Question

A racing track of curvature 9.9 m is banked at tan−1 0.5 . Coefficient of static friction between the track and the tyres of a vehicle is 0.2. Determine the speed limits with 10 % margin

Answer 1

This is the answer of above mentioned question

Answer 2

The speed limit is 5.196 m/s < v < 7.896 m/s

The curvature of the racing track (r) is = 9.9 m

The banking is = tan⁻¹ 0.5

∴ tan θ = 0.5

The friction between the track and the tyre is (μ) = 0.2

From the formula of minimum speed, we get,

v₍min₎ = $\sqrt{rg(\frac{tan\theta - \mu}{1+\mu tan \theta}) }$

∴ v₍min₎ = √27 [By putting the values from above]

⇒ v₍min₎ = 5. 196 m/s

Give, the speed limits is within 10% margin.

So, the minimum speed can be 10% higher than the calculated speed.

So, final v₍min₎ = 5.196 × (110/100) m/s

∴ v₍min final₎ = 5.716 m/s

From the formula of maximum velocity in a banking road, we get,

$v_{max} = \sqrt{rg(\frac{tan\theta + \mu}{1-\mu tan \theta}) }$

∴ v₍max₎ = √77 m/s

⇒  v₍max₎ = 8.775 m/s

Give, the speed limits is within 10% margin.

So, the maximum speed can be 10% lower than the calculated speed.

So,

v₍max final₎ = 8.775 × (90/100) m/s

⇒ v₍max final₎ = 7.896 m/s

So, the speed limit is 5.196 m/s < v < 7.896 m/s