Question

The coefficient of static friction between the road and tyres of a car is \(0.4\). The maximum permissible speed of the car is \(10\text{ ms}^{-1}\) on a curved unbanked road. Then the maximum radius of curvature of the road is
\[ \text{(acceleration due to gravity }=10\text{ ms}^{-2}\text{)} \]

• A. \(10\sqrt{5}\text{ m}\)
• B. \(25\text{ m}\)
• C. \(20\sqrt{2}\text{ m}\)
• D. \(30\text{ m}\)

Hint:

For motion on a curved unbanked road, \[ \frac{mv^2}{r}=\mu mg \] because static friction provides the necessary centripetal force.

Answer 1

The Correct Option is B

Step 1: Identify the source of centripetal force on an unbanked road.
On a flat (unbanked) curved road, static friction is the only horizontal force and provides the centripetal acceleration. Beyond the maximum safe speed, friction is insufficient and the car skids.
Step 2: Set centripetal force equal to maximum static friction.
At maximum permissible speed: $\frac{mv^2}{r} = \mu_s mg$. The mass $m$ cancels.
Step 3: Rearrange for minimum radius $r$.
\[ r = \frac{v^2}{\mu_s g} \]
Step 4: Substitute given values.
$v = 10$ ms$^{-1}$, $\mu_s = 0.4$, $g = 10$ ms$^{-2}$: \[ r = \frac{100}{0.4 \times 10} = \frac{100}{4} = 25 \text{ m} \]
Step 5: Interpret physically.
Any curve with radius less than 25 m at this speed would demand more friction than available, causing skidding.
Step 6: State the final answer.
\[ \boxed{r = 25 \text{ m}} \]