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An automobile with 0.260 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear?

Question by OpenStax is licensed under CC BY 4.0.

$5 \times 10^7 \textrm{ rev}$

Solution Video

OpenStax College Physics Solution, Chapter 6, Problem 3 (Problems & Exercises) (1:15)

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Video Transcript
This is College Physics Answers with Shaun Dychko. The distance these tires cover which is eighty thousand kilometers, is the arc length along the perimeter of the tire. So I call it delta s and that's converted into meters by multiplying by 1000 meters per kilometer. It's eight times ten to the seven meters. We keep only one significant figure here because it's not really clear how many significant figures we should have so I guess we can assume it's one. The radius of the tire is 0.260 meters and we know that the angle over which it rotates is the arc length divided by the tire's radius. So that's eight times ten to the seven meters divided by 0.260 meters which gives us 3.0769 times ten to the eight radians. Now the question asks us for the number of rotations of the tire. So, there is one rotation or revolution in every two pi radians and so this is a conversion factor we can use to convert from radians into revolutions. So that's 3.0769 times ten to the eight radians multiplied by one revolution for every two pi radians. The radians cancel leaving us with units of revolutions and that is five times ten to the seven revolutions.


Submitted by tahir.va3 on Wed, 10/21/2020 - 12:35

80000000/.260 is not equal to 307690000. It is equal to 307692307.7

Submitted by ShaunDychko on Fri, 10/30/2020 - 16:38

Hi tahir.va3, thank you for the comment. You're quite right, but since there is only going to be one significant figure in the final answer, there's no need to carry too many significant figures in the intermediate calculations. While I often mention "don't round until the final answer" in order to avoid intermediate rounding error, it seems safe enough to carry 4 additional significant figures in the intermediate calculation since surely it won't make a different to the final answer with only one sig. fig.
All the best,

In reply to by tahir.va3

Submitted by AndrewFrancis0531 on Sat, 03/20/2021 - 19:29

It's also much simpler to simply find the circumference of the tire and then divide the distance by that. I understand it's for the sake of teaching physics but it seems silly to give a question like this. Without this video I would have not known to use the arc length.

Submitted by Ley on Mon, 07/25/2022 - 07:19

Hello, I keep getting 5*10^8 and not 5*10^7.