obtain an expression for the angle of banking of curved road
Answers
Explanation:
Angle of Banking
Consider a vehicle of mass ‘m’ with moving speed ‘v’ on the banked road with radius ‘r’. Let ϴ be the angle of banking, with frictional force f acting between the road and the tyres of the vehicle.
Total upwards force = Total downward force
NcosΘ=mg+fsinΘ
Where,
NcosΘ : one of the components of normal reaction along the verticle axis
mg: weight of the vehicle acting vertically downward
fsinΘ : one of the components of frictional force along the verticle axis
therefore, mg=NcosΘ−fsinΘ (eq.1)
mv2r=NsinΘ+fcosΘ (eq.2)
Where,
NsinΘ : one of the components of normal reaction along the horizontal axis
fcosΘ : one of the components of frictional force along the horizontal axis
mv2rmg=NsinΘ+fcosΘNcosΘ−fsinΘ (after diving eq.1 and eq.2)
therefore, v2rg=NsinΘ+fcosΘNcosΘ−fsinΘ
Frictional force f=μsN v2rg=NsinΘ+μsNcosΘNcosΘ−μsNsinΘ v2rg=N(sinΘ+μscosΘ)N(cosΘ−μssinΘ) v2rg=(sinΘ+μscosΘ)(cosΘ−μssinΘ) v2rg=(tanΘ+μs)(1−μstanΘ)
therefore, v=rg(tanΘ+μs)(1−μstanΘ)−−−−−−−−√ vmax=rgtanΘ−−−−−−√ tanΘ=v2rg Θ=tan−1v2rg
Explanation:
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.
v^2/rg = ( R sin α + μR cos α)/ ( R cos α - μR sin α)
angle of banking = α = tan^(-1) [(v^2 - μrg) / (rg + μv^2)]