other points' motions via that point, if you want. But physicists find it easy exploiting a mathematical point named COM, defined by the formula M\vec{r}_{COM}=\Sigma \vec{r}\Delta m. Officially, the "translational motion" term actually comes after COM. It is only after COM is defined is the term born. Translational motion of the body, in fact, is COM's motion. Rotational motion of the body is the other points' motion relative to COM. So when talking about translational motion, we talk about COM, and when talking about COM, we talk about a mass point which has advantageous features that I said earlier.
 
hikaru's analysis is absolutely correct.

You really have to go through derivation of laws of motion for rigid bodies at least once to really see this. Try looking at Wikipedia's article on Rigid Body. It seems sufficiently detailed.
 
So we have concluded that the rolling cylinder and the sliding cylinder will have the same translational motion. If they where to race down a slope, it would be a dead one.

What about kinetic energy? It arises from both translation and rotation, meaning that the rolling cylinder will have more KE than the sliding one. Where does this energy come from, then? How is the rolling sylinder able to obtain more KE than the sliding one?
 
k4ff3 said:
Anyway, consider this: You have a rod laying down on a frictionless surface.

If you hit the rod with a force near the edges two things will happen: a) it will get translational motion in the direction of the force, b) it will get rotational motion due to the torque given by the force.

Hit the rod again: If you hit the rod in COM, only a) occurs. This means that it will translate faster than the above. The translational acceleration is greater here than above. In my own clumsy words: "The force acting in COM is more effective in giving translational acceleration of the object".
As hikaru1221 has already explained, it doesn't matter where on an object you apply the force: the acceleration of the COM will be the same.

What might be throwing you off is your intuition that it's more difficult to apply the same force if the body turns than if you push it square towards its center of mass. That's perfectly true! The point of application moves away from you, so you have to move faster to continue to exert the same force. But if you are able to apply the same force at the edge of the rod or at the center, you'll get the same acceleration of the COM.
 
  • #10
I'm not sure if I missed something here, but it seems like people are saying that whether or not there is friction, the object will accelerate down the ramp at the same speed. This isn't true. And the OP figured out probably the clearest way to show why:
k4ff3 said:
What about kinetic energy? It arises from both translation and rotation, meaning that the rolling cylinder will have more KE than the sliding one. Where does this energy come from, then? How is the rolling sylinder able to obtain more KE than the sliding one?
This is correct: when the cylinder gets to the bottom of the slope, if it is sliding, all of the KE is in the translational motion, but if it is rolling, some is in the translational and some is in the rotational. By conservation of energy, the translational KE (and therefore speed) of the rotating cylinder must be lower than that of the sliding cylinder.
 
  • #11
k4ff3 said:
What about kinetic energy? It arises from both translation and rotation, meaning that the rolling cylinder will have more KE than the sliding one. Where does this energy come from, then? How is the rolling sylinder able to obtain more KE than the sliding one?

For a body, KE = translational KE + rotational KE. Since the net forces in the 2 cases are the same, the translational speeds are the same, and thus, translational KEs are too. Again, the difference lies in rotation. So the difference has something to do with rotation. Here is what rotation does: it affects the velocity of the point A that friction acts on. Since work element done by friction is dW=\vec{F_{friction}}\vec{v_{A}}dt, the works done by friction in the 2 cases are different. |vA| in the rolling case is smaller, but friction does negative work, so eventually, KE gained in the rolling case greater.
 
  • #12
Doc Al said:
As hikaru1221 has already explained, it doesn't matter where on an object you apply the force: the acceleration of the COM will be the same.

What might be throwing you off is your intuition that it's more difficult to apply the same force if the body turns than if you push it square towards its center of mass. That's perfectly true! The point of application moves away from you, so you have to move faster to continue to exert the same force. But if you are able to apply the same force at the edge of the rod or at the center, you'll get the same acceleration of the COM.
While I understand this, I'm not sure how it answers the OP's question. The applied force in both cases is the weight of the cylinder...but the reaction forces are different in each case, so the "unbalance" of the cylinder is different. This is not a case like the one you just described.

