this video explains in a clear way what exactly is Banking of Tracks, and how to get the various vectors acting on the car — example , the Weight, Normal Reaction on car from road, Static Friction Force acting at the contact point of tyres with road surface, and finally the Centripetal Force that ensures the car is pulled towards the “centre of the circular path” .The fine balance keeps the car moving in a circle , only if its speed (Angular velocity) is right. For the given road and tyres, and Banking Angle, if the car’s angular velocity increases beyond the balancing value given by the equation, the car will skid outwards.
I am getting a query from readers and I am reproducing the explanation here:-
What about the Normal Reaction ‘N’ in terms of weight ‘mg’?
OK. The component of weight ‘mg’ along the direction of normal would be ‘mg cos theta’. Thus, Normal Reaction N= mg cos theta. For example, suppose we increase the banking angle of the road , let’s make it almost vertical, theta will tend to 90 degrees, cos 90 will tend to zero, and Normal reaction ‘N’ will tend to 0. Although the weight of the car = mg will still continue to act downwards, vertically downwards.
Using Newton’s 2nd law, and Free Body Diagram, as long as the car is not moving perpendicular to the road surface, it means all forces in that direction are perfectly balanced. That is why we have to take N= mg cos theta.
Hope that clarifies. Thankyou.
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