Skip to 0 minutes and 9 seconds Your design task for this week is to evaluate the power requirements of an electric car. Rolling friction is an important part of this, particularly at low speeds. Analysing the experiment will help you understand the concepts that you will need. And you’ll see new examples of free-body diagrams and friction around a bollard. To find rolling friction from our test, we’ll need to start with an FBD of the tractor. We’ll assume that the table is level and we can neglect air resistance. We’ll represent rolling resistance as forces opposing motion. We’ll show them at the contact between the wheel and the surface. We are interested in the total rolling resistance, which we’ll call F R.

Skip to 0 minutes and 59 seconds We’re not interested in the resistance at individual wheels. Actually, rolling resistance can be complicated. Sometimes it is just contact friction, where the wheels touch the road. Other times, it includes bearing friction at the axles. We will be including bearing friction. Because the speed v is steady, we can use some of the forces along the table equals 0. This gives us T minus F R equals 0. And as you might have guessed, the rolling resistance is equal to T, the force with which we are pulling the tractor. Now, T is generated by a falling weight, mg. But we’ll need to determine the effect of friction where the string goes over the edge of the table.

Skip to 1 minute and 56 seconds We’ll start with an FBD of the weight pan. Speed is constant again. So from the sum of the forces equals 0 vertically, we get P minus mg equals 0, where P is the force between the string and the weight, and mg is the weight of the falling object. Now we’ll find the effect of friction where the string goes over the edge of the table. Remember the test with the two weights connected by string that went over the table top? We can use this to estimate the ratio mg divided by T. This diagram shows the table and the string going over it. We’ll call the heavy load on the string P1 and the lighter load on the string P2.

Skip to 2 minutes and 48 seconds And we’ll assume the friction is the same at both ends of the table. We can use what we learned in week 5 about rope around a bollard. From that analysis, the ratio of tensions across the rounded edge of the table is T A over T B equals e to mu theta, where T A is the higher tension. We don’t know the coefficient of friction mu, but we will assume that it is the same for each end. The angle of wrapped theta is the same at each end, too. So we can replace e to the mu theta with k. It’s the same both ends.

Skip to 3 minutes and 37 seconds Calling the tension in the string on the table T, we can apply our rope around a bollard expression for each end of the table. On the right-hand end we have P1 over T equals e to the mu theta. P1 is great than T. Or P1 over T is k. So P1 equals k times T. Similarly, for the left-hand end, T over P2 equals e to the mu theta. This time, T is greater than P. So we get T over P2 equals k. And T equals k times P2. We can eliminate T from the two equations to get P1 over P2 equals k squared. We measured P1 and P2, so we can find k squared, and hence k.

Skip to 4 minutes and 36 seconds Then, given P, we can use k to find T. Neat, eh? We’ll put numbers in. Our measured data was that P1 was 68 paperclips, P2 was 27 paperclips. So P1 over P2 equals 2 and 1/2. So k squared equals 2 and 1/2, and k equals 1.6. From this, we can get mu equals 0.3, which is reasonable. Now back to finding rolling friction. T is smaller than mg, so T equals mg over k. And we can substitute mg over k for T in the equilibrium equation to get mg over k minus F R equals 0, or F R equals mg over k. And we can now put numbers in.

Skip to 5 minutes and 39 seconds Our table top was not level, so we did tests uphill and downhill and took the average. Uphill P was 8 paperclips. Downhill P was 5 paperclips. So the average was P equal 6 and 1/2 paperclips. So the average rolling resistance is 6.5 divided by k, which is 6.5 divided by 1.6 equals 4.06 clips. One clip weighs 12 and 1/4 millinewtons, from weighing 20 clip using kitchen scales. So F R equals 49.7 millinewtons, that is thousands of a newton. Engineers use a coefficient of rolling resistance to describe rolling resistance. That way, they can compare data between vehicles. We’ll try it. Coefficient of rolling resistance is defined as C R equals F R divided by mg.

Skip to 6 minutes and 47 seconds The mass of the tractor when we weighed it was 275 grams, which gives us C R equals 0.018. Typical C R from web search gives values for cars of 0.007 to 0.015, and bicycles 0.002 to 0.01. Note that sometimes the coefficient of friction includes bearing friction, like our test. Sometimes it’s just the energy lost in the contact with the road. Why not try a web search yourself? Do you remember the test where we let the weight hit the floor and the tractor kept rolling on? We’ll interpret that rolling test qualitatively later on. Now let’s look at air drag in more detail.

# Analysis: Models of rolling resistance

We’ll use the simplest model of rolling resistance, based on our dry friction approach.

Although we used this simple model, making the experiment work was not simple at all. Many subtleties arose, such as using the rope around a bollard concept and dealing with a sloping table.

**Data from the experiment are given in the Downloads section below.**

### Talking points

- Share your thoughts on how we dealt with the complications we encountered?