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Banked curve question

  1. Mar 26, 2009 #1
    Determine the maximum velocity at which a car can travel if the banking angle is 15º for the following conditions
    1) It is not to slide radially outwards.

    2) It is not to roll over.

    given:
    coefficient of friction as 0.4
    r = 60m
    width of car is 2m and height from bank to centre of car is 0.5m

    2. Relevant equations

    Looking at some work i think the equation that i need to use is
    V = square root (( tan feta + coefficent of friction/ 1-coefficent of friction tan feta) x r x g)

    3. The attempt at a solution
    my attempt at this solution gave me an answer of 22.9m/s so 23m/s
    i wanted to check if this was correct and im confused on why i have been given the lenght of the car and its distance from bank to centre of car.
     
    Last edited: Mar 26, 2009
  2. jcsd
  3. Mar 26, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi sricho! Welcome to PF! :smile:
    because it will roll over if the net torque about the inside wheel is towards the centre …

    for that, you need to know where the centre of mass is :wink:

    hmm … are you sure 2m isn't the width of the car? :confused:
     
  4. Mar 26, 2009 #3
    yes sorry that is the width of the car. so i need to work out the centre of mass. rite that makes sense with the given lenghs.
    was the equation for the velocity correct?
    pretty new to this so any help would be ausome
    thanks
     
  5. Mar 26, 2009 #4

    tiny-tim

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    Hi sricho! :smile:

    Sorry, I missed your reply. :redface:
    (have a theta: θ and a mu: µ :wink:)

    Difficult to tell, since you haven't shown how you got it …

    and do remember, the normal force won't be the usual mgcosθ, because of the centripetal acceleration :wink:
     
  6. Mar 26, 2009 #5
    well i got my answer by using the equation for velocity and subsituting the values i have.
    so
    sqr(tan15 degree + 0.4 / 1 - 0.4 tan 15 degrees) x 60 x 9.8

    but this doesnt use the normall reaction force and i dont think this answers the question
     
  7. Mar 26, 2009 #6
    Ncos θ - µNsin θ = mg
    N(cos θ - µsinθ) = mg
    N = mg / cos θ - µsinθ

    This is the equation i have for the normall reactiong force but you need the mass of the object where its not given.
    so im stuck??????
     
  8. Mar 27, 2009 #7

    tiny-tim

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    No, you don't need to know the mass: it always cancels.

    And your equations haven't taken the centripetal acceleration into account.
     
  9. Mar 31, 2009 #8
    sorry had a busy weekend
    so how do i use that acceleration in the equation i have.
    and what is the equation for the acceleration?
    thanks
    sam
     
  10. Mar 31, 2009 #9
    is this the equation i need ?
    Ac = - w^2 r
    Ac = - v^2 / r

    where w is omega

    how do i add this into the velocity equation???
     
  11. Mar 31, 2009 #10

    tiny-tim

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    you should be able to look that up :frown:

    centripetal acceleration is v2/r (= ω2r)​
     
  12. Mar 31, 2009 #11
    well im just all confused now, is there any chance you could show me how its done , i would be very happy if you could. it will help loads with the other questions i have to do.
     
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