# Do things roll on frictionless surfaces?

I was solving this question in rotational mechanics (not asking for help).
A cylinder of mass m, radius r, is released from rest from the top of the inclined plane of inclination $$\theta$$ and length l.
Find the speed of the cylinder when it reaches the bottom.
The author then goes to solve this using energy conservation, equating the final kinetic and rotational energy to the initial potential energy.

Now, he obviously assumed the object rolled on the surface, which, if I'm not wrong can only happen on a surface that offers some friction.

So my question is...
Can we really apply energy conservation here as a non-conservative force is involved?
It doesn't look like to me that the work done by the friction force will be zero.

EDIT: Oh...the thread title probably isn't the most accurate. I'm quite sure it won't roll on a frictionless surface.

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Doc Al
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Now, he obviously assumed the object rolled on the surface, which, if I'm not wrong can only happen on a surface that offers some friction.
Right.

So my question is...
Can we really apply energy conservation here as a non-conservative force is involved?
Ask yourself: Does the friction force do work? (What kind of friction is it?) Is any mechanical energy 'lost'?

Ask yourself: Does the friction force do work? (What kind of friction is it?) Is any mechanical energy 'lost'?
Yeah I've thought hard about that.
I think for the cylinder to roll, the friction would be up the plane, which combined with the component of gravitational force in the opposite direction, would form a force couple.
So it should do some work.
Though I am quite surely wrong, I can't figure out where.

Doc Al
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Hint: Is the friction static or kinetic?

Doc Al
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Another hint: Presumably we're talking about rolling without slipping, correct?

rcgldr
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If the cylinder rolls without slipping, then no energy is lost to heat due to kinetic (sliding) friction. The static friction force performs work by applying a torque to the cylinder, increasing it's angular energy. The static friction force also opposes the component of gravitational force in the direction of the plane, reducing the linear rate of acceleration of the cylinder along the plane. The end result is the sum of angular and linear kinetic energy of the cylinder will equal the change in potential energy due to gravity (m g h).

If there was slippage, then some of the energy would be lost to heat, except in the case of zero friction. In the case of zero friction, there's no loss in energy, no change in angular velocity, and all of the energy change would be due to linear kinetic energy, in which case 1/2 m v^2 would equal m g h.

Doc Al
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I kind of wanted the OP to figure that out for his or herself. So much for that idea!

rcgldr
Homework Helper
I kind of wanted the OP to figure that out for his or herself. So much for that idea!
The OP mentioned that the instructor had already shown how to solve the problem using energy based equations, so I thought I was just clarifying what the instructor probably explained in the class.

Thanks Doc Al and rcgdlr.
I did figure it out after post #4, but it was 2:00 in the morning so didn't reply.

I would like to confirm the logic behind that (static friction doing no work). That there is no relative displacement of the contact point?

Doc Al
Mentor
I would like to confirm the logic behind that (static friction doing no work). That there is no relative displacement of the contact point?
Exactly. The instantaneous speed of the point of contact is zero, so the static friction force does no work on the system. Mechanical energy is conserved. There's certainly do dissipative work done, since that would require slipping between the surfaces.

Note that you can still apply Newton's 2nd law for translation and rotation and deduce how the translational and rotational energies will change. But, nonetheless, the static friction does no work.

arildno
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Gold Member
Dearly Missed
Rolling is a very good example to dismantle the rather widespread misconception among laymen that if there is no work, then nothing "happens" in physics.

Static friction definitely makes "something" happen, namely twist the type of momentum contained within a sliding object into the type of momentum a rolling object has.
And that is no mean feat, particularly since it doesn't cost any energy to achieve that result!

rcgldr
Homework Helper
I would like to confirm the logic behind that (static friction doing no work). That there is no relative displacement of the contact point?
I'm not sure of the point here. There's no energy loss (conversion to heat) due to static friction, but there is conversion of gravitational potential energy into linear and angular kinetic energy. The net linear force in the direction of the plane is the component of gravity force (weight) in the direction of the plane, minus the opposing static friction force related to the angular inertia of the cylinder. This net force times distance rolled equals the linear work done on the cylinder, and shows up as it's linear kinetic energy. The opposing friction force, times the radius of the cylinder, results in a torque, and this torque time angular displacement (rotation) equals the angular work done on the cylinder and shows up as it's angular kinetic energy.

