Question

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

• A. \(v^2 \ge \mu R\)
• B. \(v^2 \le \mu Rg\)
• C. \(v \le \mu Rg\)
• D. \(v = \frac{\mu R}{g}\)
• E. \(v^2 \ge \frac{\mu R}{g}\)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When a vehicle takes a turn on a level (unbanked) road, the necessary centripetal force is provided by the friction between the tires and the road surface.

Step 2: Key Formula or Approach:
For no slipping:
Centripetal Force \(\le\) Maximum Static Frictional Force
\[ \frac{mv^2}{R} \le f_s(\text{max}) \]

Step 3: Detailed Explanation:
On a level road, the normal force \(N = mg\).
The maximum static friction is \(f_s(\text{max}) = \mu N = \mu mg\).
Substituting this into the condition:
\[ \frac{mv^2}{R} \le \mu mg \]
Divide both sides by mass (\(m\)):
\[ \frac{v^2}{R} \le \mu g \]
Rearrange to solve for \(v^2\):
\[ v^2 \le \mu Rg \]
Alternatively, taking the square root: \(v \le \sqrt{\mu Rg}\).

Step 4: Final Answer:
The condition for not slipping is \(v^2 \le \mu Rg\).