Question

Derive the expression for the angle of banking on a curved road with certain OC efficient of friction

Answer 1

consider a vehicle of mass 'm' moving with velocity 'v' on a banked road as shown in the 1st attached diagram,

m = mass of the vehicle

f = frictional force

R = normal force

μ = coefficient of the friction between road and vehicle

α = angle of banking

r = radius of the banked curved road

Now we can draw free-body diagram of the vehicle as shown in the 2nd diagram,

consider equilibrium of the vehicle,

• Total upward force = Total downward force

                           R cos α = mg + f sin α

                                  mg = R cos α - f sin α→ (1)

• Centripetal force = Total horizontal force

                          mv^2/r = R sin α + f cos α→ (2)

Divide (2) by (1),

(mv^2/r)/mg = ( R sin α + f cos α)/ ( R cos α - f sin α)

        v^2/rg = ( R sin α + f cos α)/ ( R cos α - f sin α)

but we know f=μR

       v^2/rg = ( R sin α + μR cos α)/ ( R cos α - μR sin α)

       v^2/rg = ( sin α + μ cos α)/ ( cos α - μ sin α)

divide right side of the equation by cos α,

     v^2/rg = [(sin α/cos α) + (μ cos α/ cos α)]/[(cos α/cos α)- (μ sin α/ cos α)]

     v^2/rg = (tan α + μ) / (1 - μ tan α)

now we can use cross multiplication,

v^2 X (1 - μ tan α) = rg( tan α + μ)

tan α ( rg + μv^2 ) = v^2 - μrg

                   tan α = (v^2 -μrg)/ (rg + μv^2)

angle of banking = α = tan^(-1) [(v^2 - μrg) / (rg + μv^2)]

Answer 2

Answer:

consider a vehicle of mass 'm' moving with velocity 'v' on a banked road as shown in the 1st attached diagram,

m = mass of the vehicle

f = frictional force

R = normal force

μ = coefficient of the friction between road and vehicle

α = angle of banking

r = radius of the banked curved road

Now we can draw free-body diagram of the vehicle as shown in the 2nd diagram,

consider equilibrium of the vehicle,

Total upward force = Total downward force

                          R cos α = mg + f sin α

                                 mg = R cos α - f sin α→ (1)

Centripetal force = Total horizontal force

                         mv^2/r = R sin α + f cos α→ (2)

Divide (2) by (1),

(mv^2/r)/mg = ( R sin α + f cos α)/ ( R cos α - f sin α)

       v^2/rg = ( R sin α + f cos α)/ ( R cos α - f sin α)

but we know f=μR

      v^2/rg = ( R sin α + μR cos α)/ ( R cos α - μR sin α)

      v^2/rg = ( sin α + μ cos α)/ ( cos α - μ sin α)

divide right side of the equation by cos α,

    v^2/rg = [(sin α/cos α) + (μ cos α/ cos α)]/[(cos α/cos α)- (μ sin α/ cos α)]

    v^2/rg = (tan α + μ) / (1 - μ tan α)

now we can use cross multiplication,

v^2 X (1 - μ tan α) = rg( tan α + μ)

tan α ( rg + μv^2 ) = v^2 - μrg

                  tan α = (v^2 -μrg)/ (rg + μv^2)

angle of banking = α = tan^(-1) [(v^2 - μrg) / (rg + μv^2)]