For a similar example, imagine an electric car on a road on the moon (no atmosphere, no drag, no rolling resistance), and ignore the energy factor from the rotating parts (motor, drivetrain, wheels, ...). There's no relative movement at the point of contact, but the horizontal force applied at the contact patch times distance moved equals the work done.

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Doc Al
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This net force times distanced rolled equals the linear work done on the cylinder, and shows up as it's linear kinetic energy. The opposing friction force, times the radius of the cylinder, results in a torque, and this torque time angular displacement (rotation) equals the angular work done on the cylinder and shows up as it's angular kinetic energy.
Here's something I wrote a while back on this issue:

You need to distinguish between "real" work and what is sometimes called "pseudowork" (though that last term never really caught on - pity!). "Real" work is the same work that appears in the first law of thermodynamics (conservation of energy) and depends on the details of how the force is applied and the movement of the point of contact. No real work is done by friction on the rolling cylinder, since the point of contact doesn't move. That's why mechanical energy is conserved as the cylinder rolls down the incline. (If friction did work, you'd get an increase in internal energy--"heat"--somewhere.)

On the other hand, friction is most definitely exerting a force on the cylinder. If you take a force acting on an object (like friction) and multiply it by the displacement of the object's center of mass, you get a quantity that looks like a work term but is better called pseudowork (or "center of mass" work)--what it determines is not the real work done on the object, but the change in the KE of the center of mass of the object. This is usually called the "Work-Energy" theorem:
$$F_{net}\Delta x_{cm}=\Delta (\frac{1}{2}m v_{cm}^2)$$

Despite the name, this is really a consequence of Newton's 2nd law, not a statement of energy conservation.
It's partly a matter of semantics, but I prefer to reserve the term 'work' for 'real' work, as in the first law of thermodynamics. When work is done on a system, mechanical energy is being transferred. That's not the case here, of course.

From that same quoted thread, here are a couple more examples of zero-work forces in action:
Don't confuse the fact that an applied force must accelerate an object--per Newton--with the thought that it must also transfer energy to the object (that's what real work means). Another example: An accelerating car. Friction of the road on the tires certainly accelerates the car, but it does no work on the car (for the same reason, since we assume no slipping). It's not the road that provides energy for the car, it's the gasoline and the engine! Of course, you can use pseudowork (if you know the friction force) to calculate the change in KE using the work-energy theorem described above.

Here's another example: You jump in the air. Your KE obviously increased, but did the ground do work on you? No! Your point of contact with the ground did not move, so the ground did no work on you. So where did the energy come from? From the chemical energy in your muscles, of course, not from the ground. Of course, you can consider the average force that the ground exerts on you and use it to calculate your KE--but that's an application of pseudowork again, not real work.

Thanks a lot for clearing that up Doc Al.

rcgldr
Homework Helper
Here's something I wrote a while back on this issue:
"pseudowork" ... "since the point of contact doesn't move"
Although the surfaces do not slide with respect to each other, the point of contact is moving, in the case of the OP, along the surface of the plane and around the surface of the cylinder. You have a force and you have movement of the point of contact. Work equals force times distance the point of contact moves, and power equals the force times speed the point of contact moves at.

I don't recall the concept of "pseudowork" versus work, in any of the physics books I have, although they are old. Granted in the case of the ramp and cylinder, gravitational potential energy is the source of energy, and in the case of the car, chemical potential energy from the fuel is the source of energy, but this doesn't conflict with the classical concept of "work" in physics or the ability to calculate power, work done, and/or changes in energy using the classical concept.

Doc Al
Mentor
Although the surfaces do not slide with respect to each other, the point of contact is moving, in the case of the OP, along the surface of the plane and around the surface of the cylinder. You have a force and you have movement of the point of contact. Work equals force times distance the point of contact moves, and power equals the force times speed the point of contact moves at.
Please tell me the instantaneous speed of the bottom of the cylinder (the point of contact) relative to the surface of the plane.

While pseudowork is not a common term, the idea of the friction force doing no work in the example of the cylinder rolling without slipping is standard stuff.

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arildno
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This is completely wrong, rcdr!

It is perfectly correct that you can rigourously define a curve that is expanding in time, and that can be called "the curve of the point of contact".

But, this is not a <i>material</i> object, or for that matter, a material point object, but a <i>geometric</i> point.

