what is banking of roads derive an exprenion for angle of banking?
Answers
Explanation:
Banking of roads : To avoid risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inwards, i.e., the outer side of road is raised above its inner side. This is called 'banking of roads'.
Consider a car taking a left turn along a road of radius r banked at an angle θ for a designed optimum speed V. Let m be the mass of the car. In general, the forces acting on the car are:
(a) Its weight mg, acting vertically down
(b) The normal reaction of the road N, perpendicular to the road surface
(c) The frictional force fs along the inclined surface of the road.
Resolve N and fs into two perpendicular components Ncosθ vertically up and fssinθ vertically down, Nsinθ and fscosθhorizontally towards the centre of the circular path.
If vmax is the maximum safe speed without skidding.
rmvmax2=Nsinθ+fscosθ
=Nsinθ+μsNcosθ
rmvmax2=N(sinθ+μscosθ)....(1)
and
Ncosθ=mg+fssinθ
=mg+μsNsinθ
∴mg=N(cosθ−μssinθ)...(2)
Dividing eq. (1) by eq. (2),
r.mgmvmax2=N(cosθ−μssinθ)N(sinθ+μscosθ)
∴rgvmax2=cosθ−μssinθsinθ+μscosθ=1−μstanθtanθ+μs
∴vmax=1−μstanθrg(tanθ+μs)...,.(3)
This is the expression for the maximum safe speed on a banked road.
At the optimum speed, the friction between the car tyres and the road surface is not called into play. Hence, by setting μs=0 in eq. (3), the optimum speed on a banked circular road is
v=rgtanθ...(4)
∴tanθ=rgv2 or θ=tan−1(rgv2)
From this eq. we see that θ depends upon v,r and g. The angle of banking is independent of the mass of a vehicle negotiating the curve.
hope this is useful
mark this as brainlist