Class 11 Physics Laws of Motion Motion of a car on a banked road

Motion of a car on a banked road

 

 

In the vertical direction (Y axis)

Ncosϴ = fsinϴ + mg --------------------(i)

In horizontal direction (X axis)

fcosϴ + Nsinϴ = mv2/r  ----------------(ii)

 

Since we know that f      μsN   

For maximum velocity, f = μsN

(i)becomes:

       Ncosϴ = μsNsinϴ + mg

Or, Ncosϴ - μsNsinϴ = mg

Or, N = mg/(cosϴ- μssinϴ)

 

Put the above value of N in (ii)

μsNcosϴ + Nsinϴ = mv2/r 

μsmgcosϴ/(cosϴ- μssinϴ) + mgsinϴ/(cosϴ- μssinϴ) = mv2/r 

mg (sinϴ + μscosϴ)/ (cosϴ - μssinϴ) = mv2/r 

Divide the Numerator & Denominator by cosϴ, we get

v2 = Rg (tanϴ +μs) /(1- μs tanϴ)

v = √ Rg (tanϴ +μs) /(1- μs tanϴ)

This is the miximum speed of a car on a banked road.

 

Special case:

When the velocity of the car = v0 ,

  • No f is needed to provide the centripetal force. (μs =0)
  • Little wear & tear of tyres take place.

             vo = √ Rg (tanϴ)

 Problem: A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the

 (a) optimum speed of the racecar to avoid wear and tear on its tyres, and

(b) maximum permissible speed to avoid slipping ?

 Solution.

R = 300m

ϴ = 15o

μ= 0.2

  • vo = √ Rg tanϴ

      = √300 * 9.8 * tan 15o

      = 28.1 m/s

  • vmax = √ Rg (tanϴ +μs) /(1- μs tanϴ)

             = √ 300 * 9.8 * (0.2 + tan 15o)/(1-0.2tan15o)

             = 38.1 m/s

 

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