Clarification on the displacement in the definition of Work

I'd like to ask a question about work. The definition of work gives us a way to calculate the work done by a force along a path but in practice it's not always clear what path to take in consideration. Moreover, this fact that work is defined along a path is not taken in consideration when applying the conservation of energy. Could someone clarify this points?

I'd like to give an example to make my position clearer. There's a ball rolling of pure rolling down a slope (v=wR) with friction. I've been told that in this case friction doesn't make work because although the ball (the object on which friction is applied) is moving, the point of contact, where friction is applied, is not moving relative to the slope. This makes me think that I have a problem understanding the definition of work :)

• The point on the ball where friction might act doesn’t move in the direction of the friction force (see cycloid). That’s all that matters; how other parts of the ball move is irrelevant. Is this what you’re asking about? Dec 13, 2021 at 21:26

The definition of work gives us a way to calculate the work done by a force along a path but in practice it's not always clear what path to take in consideration.

The line integral $$\int\mathbf F\cdot \text d\mathbf x$$ always follows the point of application of the force.

Moreover, this fact that work is defined along a path is not taken in consideration when applying the conservation of energy.

This is because with conservation of energy we are usually looking at $$\Delta K$$ and $$\Delta U$$. The former can be determined just through velocities, and the latter is path independent. Since $$W_\text{ext}=\Delta K+\Delta U$$, or stated a different way $$W_\text{net}=\Delta K$$, we don't necessarily need to think about line integrals to look at work done in many contexts. This is actually one of the benefits of considering energy.

I'd like to give an example to make my position clearer. There's a ball rolling of pure rolling down a slope (v=wR) with friction. I've been told that in this case friction doesn't make work because although the ball (the object on which friction is applied) is moving, the point of contact, where friction is applied, is not moving relative to the slope.

Yes, this is correct. The point in contact with the slope is always instantaneously at rest relative to the slope with rolling without slipping. Hence no work is done in the rest frame of the slope.

Even if you wanted to treat the friction force as doing work, you will find that... it does no work. If the ball of radius $$R$$ rolls a distance $$x$$ down the incline, then the friction force $$F$$ does work $$-Fx$$ with the translational displacement and $$FR\Delta\theta=FR\cdot(x/R)=Fx$$ with the rotational displacement. This still gives a total work of $$0$$.

• First, thanks for the answer. Second, I still miss while the work is zero. If the integral follows the point of application the work is not zero because surely the part of the rolling object touching the slope is at rest but the point of application is not, the point of application is just a point moving through space down the slope at speed v... Where am I wrong? Dec 14, 2021 at 13:07
• @strategaD Yes, the geometric point moves down the incline, but this is not the same thing as the force acting over a distance. In any case, see my edit: if you try to calculate it out as if it was moving with the geometric point you still get $0$ work. Dec 14, 2021 at 14:18

Moreover, this fact that work is defined along a path is not taken in consideration when applying the conservation of energy. Could someone clarify this points?

It's not quite clear what you are asking, but it seems you are asking about the difference between work done by a conservative force, which is path independent, and work done by a non conservative force, like friction, which is path dependent. So my answer is based on that assumption. If that's not correct, let me know and I'll either revise or withdraw it.

Mechanical energy, i.e., the sum of potential energy (PE) and kinetic energy (KE) at the macroscopic level, is conserved when the work is done by a conservative force, such as gravity, electrostatic, and elastic forces, because such work is independent of the path.

However, overall conservation of energy still applies to work done by non-conservative forces, such as kinetic friction, when you include the change in kinetic energy at the microscopic level, i.e., atomic and molecular kinetic energy. The negative work done by kinetic friction converts macroscopic KE (loss of KE of moving object) into microscopic KE (increase in molecular KE, i.e., increase in internal KE of the material at the rubbing surfaces).

In the case of a ball undergoing pure rolling down a slope with friction, as you are aware, static friction does no work. But it enables the ball to roll without slipping so that the work done by the component of the gravitational force acting down the block, which is a conservative force, converts gravitational PE into a combination of translational plus rotational KE.

Without friction, the ball would slide down the slope without rolling so that the GPE will be converted into translational KE only. It then sort of becomes the equivalent of a free falling ball in a reduced gravitational field of $$g\sin\theta$$ where $$\theta$$ is the angle of the slope.

Hope this helps.

The effect of the rolling friction is to decrease the net force: $$F_{net} = mgsin(\theta) - F_{fric}$$. $$F_{net} = ma \implies mgsin(\theta) - F_{fric} = m\frac{dv}{dt}$$

When there is no slip, the friction force can be expressed in terms of the momentum of inertia and angular acceleration $$F_{fric} R = I\frac{d\omega}{dt} \implies F_{fric} = \frac{I}{R}\frac{d\omega}{dt}$$

As $$v = \omega R$$, and multiplying both sides by dx $$dw = mgsin(\theta)dx = m\frac{dv}{dt}dx + \frac{I}{R^2}\frac{dv}{dt}dx = (m + \frac{I}{R^2})\frac{dx}{dt}dv = (m + \frac{I}{R^2})vdv$$

$$vdv = \frac{1}{2}d(v^2) \implies dw = \frac{1}{2}(m + \frac{I}{R^2})d(v^2)$$

The work of the gravity force results in an increase of the translational kinetic energy (first term) and rotational kinetic energy (second term). If there was some slip, only part of the torque would result in angular acceleration $$F_{fric} R - \Delta = I\frac{d\omega}{dt}$$ In this case, the final expression $$dw$$ would be: $$mgsin(\theta)dx - \Delta dx$$, the latter being the work of the friction force. Without slip, all the work of the gravitational force results in kinetic energy.