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A Turn of Radius 20 M is Banked for the Vehicles Going at a Speed of 36 Km/H. If the Coefficient of Static Friction Between the Road and the Tyre is 0.4, - Physics

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Question

A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Solution

Given:
Speed of vehicles = v = 36 km/hr = 10 m/s
Radius = r = 20 m 
Coefficient of static friction = μ = 0.4
Let the road be banked with an angle \[\theta\]

We have :

\[\theta = \tan^{- 1} \frac{\text{v}^2}{\text{rg}}\]

\[ = \tan^{- 1} \frac{100}{20 \times 10}\]

\[ = \tan^{- 1} \left( \frac{1}{2} \right)\]

\[ \Rightarrow \text{tan } \theta = 0 . 5\]

When the car travels at the maximum speed, it slips upward and μN1 acts downward.
Therefore we have :

\[\text{ N}_1 - \text{mg}\cos\theta - \frac{\text{mv}_1^2}{r}\sin\theta = 0 . . . \left( \text{i} \right)\]

\[\mu \text{N}_1 + \text{mg }\sin\theta - \frac{\text{mv}_1^2}{r}\cos\theta = 0 . . . \left(\text{ ii} \right)\]

On solving the above equations, we get :

\[\text{v}_1 = \sqrt{\text{rg}\frac{\mu + \tan\theta}{1 - \mu\tan\theta}}\]

\[ = \sqrt{20 \times 10 \times \frac{0 . 9}{0 . 8}}\]

\[ = 15 \text{ m/s = 54 km/hr }\]

Similarly, for the other case, it can be proved that :

\[\text{v}_2 = \sqrt{\text{rg}\frac{\text{tan}\theta - \mu}{\sqrt{1 - \mu \text{ tan }\theta}}}\]

\[ = \sqrt{20 \times 10 \times \frac{0 . 1}{1 . 2}}\]

\[ = 4 . 08 \text{ m/s = 14 . 7 km/hr}\]

Thus, the possible speeds are between 14.7 km/hr and 54 km/hr so that the car neither slips down nor skids up.

  Is there an error in this question or solution?
Solution A Turn of Radius 20 M is Banked for the Vehicles Going at a Speed of 36 Km/H. If the Coefficient of Static Friction Between the Road and the Tyre is 0.4, Concept: Circular Motion.
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