Question 179: Work, Energy and Power & Circular Motion

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 ' $\mu_{\mathrm{s}}$ ', then the minimum speed required is ( $\mathrm{g}=$ acceleration due to gravity)

Accepted Answer

Step 1: Forces on the motorcyclist
Suppose the motorcyclist is moving in a horizontal circle inside the vertical cylindrical wall of radius $R$.
The normal reaction $N$ of the wall acts
horizontally
, towards the center.
The weight $mg$ acts
downward
.
The static friction force $f$ acts
upward
, opposing the weight. For equilibrium in vertical direction, friction provides balance to gravity.
Step 2: Force equations
Vertical balance:
$ f = mg $
But maximum available friction is:
$ f \leq \mu_s N $
Thus the condition:
$ mg \leq \mu_s N $
Horizontal direction (centripetal force):
$ N = \frac{mv^2}{R} $
Step 3: Combine conditions
Substitute $N$:
$ mg \leq \mu_s \cdot \frac{mv^2}{R} $
Cancel $m$:
$ g \leq \mu_s \cdot \frac{v^2}{R} $
So:
$ v^2 \geq \frac{gR}{\mu_s} $
Step 4: Minimum speed
$ v_{\min} = \sqrt{\frac{gR}{\mu_s}} $
Final Answer:
The correct option is:
Option B: $\sqrt{	frac{Rg}{\mu_s}}$
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Answer

$\sqrt{\mu_{\mathrm{r}} \mathrm{Rg}}$

Answer

$\sqrt{\frac{\mathrm{Rg}}{\mu_{\mathrm{s}}}}$

Answer

$\sqrt{\frac{\mu_{\mathrm{s}}}{\mathrm{Rg}}}$

Answer

$\sqrt{\frac{R^2 g}{\mu_s}}$

Work, Energy and Power & Circular Motion

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_{\mathrm{s}} ', then the minimum speed required is ( \mathrm{g}= acceleration due to gravity)

Choose the correct answer: