Question

What is the banking of roads? Obtain an expression for maximum safety speed of a vehicle moving along a curved horizontal road.

Answer 1

Hint: We can use the idea for a circular motion some force to be providing as a centripetal force to act towards the centre.
We can consider all forces acting on the car which can provide sufficient centripetal force.

Complete step by step answer:
We know the banking of roads \[-\] To avoid risk of skidding the road surface at a bend is tatted inwards, i.e. the outer side of road is raised above its inner ride. This is called ‘Banking of roads’
We can consider taking a left turn along a road or banked at an angle $\theta $ for a designed optimum speed V. Let M be the man of the car. In general, the force is aching on the car one.
(a) Its weight$\overrightarrow {mg} $, acting vertically drawn.
(b) The functional reactor of the road$\overrightarrow N $, perpendicular to the road surface.
(c) The frictional force ${\overrightarrow f _1}$ along the inclined surface of the road.
Resolve $\overrightarrow N $ and $\overrightarrow {{f_1}} $ into two perpendicular components.
$N\cos \theta $ Vertically up and ${\overrightarrow f _s}\sin \theta $ vertically down, $N\sin \theta $ and ${\overrightarrow f _s}\cos \theta $ horizontally towards the center of the circular path.
If ${V_{\max }}$ is the maximum safe speed without skidding, then
$
  \dfrac{{mv_{\max }^2}}{t} = N\sin \theta + {f_s}\cos \theta \\
   = N\sin \theta + {\mu _s}N\cos \theta \\
 $
$\dfrac{{mv_{\max }^2}}{{{t_0}}} = N\left( {\sin \theta + {\mu _s}\cos \theta } \right)...(1)$
And,
$
  N\cos \theta = mg + {f_s}\sin \theta \\
   = mg + {\mu _s}N\sin \theta \\
  mg = N\left( {\cos \theta - {\mu _s}\sin \theta } \right)...(2) \\
 $
Dividing ($1$) by ($2$)
$\dfrac{{mv_{\max }^2}}{{{V_0}mg}} = \dfrac{{N\left( {\sin \theta + {\mu _s}\cos \theta } \right)}}{{N\left( {\cos \theta - {\mu _s}\sin \theta } \right)}}$
$\therefore \dfrac{{v_{\max }^2}}{{mg}} = \dfrac{{\sin \theta + {\mu _s}\cos \theta }}{{\cos \theta - {\mu _s}\sin \theta }} = \dfrac{{\tan \theta + {\mu _s}}}{{1 - {\mu _s}\tan \theta }}$
${V_{\max }} = \sqrt {\dfrac{{mg\left( {\tan \theta + {\mu _s}} \right)}}{{1 - {\mu _s}\tan \theta }}} $
This is the expression for the maximum safety speed on a curved horizontal road.

Note: The friction reaction of the road $\overrightarrow N $ perpendicular to the road surface.
And the frictional force ${\overrightarrow d _s}$ along the inclined surface of the road.
Other than banked roads in formula races we can see they bend the bike at some angle.