Question

derive equation for motion of a car on a banked road

Answer 1

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.

 

Answer 2

Answer:

see the attachment

this may help you