Specifically, the friction force causes the cylinder to spin while opposing (reducing) its translational acceleration.
 
  • #13
russ_watters said:
I'm not sure if I missed something here, but it seems like people are saying that whether or not there is friction, the object will accelerate down the ramp at the same speed. This isn't true. And the OP figured out probably the clearest way to show why:
The OP made up 2 situations:
1 - there is friction, and it's rolling
2 - there is no friction, and it's sliding, but there is another force of the same magnitude of friction, applying at the center, acting upwards.
If you understood the problem that way, you may rethink of the last conclusion you made.
 
  • #14
hikaru1221 said:
For a body, KE = translational KE + rotational KE. Since the net forces in the 2 cases are the same, the translational speeds are the same, and thus, translational KEs are too. Again, the difference lies in rotation. So the difference has something to do with rotation. Here is what rotation does: it affects the velocity of the point A that friction acts on. Since work element done by friction is dW=\vec{F_{friction}}\vec{v_{A}}dt, the works done by friction in the 2 cases are different. |vA| in the rolling case is smaller, but friction does negative work, so eventually, KE gained in the rolling case greater.
The KE gained in the two cases cannot be different because conservation of energy requires that the total energy of the system is constant throughout the entire experiment. The energy available is the potential energy at the start of the experiment and at the end of the experiment, the resulting KE must be exactly equal to that PE.

Consider the exaggerated example of a yoyo. By changing the diameter of the shaft, you can change where the force is applied to the cylinder. The smaller the diameter, the more of the energy is converted to rotational KE and the slower the yoyo will fall.
 
  • #15
I see what russ' complaint is, and if this is exactly what OP was talking about, russ has a point.

The friction force in rolling case will not be equal to friction force if you restricted cylinder to slide, rather than roll. In first case, the force will be determined by angle of the incline and the moment of inertia. In the second, by kinetic friction coefficient.

But the way OP stated it, I got the impression that he simply applies a force equivalent to friction in the rolling case to the cylinder. In which case, hikaru's logic applies.

Might be nice if OP could clarify that.
 
  • #16
K^2 said:
I see what russ' complaint is, and if this is exactly what OP was talking about, russ has a point.

The friction force in rolling case will not be equal to friction force if you restricted cylinder to slide, rather than roll. In first case, the force will be determined by angle of the incline and the moment of inertia. In the second, by kinetic friction coefficient.
Key words in the OP: "...sliding down the same slope without friction..." Obviously, if it is sliding and not rolling, there can be no friction and vice versa - if there is no friction, it will be sliding, not rolling.
 
  • #17
This is what the OP said in post #1:

"If you have an object (cylinder or not) sliding down the same slope without friction, with the same mass as the cylinder in the figure. And if you then apply a force through the center of mass equally big as the friction force f above"
 
  • #18
These two figures illustrate the rolling case and the sliding case respectively:

Rollin'
img1437.png

---
Slidin'
[PLAIN]http://img213.imageshack.us/img213/8353/frce.png


Let me break down my questions:
  • Which one wins the race?
  • If it's a dead race, as some here claim, then what about the KE?
  • If the opposite is true, then it does matter where on an object you apply the force (in contrast to what Doc Al claimed). Which I also find non-intuitive.

Im sorry if I have caused confusion. I hope this clear things up.

--
Credits go to mspaint.exe
 
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  • #19
k4ff3 said:
Let me break down my questions:
Which one wins the race?
It's a tie.
If it's a dead race, as some here claim, then what about the KE?
What about it? In the rolling case, the friction force (static friction) does no work, but it does transform gravitational PE into rotational KE. In the sliding case, the applied force does negative work.
If the opposite is true, then it does matter where on an object you apply the force (in contrast to what Doc Al claimed). Which I also find non-intuitive.
As far as the acceleration of the COM is concerned, it doesn't matter where the force f is applied. But the overall motion of the cylinder is certainly different. In one case it rotates; in the other, it doesn't.
 
