To an order of magnitude, through how many revolutions will it turn?

AI Thread Summary
The discussion focuses on calculating the number of revolutions an automobile tire makes over its lifespan of 35,000 miles. The initial calculations presented were incorrect, particularly in the final step where the number of revolutions was miscalculated. Participants emphasize the need to express the answer in terms of order of magnitude, which simplifies the result to a single significant digit. A method for estimating the answer is provided, illustrating how to convert miles to feet and perform the division for a rough estimate. The final rough estimate suggests the tire makes approximately 20 million revolutions, accurate to within an order of magnitude.
chocolatelover
Messages
238
Reaction score
0
Hi everyone,

Homework Statement



An automobile tire is rated to last for 35,000 miles. To an order of magnitude, through how many revolutions will it turn?


Homework Equations


2pir


The Attempt at a Solution



d=3f=.9144m

r=.9144/2
=.4572

n(2pi)r=35,000miles

(35,000miles)=56327040meters

n(2pi)(.4572meters)=56327040meters

409867.3=
409868

Does that look right?

Thank you very much
 
Physics news on Phys.org
I agree very much that you will need to do:

35,000 miles = (2*pi*r)n

But your last step from
n*2pi*.4572 = 56327040m
n = 409868 ? appears wrong, simply from putting those numbers into the calculator.
 
chocolatelover said:
An automobile tire is rated to last for 35,000 miles. To an order of magnitude, through how many revolutions will it turn?
...
409867.3=
409868

Does that look right?
No. First, you did something wrong. Second, your incorrect answer has too many digits. The question asks for the answer to an order of magnitude.

Work out the answer correctly to however many digits you want. After that, I will (or someone else) will show you how to calculate it to do these order of magnitude calculations.
 
Last edited:
Thank you very much

n*2pi*.4572 = 56327040m
n(2.872672322)=56327040m
n=1960788.899

Does that look right?

Could someone please show me how to convert it to the order of magintude?

Would that just be 1.96 X 10^6
 
The term order of magnitude means 1, 10, 100, etc. In other words, less than one significant digit. You are still using three significant digits.

Now that you have the correct numerical answer, here is an easy way to compute a rough estimate of the answer

N=\frac{l}{\pi d} = \frac{35000\,\text{miles}}{3\pi\,\text{feet}}

The first thing to recognize is that 3\pi feet is about 10 feet. After converting 35,000 miles to feet all we have to do is divide by 10. Multiplying 35,000 by 5,280 is a pain in the rear. Multiplying 30,000 by 6,000 is a lot easier, and will be close to correct:

(3*10^4)*(6*10^3) = 18*10^7 \approx 2*10^8

Dividing by 10 gives 20 million, which is correct to within an order of magnitude. (Actually, its even better than that; this is correct to one decimal place.)
 
Thank you very much

Regards
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top