[caption id="attachment_22100" align="aligncenter" width="300"]
Banking of road[/caption]
Banking is the inclination given to roads and rail tracks at turnings or along curved paths, by raising the outer edge of the path slightly above the inner edge towards the centre of the curvature.
Banking of road
Consider a body moving along a horizontal circular path, where no banking is provided.
A car moving on a road (or a train moving on the rails) requires a centripetal force while taking a turn. To some extent, this centripetal force is provided by the friction between the road and tyre. When the vehicles are heavy, the required centripetal force may not be provided by the frictional force, as it is limited.
[caption id="attachment_22083" align="aligncenter" width="300"]
A car moving on a road while taking a turn[/caption]
A curved road with banking[/caption]
Let the angle of banking be θ. On resolving, the normal reaction 'R' into components, we can say that normal reaction is equal to the weight of the car. Banking of road
i.e., R cos θ = mg → (1)
and R sinθ acts as centripetal force acting on the car.
i.e., $R sin θ = \frac{m v^2}{r}$ → (2)
By dividing Eq. (1) and (2)
Thus, we have
$tan θ = \dfrac{v^2}{rg}$
$⇒v= \sqrt{rg tanθ}$
Above formula is used to find expected speed at a curve. Banking of road
The roads are banked, i.e., the angle of banking is provided for the average expected speed of the vehicles. If the speed of the vehicle v is less, then the average expected speed \(v_0\) for a given banking, the vehicle would skid towards the centre. If $v >v_0$, the car would skid outwards.
Banking of road[/caption]
Banking is the inclination given to roads and rail tracks at turnings or along curved paths, by raising the outer edge of the path slightly above the inner edge towards the centre of the curvature.
Banking of road
Consider a body moving along a horizontal circular path, where no banking is provided.
A car moving on a road (or a train moving on the rails) requires a centripetal force while taking a turn. To some extent, this centripetal force is provided by the friction between the road and tyre. When the vehicles are heavy, the required centripetal force may not be provided by the frictional force, as it is limited.
[caption id="attachment_22083" align="aligncenter" width="300"] A car moving on a road while taking a turn[/caption]
As the magnitude of \(f_s\), cannot exceed \(µ_s\) R, where \(µ_s\), is the coefficient of static friction,
$f_s≤µ_s R$
$f_s≤(µ_s) mg,$ I.e., $\dfrac{m v^2 }{r} ≤µ_s$ mg
$µ_s≥\dfrac{v^2}{rg}$
If this condition is not satisfied, the car or train would skid outwards. Banking of road
Now, consider a curved road with banking. The angle made by the road with the horizontal line is called 'angle of banking'.
[caption id="attachment_22099" align="aligncenter" width="300"] A curved road with banking[/caption]
Let the angle of banking be θ. On resolving, the normal reaction 'R' into components, we can say that normal reaction is equal to the weight of the car. Banking of road
i.e., R cos θ = mg → (1)
and R sinθ acts as centripetal force acting on the car.
i.e., $R sin θ = \frac{m v^2}{r}$ → (2)
By dividing Eq. (1) and (2)
Thus, we have
$tan θ = \dfrac{v^2}{rg}$
$⇒v= \sqrt{rg tanθ}$
Above formula is used to find expected speed at a curve. Banking of road
The roads are banked, i.e., the angle of banking is provided for the average expected speed of the vehicles. If the speed of the vehicle v is less, then the average expected speed \(v_0\) for a given banking, the vehicle would skid towards the centre. If $v >v_0$, the car would skid outwards.
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