  • #20
Doc Al said:
but it does transform gravitational PE into rotational KE.
.. so the PE=mgh does not apply for rotating bodies, because PE is being transformed into KE (so that PE<mgh)??
 
  • #21
k4ff3 said:
.. so the PE=mgh does not apply for rotating bodies, because PE is being transformed into KE (so that PE<mgh)??
PE = mgh applies just the same to both cases. But in the rolling case, some of that PE is transformed into rotational KE. In the sliding case, that same energy goes into whatever is applying the force f.

In one case, mechanical energy is conserved because no external forces (besides gravity) are doing any work on the system; in the other case the external force does work to reduce the mechanical energy.
 
  • #22
I must be stupid.. I really don't get it. Sorry
 
  • #23
k4ff3 said:
I really don't get it. Sorry
Just rephrase your question and ask again. What don't you get?
 
  • #24
k4ff3 said:
.. so the PE=mgh does not apply for rotating bodies, because PE is being transformed into KE (so that PE<mgh)??
When you have an extra force added, that f you apply, you are doing work on the cylinder. I have a feeling you aren't accounting for it.
 
  • #25
Thank you for being so patient.
  • In the rolling case (r): How can the gravitational PE be transformed into rotational KE?
  • In the sliding case (s): How can the gravitational PE "go into whatever is applying the force f"?
  • How does nature know that she should transform exactly the same amount of grav. PE to rot KE in r, as she transforms grav. PE to "the force applier" in s?
  • How can an object (in rest) of the same mass as the sylinder have the same PE(=mgh) as the moving bodies (in an instant where they all are at the same height), when the moving bodies constantly are leaking Potential Energy?
  • Why is the force applied at the edge of the cylinder (i.e the friction force f) internal, but the force f applied to the COM external?
 
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  • #26
There is no such thing as rotational PE.
 
  • #27
k4ff3 said:
In the rolling case (r): How can the gravitational PE be transformed into rotational PE?
The static friction introduces a torque on the cylinder, causing it to rotate. Mechanical energy is conserved, since the point of application of the force does not move.
In the sliding case (s): How can the gravitational PE "go into whatever is applying the force f"?
Here's an example that might be easier to understand: You hold a book in your hand and lower it. What happened to the potential energy of the book? You absorbed it! You exerted a force on the book equal to its weight (in this case), so you did negative work on the book, thus transforming all of the book's gravitational PE into your internal energy. The same thing is going on when you exert a force f through the center of the cylinder.
How does nature know that she should transform exactly the same amount of grav. PE to rot KE in r, as she transforms grav. PE to "the force applier" in s?
Nature doesn't have to 'know' anything; the cylinder just responds to the forces applied. In both cases the force f is the same, so the amount of energy that is 'diverted' from going into translation KE is the same. (If there were no friction and no force f, all of the PE would go into translational KE.)
How can an object (in rest) of the same mass as the sylinder have the same PE(=mgh) as the moving bodies (in an instant where they all are at the same height), when the moving bodies constantly are leaking Potential Energy?
What do you mean by 'leaking' Potential energy? At any given height, the gravitational potential energy is the same regardless of whether the cylinder is rotating or sliding. As the cylinder lowers, the potential energy is less. The energy goes into rotation in one case or work done by the applied force in the other case.
Why is the force applied at the edge of the cylinder (i.e the friction force f) internal, but the force f applied to the COM external?
In both cases the force is external. But in the sliding case, that force does no work. Key point: Since the cylinder rolls without slipping, there is no relative motion between the bottom of the cylinder and the surface. In order for an external force to perform work, the point of application of that force must be displaced.
 
