Question

The maximum velocity to avoid overturning on a circular track of radius \(r = 100\) m, with coefficient of friction \(\mu = 0.2\), is:

• A. 0.14 m/s
• B. 140 m/s
• C. 1.4 m/s
• D. 14 m/s

Hint:

Use v equals the square root of mu times g times r, since friction supplies the centripetal force.

Answer 1

The Correct Option is D

Step 1: A vehicle moving on a flat circular track needs a centripetal force directed toward the centre of the track to keep turning instead of sliding outward. On a flat track, this force comes only from friction between the tyres and the road.

Step 2: The maximum friction force the road can supply is \(f_{max} = \mu m g\), where \(m\) is the mass of the vehicle and \(g\) is the acceleration due to gravity. The centripetal force needed to move in a circle of radius \(r\) at speed \(v\) is \(F_c = \dfrac{m v^2}{r}\).

Step 3: To just avoid sliding or overturning, the maximum available friction must equal the required centripetal force: \(\mu m g = \dfrac{m v^2}{r}\). The mass \(m\) cancels from both sides, leaving \(v^2 = \mu g r\), so \(v = \sqrt{\mu g r}\).

Step 4: Substitute the values: \(\mu = 0.2\), \(g = 9.8\) m/s^2, \(r = 100\) m. So \(v = \sqrt{0.2 \times 9.8 \times 100} = \sqrt{196} = 14\) m/s.

Answer: 14 m/s, option D.