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Centripetal force question -- car speed on an offramp

  1. Nov 15, 2015 #1
    1. The problem statement, all variables and given/known data
    If a car doubles its speed on a circular off-ramp how much more centripetal force will be required for it to make the turn


    2. The attempt at a solution
    I think that half of the centripetal force will be required for it to make the turn due to the fact that faster the car goes, the lower the centripetal force. Is this correct? Thanks in advance!
     
    Last edited by a moderator: Nov 15, 2015
  2. jcsd
  3. Nov 15, 2015 #2

    berkeman

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    Welcome to the PF.

    Please write out the equations involved for centripetal acceleration versus linear and angular speed. Base your answer on those equations, and give a quantitative answer (how much more or less...?). :smile:
     
  4. Nov 15, 2015 #3

    SammyS

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    Hello Dorthea. Welcome to PF.

    Please explain your answer further.

    How is centripetal acceleration related to speed and the radius of the circular off-ramp?
     
  5. Nov 15, 2015 #4
    Centripetal acceleration is directly related to the speed and inversely related to the radius of the off ramp
     
  6. Nov 15, 2015 #5

    SammyS

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    Not quite right .

    Do you have a formula for this?
     
  7. Nov 15, 2015 #6
    Is the formula Ac=v2/r?
     
  8. Nov 15, 2015 #7

    SammyS

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    That's right.

    So the centripetal force is directly proportional to the square of the speed and inversely proportional to the radius.

    Now, can you answer the original question in post #1?
     
  9. Nov 15, 2015 #8
    This is an application question for a lab, so we aren't given actual values and I'm just going to use hypothetical values.
    Say the original speed is 10 m/s, which would be doubled to equal 20 m/s and the radius is 10m.

    The centripetal acceleration for the first speed would be 102/10 which equals 10m/s2
    The centripetal acceleration for the second speed would be 202/10 which equals 40m/s2

    Thus, 4 times the centripetal force will be required for it to make the turn.
     
  10. Nov 15, 2015 #9

    SammyS

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    Right.

    Of course you can come to the same conclusion using only the information in post #1, without plugging in specific numbers.
     
  11. Nov 15, 2015 #10
    Ohh ok. Thank you so much for helping me understand this question!!!
     
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