Question

Derive+an+expression+for+maximum+velocity+for+the+vehicle+in+a+banked+road

Answer 1

Correct Question:

Derive an expression for maximum velocity for the vehicle in a banked road?

Answer:

$\bullet\;\boxed{\sf v=\sqrt{gr\;\bigg(\dfrac{sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\bigg)}}$

Explanation:

$\rule{300}{1.5}$

Let us consider a body of mass m is moving on a banked road. the road is banked at an angle θ, and let the Normal reaction and force of friction be denoted by N and f respectively. we can clearly see that components of normal reaction and force of friction balances the Weight of the body and centripetal force acting on the body. Simplifying further,

Case-1

$\longrightarrow\sf W_{c}=N\cos\theta - f\sin\theta\\\\\\\longrightarrow\sf Mg=N\cos\theta-f\sin\theta\\\\\\\longrightarrow\underline{\underline{\sf Mg=N\cos\theta-f\sin\theta}}\quad\dots\dots\sf (1)$

Case-2

$\longrightarrow\sf F_{c}=N\sin\theta+f\cos\theta\\\\\\\\\longrightarrow\sf \dfrac{Mv^{2}}{r}=N\sin\theta+f\cos\theta \\\\\\\\\longrightarrow\underline{\underline{\sf \dfrac{Mv^{2}}{r}=N\sin\theta+f\cos\theta}}\quad\dots\dots\sf (2)$

• Dividing 2 by 1.

$\longrightarrow\sf \dfrac{Mv^{2}}{r\times Mg}=\dfrac{N\sin\theta+f\cos\theta}{N\cos\theta-f\sin\theta}\\\\\\\\\longrightarrow\sf \dfrac{v^{2}}{rg}=\dfrac{N\sin\theta+f\cos\theta}{N\cos\theta-f\sin\theta}$

• We know f = μN, substituting,

$\longrightarrow\sf \dfrac{v^{2}}{rg}=\dfrac{N\sin\theta+\mu N\cos\theta}{N\cos\theta-\mu N\sin\theta}\\\\\\\\\longrightarrow\sf \dfrac{v^{2}}{rg}=\dfrac{N\Big(\sin\theta+\mu \cos\theta\Big)}{N\Big(\cos\theta-\mu \sin\theta\Big)}\\\\\\\\\longrightarrow\sf \dfrac{v^{2}}{rg}=\dfrac{\Big(\sin\theta+\mu \cos\theta\Big)}{\Big(\cos\theta-\mu \sin\theta\Big)}\\\\\\\\\longrightarrow\sf v^{2}=gr\;.\;\dfrac{\Big(\sin\theta+\mu \cos\theta\Big)}{\Big(\cos\theta-\mu \sin\theta\Big)}$

$\longrightarrow\sf v=\sqrt{gr\;.\;\bigg(\dfrac{\sin\theta+\mu \cos\theta}{\cos\theta-\mu \sin\theta}\bigg)}\\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf v=\sqrt{gr\;.\;\bigg(\dfrac{\sin\theta+\mu \cos\theta}{\cos\theta-\mu \sin\theta}\bigg)}}}}}$

Hence Derived!

$\rule{300}{1.5}$