what is banking angle and derive teta =v,square÷rg
Answers
Answer:
Explanation:
Banking of roads is defined as the phenomenon in which the edges are raised for the curved roads above the inner edge to provide the necessary centripetal force to the vehicles so that they take a safe turn. Now, let us recall, what is centripetal force? It is the force that pulls or pushes an object toward the centre of a circle as it travels, causing angular or circular motion. In the next few sections, let us discuss the angle of banking and the terminologies used in banking of roads.
Other terminologies used are banked turn which is defined as the turn or change of direction in which the vehicle inclines towards inside. The angle at which the vehicle is inclined is defined as the bank angle. The inclination happens at longitudinal and horizontal axis.
Banking Of Roads
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