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My question is: are these three statements correct?

(1) The net work on an object that rolls without slipping can be exactly divided into a "work on the center of mass" and a "work causing rotation about the center of mass": $W_\text{net} = W_\text{com} + W_\text{rot}$.

(2) Although it is true that the work done by static friction is zero on an object that rolls without slipping down an incline, it is not because "the point at which friction is applied does not move."

(3) When a car accelerates on a flat road, its kinetic energy is changed because of the work done by static friction on the rolling tires.

For brevity here, I will leave my detailed explanation and defense of them as an "answer". But first a quick explanation of why I am asking this.

While writing an answer for this question today, I learned some things that I had not understood before regarding the work done by static friction on rolling objects. There are many questions (and answers) on exactly this topic, but: (i) I think that many of these answers --- which echo statements directly out of most textbooks --- are incorrect, or at least misleading, and (ii) I believe the above statements are true, but they've either been omitted/deemphasized or are contradicted in various books (and here), so I would like to hear the community comment on them.

The statements above are not radical, but follow, I think, directly from the application of Newton's Laws --- axioms for the dynamics of a particle (Newton II), and the interaction of two particles (Newton III) --- extended to composite systems forming rigid bodies. I'm also not trying to claim that they are new; these ideas appear in some form in some textbooks (and in some answers here). But, I don't think they are presented very carefully or explicitly in most texts, and, given the volume of statements that are in contradiction, I don't think they are widely appreciated.

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Commented Mar 16 at 21:40

3 Answers 3

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(1) The net work on an object that rolls without slipping can be exactly divided into a "work on the center of mass" and a "work causing rotation about the center of mass": $W_\text{net} = W_\text{com} + W_\text{rot}$.

In other words, for a macroscopic object (which should be thought of as rigid body composed of $N$ connected particles) the net work on that object is well-defined as the sum of the net works on each particle, and that sum can be decomposed into two such-described parts: $$ \begin{align} W_\text{net} &= W_\text{com} + W_\text{rot} \\ \sum_{i=1}^N W_{F_\text{net,$i$}} &= \int_{t_i}^{t_f} \vec{F}_\text{net,ext} \cdot \vec{V} \, dt + \int_{t_i}^{t_f} \tau_\text{net,$z$} \, \omega_z \, dt \end{align} $$ where $\vec{F}_\text{net,ext}$ is the sum of the external forces on all particles, $\vec{V}$ is the center-of-mass velocity, $\tau_\text{net,$z$}$ is the net torque on the object about the axis through its center of mass, and $\omega_z$ is the angular velocity of the object about its center of mass. This assumes a circular cross-section, such that the rotational axis passes through the center of mass.

I have proven this at the end of my answer to the above-linked question. The question was essentially about a claim by Kleppner and Kolenkow that work should be treated this way for an object rolling down an incline (although KK did not prove it to be true). The statement is claimed in some form in many textbooks, but rarely is it applied explicitly, and I have not seen it proven, though it it quite simple to do. (Perhaps it is too advanced for an introductory textbook but too basic for a Classical Mechanics textbook.)

I should also note that a work-KE theorem, with distinct center-of-mass and rotational parts, is an immediate consequence of the above equation: $$ W_\text{net} = \Delta K_\text{com, trans} + \Delta K_\text{rot} $$ where $K_\text{com, trans} = \frac{1}{2} M V^2$, with $M = \sum m_i$, and $K_\text{rot} = \frac{1}{2} I \omega^2$, with $I = \sum_i M_i R_i^2$ and $R_i$ the distance to the axis of $m_i$.

(2) Although it is true that the work done by static friction is zero on an object that rolls without slipping down an incline, it is not because "the point at which friction is applied does not move."

Using the above expression for the net work on the rolling object, one finds that: $$ W_\text{net} = W_\text{com} + W_\text{rot} = \left[\left(F_g \sin \theta - F_{fs}\right) \ell \right] + \left[F_{fs} \ell \right] = F_g \,\ell \sin \theta $$ where $\ell$ is the distance traveled and $\theta$ is the angle of the ramp. Therefore the work done by friction is zero because its contribution to decelerating the center-of-mass exactly cancels its contribution to accelerating the rotation about the center of mass. This was the result shown by Kleppner and Kolenkow and, interestingly, highlighted in a paper published last month in Physics Education.

The claim that "the point of contact does not move with respect to the surface, and therefore $F_{fs}$ does no work" is perhaps the most widely repeated claim about rolling friction (I remember teaching it just a few months ago), but it seems, to me, to be simply false. The "point of contact" is, I think, a red herring. The net work on the object is, as written above in statement (1), the sum of the net works on all particles. So, one could say that the particle in contact with the surface is not moving at that moment and therefore there is no work done on it. But for the rigid body, the effect of static friction carries through to all other particles through their internal forces on each other. These internal forces explicitly cancel out, via Newton III, in the calculation of $W_\text{com}$, but they remain in the calculation of $W_\text{rot}$.