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  • #28
I'll borrow your picture for a while. Thanks in advance :biggrin:

k4ff3 said:
Thank you for being so patient.
  • In the rolling case (r): How can the gravitational PE be transformed into rotational PE?
  • In the sliding case (s): How can the gravitational PE "go into whatever is applying the force f"?
  • How does nature know that she should transform exactly the same amount of grav. PE to rot PE in r, as she transforms grav. PE to "the force applier" in s?
  • How can an object (in rest) of the same mass as the sylinder have the same PE(=mgh) as the moving bodies (in an instant where they all are at the same height), when the moving bodies constantly are leaking Potential Energy?
  • Why is the force applied at the edge of the cylinder (i.e the friction force f) internal, but the force f applied to the COM external?

Ooh, you are going away from the right track
Let's go from the start. We have 2 cases:

CASE 1: Rolling.
In this case, let's make another assumption: the cylinder rolls without slipping, i.e. \vec{v_A}=0. This will make things easier and concurs with what Doc Al said. Because of that assumption, friction in this case doesn't do any work on the cylinder (it's static friction!). Let's assume that the cylinder starts moving from rest.
By the energy conservation law: PE = KE = translational KE + rotational KE
This concurs with the statement of Doc Al, "some of that PE is transformed into rotational KE". In addition, friction doesn't do work, but the cylinder gains both translational and rotational KE, not translational KE only as in the "sliding" case. This is why people usually say, static friction in this case only does the job of transferring a part of PE into rotational KE.

CASE 2: Sliding.
Because now the force f is applied at the center, whose velocity is not zero, f does do work on the cylinder. Besides, we have already known that the cylinder does not rotate in this case.
Therefore: PE - (work of f) = KE = translational KE
In addition, we all agree that the translational motions in 2 cases are the same, so translational KE in case 1 = translational KE in case 2. Now do you agree with me that total KE in case 2 (sliding) is smaller than in case 1 (rolling)?

For the above questions of yours:
_ #1 and 2: Maybe you mean "rotational KE" instead of "rotational PE"?
In general, PE will transfer into any other kind of energy if the object decreases its height. The object, initially, already has an amount of PE, i.e. it has the potential to gain kinetic energy or any other kind of energy if it goes down. This is because of the law of energy conservation.

_ #3: Nature only knows static friction doesn't do any work (i.e. \vec{v}_A=0, and that's the only thing she wants. It's like an initial condition. If it's not static friction, things will change. And because it's static friction, friction doesn't do work, which leads to so many results afterward.
So why static friction? It's the nature of the interaction between the material of the ground and the material of the cylinder, and I have no answer for that.

_ #4: Now it's like you have $100. Staying at rest = no trading, the money remains the same. If you trade $100 for shares, you will have, say, 100 shares.
But if you are investing in something, buying shares for example, your money will go down as you have more shares (or more KE!). Maybe yesterday you had $101, and tomorrow you will have $99, but now, you still have $100, and you have the potential to have 100 shares if you use up your money. The same thing goes for PE. People don't call it "POTENTIAL energy" for nothing!

_ #5: If we consider the cylinder only, friction and the "f" force are never internal forces. Friction is the force by the ground, so it is external force.
 

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  • #29
I have corrected the rotational PE typo, I meant of course rotational KE.


Doc Al said:
Mechanical energy is conserved, since the point of application of the force does not move.
What does it mean that the point of application of the force do not move? Why does it move when you apply the same force on the COM?

Doc Al said:
What do you mean by 'leaking' Potential energy? At any given height, the gravitational potential energy is the same regardless of whether the cylinder is rotating or sliding. As the cylinder lowers, the potential energy is less. The energy goes into rotation in one case or work done by the applied force in the other case.

