How Do You Calculate Tire Rotation Frequency and Edge Speeds?

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Homework Statement


A car tire is 65.0 cm in diameter. The car is traveling at a speed of 20.0 m/s .

What is the tire's rotation frequency, in rpm?

What is the speed of a point at the top edge of the tire?


What is the speed of a point at the bottom edge of the tire?

Homework Equations



i've looked through my book and I'm not even sure what equations to use here.

The Attempt at a Solution




i tried taking 65(20^2) to get rpm and came up with 26000
 
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What does rpm stand for? (I know what it stands for, I want to make sure you know)
 
revolutions per minute? correct?
 
Correct. Now they want to know what the tires rotation frequency is. How does frequency relate to time? Is time in seconds or minutes or hours?
 
time is in minutes... seconds * 60
 
Are you sure its seconds * 60 ?

So if it takes me 10 seconds to eat a cheeseburger, you are telling me it takes 10s*60 = 600 minutes to eat that cheeseburger?
 
i mean... seconds /60!
 
so 20/60* .65?
 
I think you are confused with the notation here.
1 min = 60 seconds. You were right the first time. But you do not take the amount of seconds you have and multiply by 60 to get how many minutes we have.

Ok now let's imagine we take our tire and stretch it out into a straight line. How long would this straight line be? Is it it cm's or m's?

Also, rpm does mean revolutions per minute. But explain to me in words what that means. If we take a red marker and dotted the bottom of the tire, after 1 revolution, where is this red dot now?
 
  • #10
the diameter of the tire is 65 cm. so if it was stretched out then it would be longer than 65 cm.
 
  • #11
the red dot is at the bottom again. it is the amount of time that it takes for a full rotation, and how many of these rotations are present within the span of one minute.
 
  • #12
After you stretch it out, it would be longer than 65cm's. But how much longer? You should be able to tell me that with just knowing the diameter. Remember the tire is in the shape of a circle. So really, we want to know the perimeter of the circle. There is a special name for perimeters of circles, and there is also a special formula for it.

And your are correct, that is what rpm is.
 
  • #13
2pi(r)... = 2(3.14)(65) = 408 cm
 
  • #14
wait, i mean... r would be half of 65 so 408/2 = 204 cm
 
  • #15
Almost. r in that formula stands for "radius". A radius of a circle is define to be 1/2 the diameter. That is, 2r = d.

Yup. Ok now what do we want? We want to know how long it takes to go from the start of this length to the end of this length. That would mean it's completed one full revolution right? So how long does it take?
 
  • #16
so 204 cm... and it goes 20 m/s, so every second it would go 20 m. or 2000 cm. so 2000/204*60 = 588 rpm?
 
  • #17
Keep your units! 2000 what? 204 what? 60 what?

Once you put in the units and multiply through, you should be able to easily see if your answer is in the required rpm units.
 
  • #18
ok, so i would leave 2000 in meters. so 20m/.24m*60s = 5000 rps hmmm...
 
  • #19
I meant keep your units in your equations/numbers. It was perfectly fine for you to leave it in cm's, as they would cancel out afterwards.

We have rps, but we want rpm. What to do?

You might want to check your conversion from cm's to m's.

And I suggest you do things step by step. Write each step as one line. As it would appear you are losing track of what we are calculating.
 
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  • #20
2.04 m... so 20/2.04 = 9.8 rps...to get revolutions per minute we would multiply by 60 seconds again. 20/2.04*60*60, correct?
 
  • #21
Keep your units in your equations! I cannot stress that enough.
Let's think about what we did in each step.

First we calculated how "long" the circle was.
You told me it's 2pi(r)... = 2(3.14)(65)/2 = 204 cm

Then we must calculate how much time it takes to go from start of the length to the end, which is the same as saying how much time it takes to make 1 revolution.
You told me it's: 20m/2.04m*60s = 588.23s.

Which is not exactly right.

If we move at 20m/s and only need to travel 2.04 m's. Does it make sense that it would take more than 1 second to travel 2.04 m's?
 
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