Question

Derive expression for maximum permissible speed of a car on banked road​

Answer 1

Consider the attachment;

From the diagram,

N= mgcosθ + Fc sinθ   --(i)

and,

Fc cosθ = mgsinθ + μN    (ii)

Combining these two together by substituting the value, of N in the second equation,

Fc cosθ = mg sinθ + μ(mgcosθ + Fc sinθ)

Solving furthur,

Fc cosθ -  μFc sinθ= mg sin θ + μmgcosθ

Since Fc= mv²/r

Solving furthur,

Here, v = maximum velocity, r = radius of the banking, Fc= centripetal force, μ= coefficient of friction and N= normal reaction force.

Answer 2

Maximum velocity on a banked road is

V= [( rg/m) (tanθ) ]^(1/2)

◆Refer the attachment below,

◆We know that CENTRIPETAL FORCE acts on a car moving in a banked road,

◆CENTRIPETAL FORCE Fc= mv^2/r

◆Take components of Fc, and N, as it helps in easy solving

◆Thus, from the figure we see,

◆Nsinθ= mv^2/r --(1)

N being the NORMAL REACTION FORCE

◆and Ncosθ=mg. --(2)

(Both are in same direction)

◆Dividing eqn (1)/(2)

Nsinθ /Ncosθ =( mv^2)/r ÷(mg)

tanθ= mv^2 /(rg)

◆Hence, V= [ (rg/m) (tanθ) ]^(1/2)

◆Where, v = maximum velocity,

r = radius of the banking,

Fc= centripetal force

N= normal reaction force.