Question

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

Answer 1

Answer:

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,

$\frac{mv^{2} }{r} (cosθ -μsinθ)=mg(sinθ + μcosθ)\\\\\frac{v^{2} }{r} (cosθ -μsinθ) = g(sinθ + μcosθ)\\\\v^{2} =gr\frac{(sinθ + μcosθ)}{(cosθ -μsinθ)} \\\\v =\sqrt{gr\frac{(sinθ + μcosθ)}{(cosθ -μsinθ)}}$

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

Answer 2

$\textbf{Answer is in attachment!}$