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Math Help - Banked Surface Problem

  1. #1
    Super Member craig's Avatar
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    Banked Surface Problem

    Let g = 9.8 ms^{-1}

    A car of mass m = 1000kg travels on a circular circuit of radius 200m, (the circuit is rough). The circuit is banked at an angle of \theta = \frac{\pi}{6} to the horizontal to try to reduce the chance of the car skidding outwards.

    1. Find the coefficient of friction between the car and the circuit if the car starts to slip inwards when it's travelling at a speed of 80 km h^{-1}.

    2. Find the magnitude of the normal reaction force on the car when it's travelling at 80 km h^{-1}.
    I'm not too good on circular motion questions so please bare with me , here's my attempt.

    Sort a few things out first:

    80 km h^{-1} = \frac{200}{9}ms^{-1}
    In case we need it later, angular speed, \omega = \frac{v}{r} = \frac{200}{9 \times 200} = \frac{1}{9}

    At 80 km h^{-1} the car starts to slip inwards, so this means that the friction is acting up the slope, and that the friction is also equal to \mu N where N is the normal reaction of the slope on the car?

    Resolving vertically I get N = mg\cos{\theta}

    Resolving horizontally I get F = mg\sin{\theta}

    We can rewrite the second equation as \mu N = mg\sin{\theta}, and substituting in our first equation we get \mu mg\cos{\theta} = mg\sin{\theta}, therefore \mu = \tan{\theta} = \frac{\sqrt{3}}{3}?

    I have a feeling that Newton's F = ma needs to be used but I'm not sure where?

    Not sure how to approach the second question either, this has me a bit stumped.

    Thanks in advance for the help
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    Quote Originally Posted by craig View Post
    I'm not too good on circular motion questions so please bare with me , here's my attempt.

    Sort a few things out first:

    80 km h^{-1} = \frac{200}{9}ms^{-1}
    In case we need it later, angular speed, \omega = \frac{v}{r} = \frac{200}{9 \times 200} = \frac{1}{9}

    At 80 km h^{-1} the car starts to slip inwards, so this means that the friction is acting up the slope, and that the friction is also equal to \mu N where N is the normal reaction of the slope on the car?

    Resolving vertically I get N = mg\cos{\theta}

    Resolving horizontally I get F = mg\sin{\theta}

    We can rewrite the second equation as \mu N = mg\sin{\theta}, and substituting in our first equation we get \mu mg\cos{\theta} = mg\sin{\theta}, therefore \mu = \tan{\theta} = \frac{\sqrt{3}}{3}?

    I have a feeling that Newton's F = ma needs to be used but I'm not sure where?

    Not sure how to approach the second question either, this has me a bit stumped.

    Thanks in advance for the help
    How about the force acting on the car due to its cilcular motion??
    Last edited by xalk; April 1st 2010 at 02:50 PM.
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    Super Member craig's Avatar
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    Quote Originally Posted by xalk View Post
    How about the force acting on the car due to its cilcular motion??
    Hi thanks for the reply, how would I include this in my equations?

    Thanks again
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    vertical N=mg\cos{\theta}+m\frac{v^{2}}{r}\sin{\theta} ------------------------------------------------------------------------------------------------------------------------------------ ----------- horizontal F=mg\sin{\theta}-m\frac{v^{2}}{r}\cos{\theta}
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    Yes my calculations agree with those of Haytham if by F he means friction
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    Super Member craig's Avatar
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    Thanks for both replies, not got access to a pen & paper at the moment but will attempt the solution as soon as I do
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    Super Member craig's Avatar
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    Quote Originally Posted by Haytham View Post
    vertical
    N=mg\cos{\theta}+m\frac{v^{2}}{r}\sin{\theta}

    horizontal
    F=mg\sin{\theta}-m\frac{v^{2}}{r}\cos{\theta}
    So because the car starts to slip inwards, it's in limiting fraction, so am I right in saying that F = \mu N?

    If so, then \mu(mg\cos{\theta}+m\frac{v^{2}}{r}\sin{\theta}) = mg\sin{\theta}-m\frac{v^{2}}{r}\cos{\theta}

    So \mu = \frac{g\sin{\theta}-\frac{v^{2}}{r}\cos{\theta}}{g\cos{\theta}+\frac{v  ^{2}}{r}\sin{\theta}} = \frac{\frac{1}{2}g-\frac{200}{81}\cdot \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}g+\frac{200}  {81}\cdot \frac{1}{2}} \approx 0.284?

    Also for the second part of the question do I just calculate the value of the normal force?

    Thanks again for all the help.
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    Super Member craig's Avatar
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    Quote Originally Posted by Haytham View Post
    vertical N=mg\cos{\theta}+m\frac{v^{2}}{r}\sin{\theta} ------------------------------------------------------------------------------------------------------------------------------------ ----------- horizontal F=mg\sin{\theta}-m\frac{v^{2}}{r}\cos{\theta}
    Sorry could I just ask how you got the +m\frac{v^{2}}{r}\sin{\theta} and the -m\frac{v^{2}}{r}\cos{\theta}?

    I know that the \frac{v^2}{r} is the acceleration, but how did you go about including these in your equations?

    Thanks again for all the help.
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    You have 3 forces, weight, friction and the reaction.

    You chose this coordinate system


    As the movement is circular you know there is an inward net force represented in red (its the sum of the 3 forces) and its value is F_{net}=m\frac{v^{2}}{r}


    now, you just decompose the forces in the axis

    xx axis
    F_x=-\mu N
    W_x=mg\sin{\theta}
    F_{net/x}=m\frac{v^{2}}{r}\cos{\theta}

    yy axis
    N_y=N
    W_y=-mg\cos{\theta}
    F_{net/y}=m\frac{v^{2}}{r}\sin{\theta}


    and the sum of the 3 forces must be equal to the net force

    xx axis
    m\frac{v^{2}}{r}\cos{\theta}=mg\sin{\theta}-\mu N

    yy axis
    m\frac{v^{2}}{r}\sin{\theta}=N-mg\cos{\theta}

    if you wish to, you could try to solve it again using a different set of coordinates. vertical (parallel to the W) and horizontal (parallel to the net force), i find it simpler.


    Also for the second part of the question do I just calculate the value of the normal force?
    yes
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    Super Member craig's Avatar
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    Thank you so much for that! Really helps, thankyou.
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