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Angular Velocity of a car

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data
    "A car is traveling at 100 km/h and the tire of the car has a radius of 36cm. Find the number of revolutions per second."

    3. The attempt at a solution


    100 km/h * (10,000,000 cm/km) * (1h/3600 secs) = 2777.77777778 cm/s is the speed of the car.

    Θ = a/r
    Θ = (2777.78) / (36)
    Θ = 77.16

    To find number of revolutions, we must divide by 2pi.

    77.16/2pi = 12.28 revolutions/sec. That is the correct answer.

    a) I did not get that on the quiz because I do not understand the mechanics behind the operation. Can anyone walk me through each calculation and state why that step is done?
    b) And why is 2777.78 cm/s equal to the arc length? Isn't arc length a distance? I thought 2778.78 cm/s was a velocity measurement.

    Thank you.
     
  2. jcsd
  3. Nov 15, 2009 #2
    Nevermind, I thought of it quite a bit and I think I've got the concept.

    But just to make sure:

    First 100 km/h is converted into 2777.78 cm/s. That is the speed at which the tire travels. So basically, every second, it moves 2777.78 cm. This would make it the arc length.

    Now we have to look at it in a perspective of ONE SECOND intervals.

    So Θ = a/r
    Θ = 77.16 is the value of the angle in radians but for ONLY ONE SECOND.

    Then we must find out how many times it rotates in one second or how many revolutions it has so we divide that number by 2pi.

    Is my understanding correct?
     
    Last edited: Nov 15, 2009
  4. Nov 15, 2009 #3

    LCKurtz

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    Here's how I would write it using the "dimensional unit" method:

    [tex]\frac {x\ rev}{1 sec}=\frac {100\ km}{1\ hr}\times\frac{1\ hr}{3600\ sec}\times\frac {10^5\ cm}{1\ km}\times\frac {1\ rev}{2\pi 36\ cm}[/tex]

    Each conversion fraction is one expressed in different units and the unwanted units cancel out.
     
  5. Nov 15, 2009 #4
    So basically you're just converting the units for speed and then dividing by the circumference of the tire, correct?
     
    Last edited: Nov 15, 2009
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