Question

A motorcyclist rides in a horizontal circle about a central vertical axis inside a cylindrical chamber of radius \(r\). If the coefficient of friction between the tyres and the inner surface of the chamber is \(\mu\), the minimum speed of motorcyclist to prevent him from skidding is (\(g\) = acceleration due to gravity)

• A. \( \sqrt{\dfrac{\mu g}{r}} \)
• B. \( \sqrt{\dfrac{r\mu}{g}} \)
• C. \( \sqrt{\dfrac{g}{r\mu}} \)
• D. \( \sqrt{\dfrac{rg}{\mu}} \)

Hint:

In circular motion on vertical walls, friction balances weight while normal reaction provides centripetal force.

Answer 1

The Correct Option is D

Step 1: Forces acting on the motorcyclist.
The normal reaction \(N\) from the wall provides the centripetal force, while friction acts vertically upward to balance the weight.
Step 2: Writing force equations.
Centripetal force: \[ N = \frac{mv^2}{r}. \] Limiting friction to prevent slipping: \[ f_{\max} = \mu N. \]
Step 3: Condition for no skidding.
For vertical equilibrium, \[ \mu N = mg. \] Substituting \(N = \frac{mv^2}{r}\), \[ \mu \frac{mv^2}{r} = mg. \]
Step 4: Solving for minimum speed.
\[ v^2 = \frac{rg}{\mu} \quad \Rightarrow \quad v_{\min} = \sqrt{\frac{rg}{\mu}}. \]
Step 5: Conclusion.
The minimum speed required to prevent skidding is \[ \sqrt{\dfrac{rg}{\mu}}. \]