Question:medium

A cyclist speeding at a velocity \(v\) on a level road takes a sharp circular turn of radius R. If \(\mu\) is the static friction between the tyres and road, then the condition for the cyclist not to slip is

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This formula tells you the maximum safe speed for a turn: \(v_{max} = \sqrt{\mu Rg}\).
Slowing down (lower \(v\)) or increasing the radius (\(R\)) makes the turn safer.
Updated On: Apr 29, 2026
  • \(v^2 \ge \mu R\)
  • \(v^2 \le \mu Rg\)
  • \(v \le \mu Rg\)
  • \(v = \frac{\mu R}{g}\)
  • \(v^2 \ge \frac{\mu R}{g}\)
Show Solution

The Correct Option is B

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