Integrating Friction: Net Force and Torque

1. The Classic Case: Frictional Torque on a Rotating Disc

Let's begin with the most direct application of this concept: calculating the torque generated by friction between two rotating surfaces. This is the principle behind clutches, disc brakes, and collar bearings.

Imagine a flat, circular disc being pressed against a stationary surface with a total normal force $F_N$ and rotating. Kinetic friction will oppose this motion, creating a resistive torque.

The key challenge is that while the frictional force per unit area might be uniform, the torque it generates is not. A frictional force element $d{F}_f$ acting at a radius $r$ from the center contributes a torque $d\tau =rd{F}_f$. Elements farther from the center have a larger lever arm and thus contribute more to the total torque. To find the total torque, we must integrate these contributions over the entire contact area.

The following resource provides a clear, step-by-step derivation for this exact scenario. It starts by considering a hollow circular contact area, which is the general case.

As you've read, the fundamental steps are:

1. Pressure: Assume the total normal force $F_{load}$ is spread uniformly over the contact area $A$, giving a constant pressure $p=F_{load}/A$.
2. Elemental Area: Consider an infinitesimally thin ring (an annulus) at radius $r$ with thickness $dr$. Its area is $dA=2\pi rdr$.
3. Elemental Normal Force: The normal force on this ring is $dN=pdA=p(2\pi rdr)$.
4. Elemental Frictional Force: The kinetic friction on this ring is $d{F}_f={\mu }_{k}dN={\mu }_{k}p(2\pi rdr)$.
5. Elemental Torque: The torque from this ring's friction is $d\tau =rd{F}_f=r\left[{\mu }_{k}p(2\pi rdr)\right]=2\pi {\mu }_{k}p{r}^{2}dr$.
6. Total Torque: Integrate from the inner radius $R_i$ to the outer radius $R_o$. $$\tau ={\int }_{R_i}^{R_o}2\pi {\mu }_{k}p{r}^{2}dr$$

Substituting $p=F_{load}/(\pi (R_o^2-R_i^2))$ and integrating gives the final result shown in the reading:

Special Case: Solid Disc

A common scenario involves a solid disc, like a simple end bearing or the sander shown in the image. This is a special case of our general formula where the inner radius $R_i=0$.

For a solid disc of radius $R$, we set $R_i=0$ and $R_o=R$. The formula simplifies nicely:

This result is derived from scratch in the following short video, which provides an excellent visual recap of the integration process for the solid disc.

2. A More Complex Case: Friction's Balancing Act

The disc friction problem is relatively straightforward because the direction of the kinetic friction force is always tangential, opposing the rotation. Now, let's consider a more subtle problem where the direction of the static friction force can vary with position to maintain equilibrium. This kind of thinking is essential for solving olympiad-level problems.

Consider a thin, uniform bar of mass $m$ and length $L$ lying on a floor with a coefficient of static friction ${\mu }_s$. What is the minimum horizontal force required to make the bar start moving?

A naive guess might be that you just need to overcome the maximum total static friction, $F_{f,max}={\mu }_s{N}_{total}={\mu }_{s}mg$. However, the answer is significantly less than this. Why?

The reason is torque. When you apply a force $F$ at one end of the bar, you're not only trying to translate it but also to rotate it. To prevent rotation, the static friction forces along the bar must arrange themselves to create a counter-torque. This means that not all friction can be pointing backward to oppose your push; some of it must point forward to create the necessary balancing torque.

The following problem from a physics olympiad handout explores this exact idea.

This problem reveals a deeper level of analysis. At the moment the bar is about to slip, the static friction is at its maximum value everywhere, but its direction is position-dependent.

• Force Balance: The applied force $F$ must equal the net friction force, which is the integral of the backward-pointing friction minus the integral of the forward-pointing friction.
• Torque Balance: The torque from the applied force must be balanced by the torque from the distributed friction forces.

As the solution shows, by solving the force and torque balance equations (which are themselves integrals over the distributed friction), you find that the minimum applied force is $F=(\sqrt{2}-1){\mu }_{s}mg$, which is less than half the naive guess! This demonstrates that correctly formulating integrals for distributed forces is key to solving such advanced problems.

3. Generalizing the Skill: An Irodov-style Problem

The technique of integrating a position-dependent force to find its net effect is not limited to friction. It's a general tool for rigid body dynamics. Let's look at an example similar to a problem from I.E. Irodov, which involves integrating a centrifugal force.

Consider a rod of length $L$ pinned at one end to the center of a rotating disc, with its other end touching the disc, as shown in the video. As the disc rotates with angular velocity $\omega$, the rod rotates with it. In the rotating frame of reference, every element of the rod experiences an outward centrifugal force. This force depends on the element's distance from the axis of rotation, so it is position-dependent.

To find the total centrifugal force and its torque about the pivot, one must integrate along the length of the rod.

This example powerfully illustrates that the skill you've developed today is a general-purpose tool. The process of identifying an elemental force dF that depends on position and then integrating dF to get the total force and dτ = r_perp dF to get the total torque is fundamental to advanced mechanics.

Conclusion

In this lesson, we have extended your calculus skills to handle distributed forces in rigid body dynamics. You are now equipped to move beyond simple point forces and analyze more realistic scenarios where forces like friction are spread over a surface.

Key takeaways from today include:

• The Integration Strategy: To find the net force or torque from a distributed load, you must integrate the infinitesimal contributions from each part of the body. The general process is to define an element, find the force on it, find the torque from it, and sum (integrate) over the whole body.
• Disc Friction: For a rotating disc, the frictional torque is found by integrating $d\tau =rd{F}_f$ over the contact area, leading to the formula $\tau =\frac{2}{3}{\mu }_k{F}_{load}R$ for a solid disc.
• Force and Torque Balance: In more complex static problems, the distribution of friction is determined by the need to balance both external forces and external torques simultaneously. This can lead to non-intuitive friction patterns, as seen in the bar problem.
• General Applicability: The method of integrating position-dependent forces is a versatile tool, applicable not just to friction but to other effects like centrifugal or hydrostatic forces, which you will encounter later.

In our next lesson, we will continue our exploration of friction by tackling problems involving rolling with slipping. We'll see how kinetic friction provides the force and torque necessary to change a body's linear and angular velocities until the condition for pure rolling is achieved.