Question

A motor cyclist has to rotate in horizontal circles inside the cylindrical wall of inner radius 'R' metre. If the coefficient of friction between the wall and the tyres is '$\mu_s$', then the minimum speed required is ( g = acceleration due to gravity)

• A. $\sqrt{\mu_s Rg}$
• B. $\sqrt{\frac{Rg}{\mu_s}}$
• C. $\sqrt{\frac{\mu_s}{Rg}}$
• D. $\sqrt{\frac{R^2 g}{\mu_s}}$

Hint:

In a vertical cylinder, the normal force acts horizontally as the centripetal force, while friction acts vertically to oppose gravity.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
In a "Death Well" (cylindrical wall), a motorcyclist moves in horizontal circles. To prevent the rider from falling down, the frictional force must balance the gravitational weight.
Step 2: Key Formula or Approach:
The normal reaction $N$ from the wall provides the centripetal force:
\[ N = \frac{mv^2}{R} \]
The frictional force $f_s$ must balance the weight $mg$:
\[ f_s = mg \]
Also, for the rider not to slip, $f_s \leq \mu_s N$.
Step 3: Detailed Explanation:
Substituting the expressions into the inequality:
\[ mg \leq \mu_s \left(\frac{mv^2}{R}\right) \]
Cancelling mass $m$ from both sides and rearranging for velocity $v$:
\[ g \leq \frac{\mu_s v^2}{R} \implies v^2 \geq \frac{Rg}{\mu_s} \]
\[ v \geq \sqrt{\frac{Rg}{\mu_s}} \]
Thus, the minimum speed required is $v_{min} = \sqrt{\frac{Rg}{\mu_s}}$.
Step 4: Final Answer:
The minimum speed required is $\sqrt{\frac{Rg}{\mu_s}}$.