Skidding on a level circular path A vehicle moves on a level circular track. Skidding can be avoided if the available frictional force between the tyres and the road is __________ the required centrifugal (outward) force for the turn.

Introduction / Context:Cornering safety depends on whether tyre–road friction can supply the lateral (centripetal) force needed for circular motion. If insufficient, the vehicle skids outward.

Given Data / Assumptions:

• Level road (no banking), so lateral force is provided solely by friction.
• Centripetal force required: m * v^2 / r directed toward the centre of the curve.
• Maximum available static friction: F_max = μ_s * N = μ_s * m * g.

Concept / Approach:For no skidding, the maximum available static friction must meet or exceed the required centripetal force. If the requirement exceeds what friction can provide, tyres will slide, producing skid.

Step-by-Step Solution:

Verification / Alternative check:If μ_s * g = v^2 / r exactly, the tyres are at the limit of adhesion (impending skid). If μ_s * g is larger, there is a safety margin and no skidding occurs.

Why Other Options Are Wrong:

• Less than / much less than: Guarantees skidding because friction cannot supply the needed centripetal force.
• Equal to: Only the limiting case; while it avoids skidding, the general safe condition is 'greater than or equal to'.
• Greater than: True but excludes the limiting (equality) case where skidding just does not occur.

Common Pitfalls:Confusing 'centripetal' and 'centrifugal' descriptions. In vehicle dynamics we require sufficient inward (centripetal) frictional force; outward 'centrifugal' is an apparent effect in the rotating frame.

Final Answer:greater than or equal to