Question

A car of 800 kg is taking a turn on a banked road of radius 300 m and angle of banking 30°. If the coefficient of static friction is 0.2, then the maximum speed with which the car can negotiate the turn safely: (Given $g = 10 \, \text{m/s}^2$, $\sqrt{3} = 1.73$).

• A. 70.4 m/s
• B. 51.4 m/s
• C. 264 m/s
• D. 102.8 m/s

Answer 1

Answer: (B)

51.4 m/s

Answer 2

Given:
$m = 800 \, \text{kg}, \quad r = 300 \, \text{m}, \quad \theta = 30^\circ, \quad \mu = 0.2$
The maximum speed $V_{\text{max}}$ for safe negotiation of the turn is given by:
$V_{\text{max}} = \sqrt{r g \frac{\tan \theta + \mu}{1 - \mu \tan \theta}}$
Substitute the values:
$V_{\text{max}} = \sqrt{300 \times 10 \times \frac{\tan 30^\circ + 0.2}{1 - 0.2 \times \tan 30^\circ}}$
$= \sqrt{300 \times 10 \times \frac{0.57 + 0.2}{1 - 0.2 \times 0.57}}$
$= \sqrt{3000 \times \frac{0.77}{0.886}}$
$36$
$V_{\text{max}} = 51.4 \, \text{m/s}$