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Linear and Angular Motion and Friction Q.2a & b

  1. Jun 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Determine the maximum speed at which a vehicle can negotiate a curve of
    20m radius:
    The vehicle has a track width of 1.3m and a centre of gravity midway between the wheels and 0.6m above the ground. The coefficient of friction between the tyres and the road is 0.5.

    a) without skidding

    b) without overturning

    2. Relevant equations

    a) mv^2/r = μmg

    b) v^2 = gdr/2h

    3. The attempt at a solution

    a) v = sqrt μrg

    v = sqrt 0.5 x 20m x 9.81m/s^2

    = sqrt 98.1 m^2/s^2

    = 9.9 m/s

    b) v = sqrt gdr/2h

    = sqrt 9.81m/s^2 x 1.3m x 20m/2 x 0.6m

    = sqrt 255.06m^3/s^2 / 1.2m

    = 212.55m^2/s^2

    = 14.58 m/s

    Again please can someone confirm that i have done these correctly and if not could you point me to which part i have gone wrong. I am not looking for the answer just direction.

    Many thanks
     
  2. jcsd
  3. Jun 19, 2007 #2
    either this makes no sense or i am more clever than i thought :p
     
  4. Jun 19, 2007 #3
    The first part is correct, of that I'm sure, but the second part, I don't know how you arrived at that relation.
     
  5. Jun 19, 2007 #4
    This is what I've been able to do so far, perhaps I'm missing something, if so, maybe you can point it out:

    Lets split the wheels on the car to that on the left side and the right side.
    The forces:

    The normal reaction on the left side be N1, and on the right N2.
    The centrifugal force being mv^2/r
    Frictional force being [tex]\mu N_1[/tex] on the left side and [tex]\mu N_2[/tex] on the right side.

    At equilibrium, just before the car is about to overturn, all the torques about the center of gravity must sum to zero.

    x is the horizontal distance of either of the wheels from the center of gravity.

    This gives [tex]N_1+\frac{mv^2}{r}\frac{h}{x} =N_2[/tex]. Also [tex]N_1+N_2=mg[/tex].

    Solving this, you get [tex]N_1=\frac{m}{2}(g-\frac{v^2h}{rx})[/tex] and [tex]N_2=\frac{m}{2}(g+\frac{v^2h}{rx})[/tex].

    Now, I'm missing something which relates N1, N2 and the velocity in another manner... any ideas?
     
  6. Jun 19, 2007 #5
    Got it. When the wheels leave the ground, N1=0. This gives [tex]g=\frac{v^2h}{rx}[/tex]. This gives [tex]v=\sqrt{\frac{grx}{h}}[/tex]
     
  7. Jun 19, 2007 #6
    hi, could you confirm your thoughts as to whether you feel that i have both parts correct? as you appear to have only h under the fraction as i have 2h, this looks to be the only difference in the given equation.
     
  8. Jun 21, 2007 #7
    It should only be h and not 2h.
     
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