(3) When a car accelerates on a flat road, its kinetic energy is changed by the work done by static friction on the rolling tires.

I don't see how this can be a controversial statement. The only horizontal external force on a car (neglecting air resistance, and assuming the tires roll without slipping) is static friction. Therefore, when it speeds up or slows down it is because of the static friction force. But, if one accepts that "static friction can do no work because the point of contact is not in motion with respect to the ground", then I don't understand how one can explain the change in kinetic energy of a car when it accelerates.

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    $\begingroup$ Regarding 3), work doesn't have to come from external forces. The work is done by the engine (or by the fuel if you want to look at it that way). $\endgroup$
    – Javier
    Commented Mar 16 at 3:30
  • $\begingroup$ For the car to accelerate then, yes, it will need an external force to do so. The engine spins the tires which exert a force on the road, which then exerts a reaction force on the car. If the car is on a frictionless surface the engine and the spinning tires will have no effect. $\endgroup$
    – Ben H
    Commented Mar 16 at 3:34
  • $\begingroup$ A person walks because of static friction, is no work done here? How does kinetic energy change then? These questions need to be answered , it should be more of an argument; explaining both points , because after all work done is $F.ds$, isn't it? $\endgroup$
    – PinkAura
    Commented Mar 16 at 5:18
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    $\begingroup$ Work is done when energy is transferred. The road is not transferring energy to the tyres so it is not doing work on the tyres. It's the engine that is transferring energy to the rotating crankshaft and thereby doing work. $\endgroup$ Commented Mar 16 at 6:17
  • $\begingroup$ work- energy theorem for extended objects seems to be a bit flawed because it was actually derived for a particle... Meanwhile the work-energy equation is a fundamental rule and it cannot be derived from Newton's Second Law (more fundamental than Newton's Law I believe)... $\endgroup$
    – PinkAura
    Commented Mar 16 at 10:17
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Before commenting on the three statements, a couple of points on work:

  1. Work is one of two possible forms of energy transfer, the other being heat (energy transfer due solely to temperature difference). When A does positive work on B, A transfers energy to B. Considering A and B as an isolated system, conservation of energy dictates that the energy gained by B equals the energy lost by A, the source of the work being performed.

  2. In order for A to transfer energy to B the force $F$ that A exerts on B must cause the displacement of a material point of B at the point of application of the force $F$. That displacement is the $d\vec s$ in the equation for work $W=\vec F\cdot d\vec s$

(1) The net work on an object that rolls without slipping can be exactly divided into a "work on the center of mass" and a "work causing rotation about the center of mass": $W_\text{net} = > W_\text{com} + W_\text{rot}$.

Better to say "that causes rolling without slipping" since work is not necessarily required to maintain rolling without slipping.

(2) Although it is true that the work done by static friction is zero on an object that rolls without slipping down an incline, it is not because "the point at which friction is applied does not move."

The work done by static friction is zero because the source of that force (the incline) does not transfer energy to the rolling object, as per by my point 1 above. The reason why the static friction does not transfer energy to the object is that the material at the point of application of the force is not displaced, as per my point 2 above.

All the work done on the object rolling down the incline is done by gravity. If the incline were frictionless all the work done by gravity goes into increasing the translational kinetic energy of the object, or $mgh=\frac{1}{2}mv_{COM}^2$. If static friction is present and enables the object to roll without slipping the work done by gravity is split between translational and rotational KE, or $mgh=\frac{1}{2}mv_{COM}^{2}+\frac{1}{2}I\omega ^2$. The role of static friction here is to enable the object to acquire rotational KE by not allowing slipping to occur, not to give it rotational KE. For more about "enabling" forces see the examples below.

(3) When a car accelerates on a flat road, its kinetic energy is changed because of the work done by static friction on the rolling tires.

Its kinetic energy comes from the conversion of chemical potential energy of the fuel into mechanical energy.

The equal and opposite static friction force applied by the road to the wheel, per Newton's 3rd law, enables the wheel to rotate without slipping. There is no energy transfer from the road to the car as there is no displacement of the material of the wheel at the point of application of the force. Thus static friction does no work on the car. You can think of the role of static friction as enabling the car to accelerate by "re-directing" the backward force the wheel applies to the road, forward.

