Question

why banking of road is necessary ? write an expression for optimum speed on a banked road.​

Answer 1

●Please refer the attachment for the expression of optimum speed.

Q1. Why is banking of roads necessary?

answer➡️Raising the outer edge of a circular track slightly above the inner edge refers to banking of roads.

●Necessary centripetal force is provided due to the banking of roads by the component of the normal reaction.

●Also during rainy reason, roads become wet and slippery and thus affecting the frictional force which can disbalance the car movement. So banking reduces chances of slippering.

Answer 2

$\textsf{\underline {\Large {Banking\:of\:a\:road}}}$ :

The system of raising the outer edge of a curved road above its inner edge is called $\bold{\underline {banking\:of\:a\:curved\:road}}$.

$\textsf{\underline {\large {Necessity}}}$ :

⚫To avoid large amount of $\bold{\underline {friction}}$ between the tyres and road.

⚫Banking of road provides necessary $\underline {\bold{Centripetal \:force}}$ .

⚫Avoids $\bold{\underline {wearing\:and\:tearing}}$.

$\textsf{\underline {\Large {Expression\:for\:optimum \:Speed}}}$ :

At equilibrium,

R cos $\theta$ - f sin $\theta$ = mg --> ( i )

R sin $\theta$ + f cos $\theta$ = m$\frac{v ^{2}}{r}$ --> ( ii )

Dividing equation ( ii ) to ( i ),

$\frac{R\: sin\:\theta\:+\: f \:cos \:\theta} {R\: cos\:\theta\:+\: f \:sin \:\theta}$ = ${\frac{\frac{mv^{2}}{r}}{mg}}$

Force of friction, f = $\mu{R}$

$\frac{R\: sin\:\theta\:+\: \mu{R} \:cos \:\theta} {R\: cos\:\theta\:+\: \mu{R} \:sin \:\theta}$ = $\frac{v^{2}}{rg}$

$\frac{R\:Cos \:\theta}{ R\:Cos \:\theta}$ $\frac{(tan\:\theta\:+\:\mu} {1\:-\:\mu\:tan\:\theta)}$ = $\frac{v^{2}}{rg}$

➡️ $\textsf{\underline{Optimum Speed}}$, v = $\sqrt{\frac{rg(\:tan\:\theta\:+\:mu\:)} {(1\:-\:\mu\:tan\:\theta)}}$