By leaking I meant that the body was constantly losing PE, i.e PE was transformed. Anyway, I understood you explanation - see if I got this right:

(Arrow here means "is being transformed to")
a) Rolling case: PE -> translational KE & rotational KE
b) Sliding case: PE -> translational KE & internal E to the "thing" that exerts the "braking force" f
c) Sliding case without any other force than gravity acting: PE -> KE

In a) and b) the translational KE is the same. However, in a) the PE transforms to energy that stays within the system, namely rotational energy. Thus, the total Mechanical Energy in a) will be conserved - as opposed to what happens in b): here energy is being transferred out to the environment (like the book). Therefore: Mechanical Energy is conserved in a), not in b) - but the translational KE are in both cases the same! The race will be a tie.
In c) the Mechanical Energy is clearly conserved, the sliding body will gradually lose PE in favor of gaining more (trans.) KE. No potential energy is lost to the enviroment, or to making rot. KE. Since the trans. KE and trans. velocity are connected, the cylinder in c) obviously wins the race!

Doc Al said:
In both cases the force is external. But in the sliding case, that force does no work. Key point: Since the cylinder rolls without slipping, there is no relative motion between the bottom of the cylinder and the surface. In order for an external force to perform work, the point of application of that force must be displaced.

Can you please elaborate on this? My English isn't too good.. "In order for an external force to perform work, the point of application of that force must be displaced". Can you say this sentence with other words?

Thank you very much! I feel like I'm on the verge of an epiphany :)
 
  • #30
The force f in rotating case is what converts potential energy to rotational.
 
  • #31
hikaru1221 said:
I'll borrow your picture for a while. Thanks in advance :biggrin:
It's Okay - you can keep it ;)

hikaru1221 said:
CASE 1: Rolling.
(...)
By the energy conservation law: PE = KE = translational KE + rotational KE
If this is a typo, and that you really mean PE => KE = (...) where arrow means "is being transformed to", I agree. Because PE = KE does not follow from the energy conservation law.

hikaru1221 said:
CASE 2: Sliding.
(...)
Now do you agree with me that total KE in case 2 (sliding) is smaller than in case 1 (rolling)?
Yes, I do :)

hikaru1221 said:
For the above questions of yours:
_ #1 and 2: Maybe you mean "rotational KE" instead of "rotational PE"?
Yes, I did :)

hikaru1221 said:
_ #3: Nature only knows static friction doesn't do any work (i.e. \vec{v}_A=0)
That's true! No energy is therefore lost to the enviroment. Mech. Energy is conserved.Thanks a bunch! :)
 
  • #32
k4ff3 said:
I have corrected the rotational PE typo, I meant of course rotational KE.
Yes, I caught that.

What does it mean that the point of application of the force do not move? Why does it move when you apply the same force on the COM?
The instantaneous speed of the bottom of the cylinder as it rolls down the incline is zero. (Otherwise it would be sliding.) The speed of the COM of the cylinder is not zero, of course.

By leaking I meant that the body was constantly losing PE, i.e PE was transformed. Anyway, I understood you explanation - see if I got this right:

(Arrow here means "is being transformed to")
a) Rolling case: PE -> translational KE & rotational KE
b) Sliding case: PE -> translational KE & internal E to the "thing" that exerts the "braking force" f
c) Sliding case without any other force than gravity acting: PE -> KE

In a) and b) the translational KE is the same. However, in a) the PE transforms to energy that stays within the system, namely rotational energy. Thus, the total Mechanical Energy in a) will be conserved - as opposed to what happens in b): here energy is being transferred out to the environment (like the book). Therefore: Mechanical Energy is conserved in a), not in b) - but the translational KE are in both cases the same! The race will be a tie.
In c) the Mechanical Energy is clearly conserved, the sliding body will gradually lose PE in favor of gaining more (trans.) KE. No potential energy is lost to the enviroment, or to making rot. KE. Since the trans. KE and trans. velocity are connected, the cylinder in c) obviously wins the race!
Sounds good to me!

Can you please elaborate on this? My English isn't too good.. "In order for an external force to perform work, the point of application of that force must be displaced". Can you say this sentence with other words?
Work requires force X displacement, not just force. Since, as pointed out above, there's no instantaneous motion of the point of contact of the friction force, there's no instantaneous displacement--and no work done. This is a subtle point.