Instantaneously, that geometric point coincides with a material point (on both surfaces), but forces act only upon the material points.

Thus, in order to use work=force*distance, you must be sure the distance travelled is done by a material point, rather than a geometric one.

rcgldr
Homework Helper
Please tell me the instantaneous speed of the surface of bottom of the cylinder relative to the surface of the plane.
There's no slippage, so speed between the surfaces is zero. The surface speed relative to the center of the cylinder is equal to the speed of the plane relative to the center of the cylinder (and equal and opposite to the speed of the cylinder relative to the plane).
Please tell me the instantaneous speed of the point of contact relative to the surface of the plane.
The point of contact moves along the plane at the same speed as (the center of) the cylinder. The instantaneous speed of the point of contact is the same as the instantaneous speed of the cylinder. If a car is moving at 100 kmh, then the 4 points of contact between the tires and pavement are also moving at 100 kmh (relative to the road).

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Doc Al
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There's no slippage, so relative surface speed is zero.
That's true.

The point of contact moves along the plane at the same relative speed as (the center of) the cylinder. The instantaneous speed of the point of contact is the same as the instantaneous speed of the cylinder.
Huh? This statement directly contradicts your first sentence above. The 'point of contact' is the bottom of the cylinder.

For any rolling object, the relative surface speed is zero, but the point of contact is moving at the same speed as the rolling object. If a car is moving at 100 kmh, then the 4 points of contact between the tires and pavement are also moving at 100 kmh (relative to the road).
I have no idea why you think this.

rcgldr
Homework Helper
Thus, in order to use work=force*distance, you must be sure the distance travelled is done by a material point, rather than a geometric one.
OK, I'll mark a spot at the surface of the tire, and note how far that spot has moved forwards for each revolution of the tire as it rolls, and also note the time it took for that spot to move that far. The average speed of that spot = the distance traveled between where the spot met the road on each revolution divided by the time it took for the tire to rotate 1 revolution. The path of the material point is a cylcoid.

If a car is moving at 100 kmh, then the 4 points of contact between the tires and pavement are also moving at 100 kmh (relative to the road).
I have no idea why you think this.
If I observe a car moving by, I can see that the 4 points of contact are moving at the same speed as the car. I consider the point of contact to be the continuous geometrical point where the tire and pavement meet, not a specific point on either surface. If I use the car as a frame of reference, than the surface speed at the point of contact, is the same as the road speed (relative to the car).

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Doc Al
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OK, I'll mark a spot at the surface of the tire, and note how far that spot has moved forwards for each revolution of the tire as it rolls, and also note the time it took for that spot to move that far. The average speed of that spot = the distance traveled between where the spot met the road on each revolution divided by the time it took for the tire to rotate 1 revolution. The path of the material point is a cylcoid.
Well, sure, the average speed along the ground of any part of the tire must equal the speed of the center of the tire. So what? What matters is the instantaneous speed of the tire patch that is in contact with the ground while it's in contact with the ground. That speed is zero. Once it breaks contact there's certainly no force on it.

rcgldr
Homework Helper
Apparently the issue here is what is considered the point of contact, and what is to be used for the frame of reference. If the center of the rolling object is used as a frame of reference, then the surface moves in a circle, and the surface speed at the contact point is the same as the speed of the plane relative to the rolling object.

As an example of how this concept is used, note the discussion on power input versus power output of a DDWFTTW cart. The cart is used as the frame of reference. The power input = ground force x ground speed (at the point of contact). The power output equals prop thrust force x prop thrust speed.

As another example, discussions of contact patches between tires and pavement on vehicles consider the contact patch to be moving at the same speed as the vehicle. Slip angle can be defined as the difference between the angle the tire is oriented, versus the the angle the tire (and it's contact patch) is actually moving.

update #1 - To avoid confusion, some texts use a phrase similar to "the point of the rolling object in contact with the surface of the plane". This defines a specific point on the rolling object. To me the the point of contact is more abstract, it's the continously moving point where a rolling object and plane are in contact.

update #2 - Perhaps my take on this could be better expressed as "the point of application of force", which would move as the rolling object moves relative to the plane. In the case of an object rolling down an inclined plane, there is a continous force, and it's point of application is moving at the same speed as the rolling object (relative to the plane). In order to do power and work calculations, it makes sense to have some convention for noting the amount of force and the speed or distance involved with the point of application of that force.

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