At some point in your life, you must have driven on a highway. You must have noticed that the roads are elevated at certain angles, where they curve. Why is it so? We will find out the answer in this article. Please read the article till the end where you will get a special gift from me!!
Circular roads are ‘banked’ so that the drivers may drive safely at permissible speed. I hope that everyone knows about centripetal force, which is directed towards the center of the circular path and responsible for circular motion and is mathematically represented as (mv2/r) where ‘m’ stands for the mass of the object, ‘v’ stands for velocity and ‘r’ stands for the radius of the circular trajectory. Friction, the ‘necessary evil,’ plays a great role. It has a limiting value of (µN), where µ is the coefficient of friction, and ‘N’ is the normal reaction force, where µ is a characteristic property of each substance. Imagine a horizontal unbanked circular road of radius ‘r,’ where a vehicle is moving with speed ‘v.’ The necessary centripetal force is provided by friction. So we have,
f = (mv2)/r…..i )
But, f ≤ µ N……ii )
Substituting Eqn.i) in Eqn.ii )
(mv2)/r ≤ µmg ( N=mg )
v ≤ √(µrg)
∴vmax = √(µrg)
Here ‘f’ refers to the frictional force. So the maximum speed for taking a safe turn is√(µrg). Now the problem is that the maximum speed is dependent on µ, which in turn depends on the tire, condition of the road, and the season. Whenever it rains, the road becomes slippery, µ decreases, and hence vmax also decreases.
Besides, after long use of the tyres, the tread gets worn out, and once again µ decreases, vmax decreases. To be independent of µ, the circular horizontal turning is banked. But how is it done? Before that we must know that any force can be divided into components whose resultant, i.e. their vector sum yields the same force. If the roads are banked, the component of normal reaction (N) helps in achieving the necessary centripetal force. Let us see how,
N sinθ = (mv2)/r…..iii )
N cosθ = mg…..iv )
Dividing iii ) by iv )
tanθ = v2/rg
∴ θ = tan-1[v2/rg]
Here the vertical component, i.e. Ncosθ balances the weight of the car and the horizontal component, i.e. Nsinθ is responsible for the centripetal force. The circular roads are banked in this manner at a small angleθ. Now the eqn. for the maximum permissible speed is,
vmax = √(rg tanθ ).
Here vmaxis independent of µ which allows the driver to drive at speeds not related to tire conditions. So now you know the reason behind the banking of circular roads.
When vehicles go through banked circular roads with maximum permissible speed Vmax then limiting friction acts down the plane. Let us see how
Nsinθ+µNcosθ = (mv2max)/r …….v )
Ncosθ = mg+µNsinθ……….vi )
Adding v ) and vi )-
mv2max+mg+µNsinθ = Ncosθ+Nsinθ+µNcosθ
vmax = √(gr((tanθ+µ)/(1-µtanθ)))
Next time you drive on a highway, remember that a lot of mathematics and physics are involved in it. Now as I had promised I will give you a special gift. I will share a link where you will provided with a video explanation of this concept.
PLEASE CLICK HERE- https://youtu.be/eGZWVwcaq0U
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