Here's another, totally different, example where a force acts to change the motion of something yet no work is done. Crouch down, then leap into the air. The ground exerts a force on your feet, propelling you upward, yet the ground doesn't move and no work is done. So where does your KE come from? (Answer: You've transformed internal energy--chemical energy in your muscles--into translational KE.)

Another example, involving static friction. Driving along, you step on the gas to accelerate your car. Assuming no slipping of the tires, the friction is static friction. And, again, the instantaneous speed of the tire patch in contact with the ground is zero so no work is done, even though, obviously, the ground is what propels the car forward. Where does the energy come from? The car converts chemical energy into mechanical energy--it burns gas.

Thank you very much! I feel like I'm on the verge of an epiphany :)
I think you are very close. As I said, some of these concepts are subtle, so don't be so hard on yourself.

Just for fun, why not use Newton's law to fully analyze the motion of the rolling cylinder? You can solve for the static friction required and the acceleration of the cylinder. And you can use the 'rolling without slipping' condition to relate rotation to translation.
 
  • #33
hikaru1221 said:
This is what the OP said in post #1:

"If you have an object (cylinder or not) sliding down the same slope without friction, with the same mass as the cylinder in the figure. And if you then apply a force through the center of mass equally big as the friction force f above"
Yeah...that second part was confusing to me. Really, it seems contradictory: if it is sliding without friction, then the friction force "f" must be zero.

In order to make it non-contradictory, we would have to completely eliminate the sliding without friction case. I see these cases right now:

1. Sliding without friction.
2. Rolling under the influence of gravity alone.
3. Rolling with a force equal to the friction force applied to the cog along the slope.

#1 & #3 are a tie, #2 is slower.

If #3 (vs #1) really is what the OP was looking for, it seems like a confusing and highly contrived scenario to me, but in that case it seems the other answers are correct.

I'm thinking it could also be a reading comprehension issue for me or organizational issue with the OP: the two sentences looked to me to be part of the same case, but maybe they are intended to be the two separate case #1 & #3.
 
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  • #34
russ_watters said:
Yeah...that second part was confusing to me. Really, it seems contradictory: if it is sliding without friction, then the friction force "f" must be zero.
In the sliding case, the force "f" is not friction, but a force equal in magnitude to the friction that exists in the rolling case that acts through the center. The point being to compare situations where the same force acts at different points.

In order to make it non-contradictory, we would have to completely eliminate the sliding without friction case. I see these cases right now:

1. Sliding without friction.
2. Rolling under the influence of gravity alone.
3. Rolling with a force equal to the friction force applied to the cog along the slope.
The three cases are:
1. Sliding without friction and no force besides gravity (and the normal force).
2. Rolling without slipping; the friction force happens to equal "f".
3. Sliding without friction and with an applied force equal to "f" acting through the center of mass.

The main comparison is between cases #2 and #3.
 
  • #35
k4ff3 said:
If this is a typo, and that you really mean PE => KE = (...) where arrow means "is being transformed to", I agree. Because PE = KE does not follow from the energy conservation law.

Okay, you may write PE => KE, no problem. But people usually write: PE = KE. Initially, it has PE, and finally it gets KE. No external work done on it, so energy is conserved: PE = KE. Of course, writing in your way shows much about the physics, but when dealing with math and calculation, you cannot avoid PE = KE.
 
  • #36
You could also have rolling with slipping... But I suppose there is enough confusion going about without it.
 
  • #37
russ_watters said:
(...)
If #3 (vs #1) really is what the OP was looking for, it seems like a confusing and highly contrived scenario to me, but in that case it seems the other answers are correct.
.

Doc Al's explanation of the cases is correct. Anyway, since I'm so good at asking questions, what is "a highly contrived scenario"? I've tried to look it up, but I can't make sense out of it in my native language. An online dictionary says: "Contrived = Obviously planned or calculated; not spontaneous or natural; labored". A "highly calculated scenario" just sounds wrong to me..
 
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