Question

for a vehicle moving on a banked curved road obtain the formula for the maximum safe speed​

Answer 1

Max Velocity on Banked Road :

Centripetal force is :

• mv²/R = $f\cos(\theta) + N\sin(\theta)$

Also, weight will be :

• mg = $N\cos(\theta)-f\sin(\theta)$

Now, dividing the Equations:

$\rm \: \dfrac{( \frac{m {v}^{2} }{R} )}{mg} = \dfrac{N \sin( \theta) + f \cos( \theta) }{N \cos( \theta) - f \sin( \theta) }$

$\rm \implies\: \dfrac{ {v}^{2} }{Rg} = \dfrac{N \sin( \theta) + f \cos( \theta) }{N \cos( \theta) - f \sin( \theta) }$

$\rm \implies\: \dfrac{ {v}^{2} }{Rg} = \dfrac{ \tan( \theta) + \frac{f}{N} }{1 - \frac{f}{N} \tan( \theta) }$

For max speed : f = $\mu_{s}N$

$\rm \implies\: \dfrac{ {(v_{max})}^{2} }{Rg} = \dfrac{ \tan( \theta) + \frac{f}{N} }{1 - \frac{f}{N} \tan( \theta) }$

$\rm \implies\: \dfrac{ {(v_{max})}^{2} }{Rg} = \dfrac{ \tan( \theta) + \mu_{s} }{1 - \mu_{s} \tan( \theta) }$

$\rm \implies\: {(v_{max})}^{2} = Rg \bigg[\dfrac{ \tan( \theta) + \mu_{s} }{1 - \mu_{s} \tan( \theta) } \bigg]$

$\rm \implies\: v_{max} = \sqrt{ Rg \bigg[\dfrac{ \tan( \theta) + \mu_{s} }{1 - \mu_{s} \tan( \theta) } \bigg]}$

Hope It Helps

Answer 2

Explanation:

Max Velocity on Banked Road :

Centripetal force is :

mv²/R = f\cos(\theta) + N\sin(\theta)fcos(θ)+Nsin(θ)

Also, weight will be :

mg = N\cos(\theta)-f\sin(\theta)Ncos(θ)−fsin(θ)

Now, dividing the Equations:

\rm \: \dfrac{( \frac{m {v}^{2} }{R} )}{mg} = \dfrac{N \sin( \theta) + f \cos( \theta) }{N \cos( \theta) - f \sin( \theta) }

mg

(

R

mv

2

)

=

Ncos(θ)−fsin(θ)

Nsin(θ)+fcos(θ)

\rm \implies\: \dfrac{ {v}^{2} }{Rg} = \dfrac{N \sin( \theta) + f \cos( \theta) }{N \cos( \theta) - f \sin( \theta) } ⟹

Rg

v

2

=

Ncos(θ)−fsin(θ)

Nsin(θ)+fcos(θ)

\rm \implies\: \dfrac{ {v}^{2} }{Rg} = \dfrac{ \tan( \theta) + \frac{f}{N} }{1 - \frac{f}{N} \tan( \theta) } ⟹

Rg

v

2

=

1−

N

f

tan(θ)

tan(θ)+

N

f

For max speed : f = \mu_{s}Nμ

s

N

\rm \implies\: \dfrac{ {(v_{max})}^{2} }{Rg} = \dfrac{ \tan( \theta) + \frac{f}{N} }{1 - \frac{f}{N} \tan( \theta) } ⟹

Rg

(v

max

)

2

=

1−

N

f

tan(θ)

tan(θ)+

N

f

\rm \implies\: \dfrac{ {(v_{max})}^{2} }{Rg} = \dfrac{ \tan( \theta) + \mu_{s} }{1 - \mu_{s} \tan( \theta) } ⟹

Rg

(v

max

)

2

=

1−μ

s

tan(θ)

tan(θ)+μ

s

\rm \implies\: {(v_{max})}^{2} = Rg \bigg[\dfrac{ \tan( \theta) + \mu_{s} }{1 - \mu_{s} \tan( \theta) } \bigg]⟹(v

max

)

2

=Rg[

1−μ

s

tan(θ)

tan(θ)+μ

s

]

\rm \implies\: v_{max} = \sqrt{ Rg \bigg[\dfrac{ \tan( \theta) + \mu_{s} }{1 - \mu_{s} \tan( \theta) } \bigg]}⟹v

max

=

Rg[

1−μ

s

tan(θ)

tan(θ)+μ

s

]

Hope It Helps