Question

Derive the formula af motion of a Car m a banked road​

Answer 1

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A car running with a velocity 72 kmph on a level road, is stopped after travelling a distance of 30 m after disengaging its engine (g=10ms

−2

). The coefficient of friction between the road and the tyres is.

Answer 2

Explanation:

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.