Question

derive an expression for maximum safe speed of a vehicle on a level road and express the symbol​

Answer 1

Explanation:

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Answer 2

Explanation:

For the vehicle to go round the curved track at a reasonable speed without skidding, the greater centripetal force is managed for it by raising the outer edge of the track a little above the inner edge. It is called banking of circular tracks.

Consider a vehicle of weight Mg, moving round a curved path of radius r, with a speed v, on a road banked through angleθ.

• The vehicle is under the action of the following forces:  
• The weight Mg acting vertically downwards

The reaction R of the ground to the vehicle, acting along the normal to the banked road OA in the upward direction

The vertical component R cos θ of the normal reaction R will balance the weight of the vehicle and the horizontal component R sin θ will provide the necessary centripetal force to the vehicle. Thus,

R cosθ = Mg …(i)

R sinθ = $\frac{Mv^2}{r}$ …(ii)

On dividing equation (ii) by equation (i), we get

$\frac{rsin\theta}{rcos\theta} = \frac{\frac{Mv^2 }{2}}{Mg}$

$tan\theta = \frac{v^2}{rg}$

As the vehicle moves along the circular banked road OA, the force of friction between the road and the tyres of the vehicle, F = μR, acts in the direction AO.

The frictional force can be resolved into two components:

μ R sinθ in the downward direction

μ R cosθ in the inward direction

Since there is no motion along the vertical,

R cos θ = Mg + μ R sinθ ……. (iii)

Let $v_m_a_x$ be the maximum permissible speed of the vehicle. The centripetal force is now provided by the components R sinθ and μ Mg cosθ, i.e.,  

R sin θ + μ R cosθ = $\frac{Mv_m_a_x}{r}$

From equation(iii),we have  

Mg = R cosθ (1−μ tanθ)…(v)  

Again from equation (iv), we have  

$\frac{Mv^2_m_a_x}{r}$= R cosθ (μ + tanθ) …(vi)  

On dividing equation (iv) by (v), we have

$\frac{v^2_m_a_x}{gr} = \frac{\mu + tan\theta}{1 - tan\theta}$

$v_m_a_x = \sqrt{ \frac{gr(\mu + tan\theta)}{1 - \mu tan\theta}$