OTHER EXAMPLES OF "ENABLING" FORCES:

  1. An ice skater applies a force to a wall and the wall applies an equal and opposite force to the skater, per Newton's 3rd law, pushing the skater away from the wall. The skater acquires kinetic energy. While pushing on the wall there is no displacement of the skater's hand at in contact with the wall. The wall does no work. The wall does not transfer energy to the skater. The skater does the work by converting chemical potential energy to mechanical work by the arm muscles. The wall enables the skater to acquire KE by redirecting the force the skater applies to the wall back onto the skater.

  2. A person climbs up a rope without slipping. The persons hands pull down on the rope. Per Newton's 3rd law, the rope applies an equal and opposite static friction up on the hands. During this pulling on the rope which elevates the climber there is no displacement of the person's hands at the point of contact with the rope. The rope does no work. The rope transfers no energy to the person. The climber does work converting chemical potential energy into mechanical energy. By preventing slipping, the static friction force of the rope enables the climber to move up by redirecting the downward force of the climber upward.

Hope this helps.

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  • $\begingroup$ While reading your answer a doubt came in my mind, hope you would like to address it, "can impulsive force do work?" (such as flicking a coin). Thanks. $\endgroup$
    – PinkAura
    Commented Mar 16 at 19:21
  • $\begingroup$ Thanks Bob D for adding to the conversation! I have to say that I don't really agree with most of your points here; many are exactly what I am trying to say is incorrect. I'd like to address some of your points individually. For your first point (1.) I only want to consider mechanical work (thermodynamic work and heat are unnecessary to consider here). So I'm talking about the work that is defined in the context of Newtonian Mechanics. This is a derived quantity based on the axioms of Newton's Laws and is therefore a statement about the work done by a force on a particle. $\endgroup$
    – Ben H
    Commented Mar 16 at 19:26
  • $\begingroup$ On your first point (2.), the law you are quoting simply does not exist in the theory. As I said in my first comment, work is derived from Newton's axioms, and $W_F = \vec{F} \cdot \Delta \vec{r}$ is defined specifically as the work done on a particle by a force. If you want to make a statement about the work done on an object (a collection of particles), then you must derive an equation from the fundamental rule for particles. This is what I have done (for a rolling rigid body) in (1). But you are claiming without proof that the same law defined for particles applies to any object $\endgroup$
    – Ben H
    Commented Mar 16 at 19:29
  • $\begingroup$ I'm not saying that you have invented a law here; these statements (or similar) appears in many textbooks and it is considered general knowledge. I'm just saying that it is not correct, and it leads to the incorrect conclusion that "a force applied to an object at a point that does not move can do no work". $\endgroup$
    – Ben H
    Commented Mar 16 at 19:33
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    $\begingroup$ @PinkAura Static friction does not "dissipate" energy. All the energy kinetic energy of the cylinder rolling down the incline without slipping is accounted for by the equal loss of gravitational potential energy the cylinder-Earth system had at the top of the incline. $\endgroup$
    – Bob D
    Commented Mar 16 at 20:17
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Consider a spherical body of mass $m$ with a radius $r$ which is rolling without slipping. Let $F$ be the force acting on the centre of mass and $f$ be the frictional force acting on the body.

Considering translational motion, we get eq1 $$ F-f=ma $$

Considering rotational motion, we get eq2 $$ f.r=I\alpha $$ where I is its moment of inertia.

now, lets look at the friction required for pure rolling $$f=\frac{I\alpha}{r}=\frac{mk^2\alpha}{r}$$ (where k is the radius of gyration)

since $\alpha=\frac{a}{r}$ we get $$f=\frac{mk^2a}{r^2}=ma \frac{k^2}{r^2}$$

substituting the value of ma in eq1 we see that $$F=f(\frac{r^2}{k^2}+1)$$ which means that for pure rolling the minimum amount of static friction should be $$f=\frac{F}{(1+\frac{r^2}{k^2})}$$

Let us look at the energies that the body possesses in pure rolling motion.

The final translational kinetic energy, (supposing $KE_it$ as zero)

$$ KE_ft = \frac{1}{2} mv^2$$

The final rotational kinetic energy would be (taking $KE_ir$ as zero) $$ KE_fr = \frac{1}{2}I\omega^2$$

So the total final $KE$ would be $$ KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Even though the body possesses a rotational KE, friction is not doing any work here. Friction prevents relative motion between the surfaces in contact. In pure rolling, the frictional force balances the force that is pushing the body forward at that point and so the point does not have any relative motion with the surface in contact. The point coming in contact with the surface keeps on changing and the frictional force is only helping to sustain pure rolling but it does no work in doing so. It helps distribute the total energy into rotational and translational KE.

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Commented Mar 16 at 21:40

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