f_{s} =< μ_{s}N

mv^{2}/r =< μ_{s}N

v^{2} =< μ_{s}NR/m = μ_{s}mgR/m = μ_{s}Rg

v =< √ μ_{s}Rg

This is the maximum speed of a car in circular motion on a level road.

In the vertical direction (Y axis)

Ncosϴ = fsinϴ + mg ——————–(i)

In horizontal direction (X axis)

fcosϴ + Nsinϴ = mv^{2}/r —————-(ii)

Since we know that f μ_{s}N

For maximum velocity, f = μ_{s}N

(i)becomes:

Ncosϴ = μ_{s}Nsinϴ + mg

Or, Ncosϴ – μ_{s}Nsinϴ = mg

Or, N = mg/(cosϴ- μ_{s}sinϴ)

Put the above value of N in (ii)

μ_{s}Ncosϴ + Nsinϴ = mv^{2}/r

μ_{s}mgcosϴ/(cosϴ- μ_{s}sinϴ) + mgsinϴ/(cosϴ- μ_{s}sinϴ) = mv^{2}/r

mg (sinϴ + μ_{s}cosϴ)/ (cosϴ – μ_{s}sinϴ) = mv^{2}/r

Divide the Numerator & Denominator by cosϴ, we get

v^{2} = Rg (tanϴ +μ_{s) }/(1- μ_{s }tanϴ)

**v = √ Rg (tanϴ +μ _{s) }/(1- μ_{s }tanϴ)**

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

**Special case:**

When the velocity of the car = v_{0 },

- No f is needed to provide the centripetal force. (μ
_{s }=0) - Little wear & tear of tyres take place.

** v _{o} = √ 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 = 15^{o}

μ_{s }= 0.2

- v
_{o}= √ Rg tanϴ

= √300 * 9.8 * tan 15^{o}

= 28.1 m/s

- v
_{max}= √ Rg (tanϴ +μ_{s) }/(1- μ_{s }tanϴ)

= √ 300 * 9.8 * (0.2 + tan 15^{o})/(1-0.2tan15^{o})

= 38.1 m/s