Two curves on a highway have the same radii. However, one is unbanked and the other is banked at an angle ɵ. A car can safely travel along the unbanked curve at a maximum speed v under conditions when the coefficient of static friction between the tires and the road is μs = 0.787. The banked curve is frictionless, and the car can negotiate it at the same maximum speed v. Find the angle ɵ of the banked curve.

For unbanked curve , centripetal force is provided by friction and for banked curve , centripetal force is provided by component of reaction from the ground.

centripetal force = m v² / r , m is mass , v is velocity and r is radius of curve.

friction force ( maximum ) = μ mg , μ is coefficient of friction ,

for motion on unbanked curve

m v² / r = μ mg

v² = μ gr

= .787 x 9.8 x r

= 7.712 r

In case of motion on banked curve having angle of banking θ

If R be the ground reaction

R cosθ = mg

Rsinθ = centripetal force

= m v² / r

Dividing the two

Tanθ = v² / rg

= 7.712 r / 9.8 r

tanθ = .787

θ = 38 degree .

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-12-09T16:56:00+01:00

The angle ɵ of the banked curve of the given highway is 38.2⁰.

The net force on the car for the unbanked curve is calculated as follows;

[tex]\mu mg = \frac{mv^2}{r} \\\\\mu g r = v^2[/tex]

The normal force on the car for the banked curve is calculated as follows;

[tex]Tcos (\theta) = mg \ \ --(1)[/tex]

The horizontal force on the car for the banked curve is calculated as follows;

[tex]Tsin(\theta) = \frac{mv^2}{r} \ \ ---(2)[/tex]

Divide 2 by 1;

[tex]tan(\theta) = \frac{mv^2}{rmg} \\\\tan (\theta) = \frac{v^2}{rg} \\\\tan(\theta) = \frac{\mu gr}{rg} \\\\tan(\theta) = \mu\\\\\theta = tan^{-1} (\mu)\\\\\theta = tan^{-1} (0.787)\\\\\theta = 38.2 \ ^0[/tex]

Thus, the angle ɵ of the banked curve of the given highway is 38.2⁰.

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