Vehicle on a level circular track — simultaneous slip and overturn: Let μ be the coefficient of friction. A wheeled vehicle on a level circular track will slip and overturn at the same limiting condition if the ratio of wheel base distance (track width) to the height of its centroid is equal to what?

Introduction / Context:
While cornering on a level circular track, a vehicle may either skid (slip) due to inadequate lateral friction or overturn due to the resultant line of action shifting beyond the outer wheel. The limiting design often checks both criteria simultaneously to determine safe speed and proportions.

Given Data / Assumptions:

• Level circular track, no superelevation.
• Coefficient of friction μ.
• Track width = distance between wheel contact lines = 2b (so half-track b).
• Height of centroid (centre of gravity) above ground = h.

Concept / Approach:

Skid (slip) limit occurs when lateral acceleration a_y satisfies a_y / g = μ. Overturn limit occurs when the overturning moment due to lateral inertia equals restoring moment: (m * a_y * h) = (m * g * b) → a_y / g = b / h. Simultaneous slip and overturn means these two nondimensional ratios are equal.

Step-by-Step Solution:

Verification / Alternative check:

Express in terms of v^2 / (g r). At both limits v^2 / (g r) equals μ and equals b / h; equality confirms the ratio above.

Why Other Options Are Wrong:

μ (a) omits the factor 2 from wheel distance = 2b; 3μ (c), μ/2 (d), and 1/μ (e) have no basis in the moment balance with the standard track-width definition.

Common Pitfalls:

Using half-track instead of full wheel distance in the requested ratio; confusing skid and overturn criteria; neglecting that the question asks for the ratio using the full wheel distance.

Final Answer:

2μ