Q3.10 Banking on friction (S)

A car travelling at 60 km hr-1 rounds a bend of radius of curvature 150 m.
  1. If the road is flat what is the minimum value of the coefficient of static friction required to prevent skidding?
  2. At what angle should the road be banked to minimise the risk of skidding?
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Hint

You need to think about the origins of the centripetal force required to sustain the circular motion. In the first case it can come only from friction. In the second there is another possiblity, as you will see when you realise that the normal contact force exerted by the road is no longer vertical.
 

Solution

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Solution

  1. Consider the vehicle (mass m say) rounding the bend (towards the left say), as seen from behind.

    If the car is on the point of slipping (outwards) the frictional force acting inwards, towards the centre of the circle, has magnitude
    FF=μsFN=μsmg
    Newton’s 2nd law applied to the vehicle then gives
    \[ \mu_s mg = \frac{mv^2}{r} \]
    Graphic - No title - roundthebenda

    The lhs is the force towards the centre (the centripetal force); the rhs is the mass times the acceleration towards the centre (the centripetal acceleration). See Key Point 1.8 …if necessary. It follows that

    \[ \mu_s= \frac{v^2}{rg} = 0.19 \hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]

  2. If the road is banked the need for a frictional force to supply the road-holding (the centripetal force) can be reduced, in principle even eliminated.

    Again consider the vehicle as seen from behind. Given the slope of the road the normal contact force (exerted by the road) now has a component in the horizontal direction, towards the centre of the circle. So this component can serve as the centripetal force.

    In the vertical direction (ie perpendicular to the plane of the circular motion) the car has zero acceleration. Newton’s 2nd law (applied to the forces in this direction: note this is not perpendicular to the surface itself!) then implies
    FNcosθ-mg=0
    Graphic - No title - roundthebendb

    Now applying Newton’s 2nd law along the centripetal direction we have

    \[ F_N\\sin{\theta} = \frac{mv^2}{r} \]

    Combining these two results we conclude

    \[ \tan \theta = \frac{v^2}{rg} = 0.19 \]
    implying
    \[ \theta = 11^{\circ} \hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]