Question

Q] Obtain an expression for maximum safety speed with which a vehicle can be safely driven along a curved banked road

And

Q] Show that the angle of banking is independent of mass of vehicle.​

Answer 1

✫ Question Given : 1

• Q] Obtain an expression for maximum safety speed with which a vehicle can be safely driven along a curved banked road ?

✫ Required Solution :

✠ From Given Diagram :-

• ⇒ Rcosθ = fsinθ + mg

• ⇒ R cos θ - μRsinθ = mg

• ⇒ R ( cosθ -μsinθ) = mg ___(eq1)

✫ Now :

• ⇒ fcosθ + R sinθ = mv²/R

• f = μR ( From law of limiting friction)

• ⇒ μRcosθ + R sinθ = mv²/R

• ⇒ R ( μcosθ+ sinθ) = mv²/R ____(eq2)

✠ Dividing equation (2) by (1)

• ⇒R ( μcosθ+ sinθ) = mv²/R / R ( cosθ -μsinθ) = mg

• By cancelling up and downwards

• ⇒ μcosθ + sinθ /cosθ - μsinθ = V²/Vg

• ⇒√ (μCosθ + Sinθ)/( Cosθ - μsinθ ) rg = V

• Here underoot(√) over expression

• Dividing all expression by Cosθ

✠ Finally Derived Out !

• ➼ √ (μ + tanθ )/( 1 - μtanθ) rg = V

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✫ Question Given : 2

• Q] Show that the angle of banking is independent of mass of vehicle.

✫ Required Solution :

• From Given Diagram :-

• ⇒ Rsinθ = mv²/ R (Centrifugal force) __(eq1)

• ⇒ Rcosθ = mg ___(eq2)

• Dividing equation (1) by (2)

✠ we have ;

• ⇒ tanθ = V² / Rg

• Finally derived out !

• ➼ tanθ = V² / Rg

✫ Note :

• All required Diagram are in attachment :D

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Answer 2

Refer to the attachment first for better understanding!

(1)

Net force acting on the vehicle,

$\to\sf{N cos \theta = mg + f sin \theta }$

$\to\sf{N cos \theta- f sin \theta = mg ....(1)}$

$\to\sf{N sin \theta + f cos \theta =\dfrac{ mv^2 }{ r} ....(2)}$

Dividing (2) and (1) we get,

$\to\sf{\dfrac{N sin \theta + f cos \theta }{N cos \theta - f sin \theta } = \dfrac{v^2}{ rg}}$

Now let's divide the numerator and denominator of the L.H.S by N cos θ,

$\to\sf{\dfrac{\dfrac{N sin \theta }{N cos \theta } + \dfrac{f cos \theta }{N cos \theta }}{\dfrac{N cos \theta }{N cos \theta } - \dfrac{f sin \theta }{N cos \theta }} = \dfrac{v^2}{ rg}}$

$\to\sf{\dfrac{{tan \theta + \dfrac{f}{N} }} {1- \dfrac{f}{N} tan \theta }= \dfrac{v^2}{ rg}}$

We know that,

$\to\sf{ f = \mu N }$

$\to \sf{\mu = f / N }$

Hence,

$\to\sf{\dfrac{{tan \theta + \mu }} {1-\mu tan \theta }= \dfrac{v^2}{ rg}}$

$\to\boxed{\sf{v = \sqrt{rg\bigg(\dfrac{{tan \theta + \mu }} {1- \mu tan \theta } \bigg)}}}$

When there is no friction acting between the road and the tire the coefficient of friction (μ) will be zero.

$\to\sf{v = \sqrt{rg\bigg(\dfrac{tan \theta +0 } {1-0 }\bigg) }}$

Hence the maximum safety speed with which a vehicle can be safely driven along a curved banked road is,

$\to\boxed{\sf{v = \sqrt{rg tan \theta }}}$

(2)

On rearranging the above equation we get,

$\to\sf{\theta = tan^{-1}\bigg(\dfrac{v^2}{rg}\bigg)}$

From the above equation, we can conclude that the angle of banking doesn't depend upon the mass of the vehicle.