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Find the acceleration of a race car traveling up a curve

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A 500 kg race car starts at rest on a curve with a radius of 50 m. If it goes 20 degrees in 4 seconds with a constant angular acceleration, what is its total acceleration at the end of this time?

    m=500kg
    r = 50 m
    angle = pi/9 radians
    time = 4 seconds

    3. The attempt at a solution

    angular velocity=∆radians/∆time
    angular accel=angular velocity/time
    tangential accel=radius*angular accel
    centripetal accel=angular velocity^2*r

    angular velocity=(pi/9)/4 seconds = 0.0873 radians/second
    angular accel=(0.0873 radians/second)/4 seconds = 0.0218 radians/second^2
    tangential accel=(50m)(0.0218 radians/second^2)=1.09 m/s^2
    centripetal accel=(0.0873 radians/second)^2(50 m) = 1.52 X 10^-4 m/s^2

    TOTAL ACCELERATION = tangential acceleration + centripetal acceleration = 1.09 m/s^2 + 1.52 X 10^-4 m/s^2 = 1.09 m/s^2

    I have a feeling I didn't do this right...
     
  2. jcsd
  3. Nov 20, 2008 #2

    PhanthomJay

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    Almost perfect. The tangential and centripetal accelerations are correct, But they are vectors acting at right angles to each other, so you can't just add them algebraically. Instead, you must______?

    Edit: Except you also have a math error in your centripetal acceleration equation. Not quite almost perfect.
     
    Last edited: Nov 20, 2008
  4. Nov 20, 2008 #3
    sqrt(1.09^2 + 0.000152^2) = 1.09 m/s^2

    That's the same acceleration I got before. Is it correct?
     
  5. Nov 20, 2008 #4

    PhanthomJay

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    correct your math error for centripetal acceleration, then it should be correct, unless i, too, have made a math error. It comes out to 0.38m/s^2, and I get the total acceleration equal to 1.15m/s^2.
     
  6. Nov 20, 2008 #5
    The tangential acceleration is not quite correct. Imagine a car that started from rest and travelled 20 metres in 4 seconds. How would you calculate its linear acceleration? The angular acceleration must be calculated the same way.
     
  7. Nov 20, 2008 #6

    PhanthomJay

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    It's 20 degrees in 4 seconds, not 20 meters.
     
  8. Nov 20, 2008 #7
    Yes, I realize that. I was trying to draw an analogy, admittedly I wasn't too successful at that.

    Let me try again.

    Angular velocity = {Change in angle}/{change in time} is not correct for uniformly accelerated angular motion. The way you work this out is the same as the way you work with uniformly accelerated linear motion.
     
  9. Nov 21, 2008 #8

    PhanthomJay

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    Oh, yes, sorry, you are correct. What has been calculated is the average angular velocity. What is needed is the instantaneous angular velocity. Good catch, thanks. That'll change all the numbers.
     
  10. Jan 15, 2009 #9
    What is the correct formula? I have this on a test review, and have no idea what I am doing.
     
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