Question

A circular frictionless track having radius of 150 m is to be designed for vehicles at an average speed of 30 m/s. The ideal angle of bank is (g = 10 m/s)​

Answer 1

SOLUTION

GIVEN

A circular frictionless track having radius of 150 m is to be designed for vehicles at an average speed of 30 m/s.

TO DETERMINE

The ideal angle of bank

EVALUATION

Here by the given -

v = Speed = 30 m/s

r = Radius = 150 m

g = 10 m/s

Let the ideal angle of bank = θ

Thus we get

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

$\displaystyle \sf{ \implies \: \tan \theta = \frac{ {30}^{2} }{150 \times 10} }$

$\displaystyle \sf{ \implies \: \tan \theta = \frac{ 900 }{1500} }$

$\displaystyle \sf{ \implies \: \tan \theta = 0.6}$

$\displaystyle \sf{ \implies \: \tan \theta = \frac{3 }{5} }$

$\displaystyle \sf{ \implies \:\theta = { \tan}^{ - 1} \bigg( \frac{3 }{5} \bigg) }$

$\displaystyle \sf{ \implies \:\theta = {30.96}^{ \circ} }$

FINAL ANSWER

The ideal angle of bank = 30.96°

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Answer 2

Given info : A circular frictionless track having radius of 150 m is to be designed for vehicles at an average speed of 30 m/s.

To find : the ideal angle of bank is ...

solution : see the diagram, here you see that a vehicle moving along the road is inclined at θ with the horizontal direction.

let N is the normal reaction acting on the vehicle.

component of normal reaction along horizontal direction = Nsinθ

component of normal reaction along vertical direction = Ncosθ

at equilibrium,

  horizontal force = Nsinθ = mv²/r ...(1)

vertical force = Ncosθ = mg ...(2)

from equations (1) and (2) we get,

tanθ = v²/rg

here, v = 30 m/s , r = 150 m and g = 10 m/s²

so, tanθ = (30)²/(150 × 10) = 3/5 ⇒ θ = tan⁻¹ (3/5) ≈ 31°

therefore the ideal angle of bank is 31°