Q. What is need the of banking obtain an expression for maximum speed with which a vehicle can safely negotiate a curved road banked at an angle theta, the coefficient of friction b/w wheel and road is μ. CLASS 11TH PHYSICS.
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Answer:
When a sharp turn comes, the surface of the road does not remain horizontal. This is called banking of the roads.
Purpose of Banking:
- To contribute in providing necessary centripetal force.
- To reduce frictional wear and tear of tryes.
- To avoid skidding.
- To avoid overturning of vehicle.
Case I: If coefficient of friction, μ = 0:-
What we really wish is that even if there is no friction between the tyres and the road, yet we should be able to take a round turn. In the given figure Vertical N cos θ component of the normal reaction N will be equal to mg and the horizontal N sin θ component will provide for the necessary centripetal force. [Please note that as we are assuming μ to be zero here, the total reaction of the road will be the normal reaction.] Frictional forces will not act in such a case.
Thus, N cos θ = mg .......... (i)
N sin θ = mv2/r .......... (ii)
Dividing equation (ii) by (i), we get
tan θ = v2/rg
Dividing equation (ii) by (i), we get
tan θ = v2/rg
where θ is the angle of banking.
Case - II : If coefficient of friction, μ ≠ 0:-
a section of the banked road and the view of the vehicle form the rear end.
vehicle-form-the-rear-end
The total forces acting are,
N1 and N2 = normal reactions
f1 and f2 = frictional forces
mg = weight
r = radius
θ = angle of banking
Let N = Resultant of N1 and N2.
f = Resultant of f1 and f2.
Let us resolve all the forces horizontally and vertically. As the vehicle has Equilibrium in vertical direction.
so, N cos θ + f sin θ = mg ............ (i)
The resultant of horizontal components i.e., (f cos θ + N sin θ), however, this becomes the net external force acting on the vehicle in the radially inward direction of the round-turn. This thus provides for the necessary centripetal force (mv2/r).
Therefore, f cos θ + N sin θ = mv2/r ............ (ii)
Further, if μ is the coefficient of friction, we have
f = μN ............ (iii)
These are the three basic equations from which, we can find out whatever we want to find out.'
Putting (iii) in (i) gives
N cos θ = μN sin θ + mg
=>N(cos θ - μsin θ) = mg
=> N = mg/cos θ - μsin θ ...............(iv)
Putting (iii) and (iv) in (ii) gives
μ × mgcos θ/(cos θ - μsin θ) + mgsinθ/(cos θ - μsin θ) = mv2/r
=> μ mgr cos θ + mgr sin θ = mv2 cos θ - μmv2 sin θ
Thus, tan θ = (v2 - μrg)/(rg + μv2) ............... (A)
or v2 = rg(μ+ tanθ)/(1-μtanθ) ................ (B)
The answer is clearly shown in the attachment.
The Diagram is made in 1 dimensional and not in 3-d so to avoid the confusion.
The Mathematical expression is shown in the diagram with the derivation of it. Also, two Important conclusions are shown.
The Maximum velocity is √rgtanθ, where r is the radius of the curved path, g is the acceleration due to gravity.
In this case, Friction is totally neglected. But this can also be taken in the given condition but for Maximum velocity we don't need to assume that.
