Question

A curve in a road forms an arc of radius $800\text{m}$ If the road is $39.2\text{m}$ wide, calculate the safe speed for turning, if the outer edge of the road is $0.5\text{m}$ higher than the inner edge. Take $g=9.8{\text{m/s}}^2$.

• A. $5\text{m/s}$
• B. $10\text{m/s}$
• C. $15\text{m/s}$
• D. $20\text{m/s}$

Answer 1

The correct option is B $10\text{m/s}$

Here, radius of circular arc $r=800\text{m}$, Distance between the edges of the road i.e $AB=39.2\text{m}$
Height of the outer edge with respect to the inner edge $h=0.5\text{m}$.


Applying the condition of banking, for maximum safe speed ($v$) :
$\tan \theta =\frac{v^2}{rg}.....\left(i\right)$

From triangle $ABC$,
$\sin \theta =\frac{0.5}{39.2}=0.012\tan \theta =\frac{BC}{AC}=\frac{0.5}{\sqrt{{39.2}^2-{0.5}^2}}=0.012$
We can see that values of $\sin \theta$ and $\tan \theta$ are very small.

For very small angle:
$\tan \theta \simeq \theta \simeq \sin \theta =\frac{BC}{AB}...\left(ii\right)$

From Eq.($i$) and Eq. ($ii$)
$\Rightarrow \frac{v^2}{rg}=\frac{BC}{AB}=\frac{h}{AB}$
$\therefore v=\sqrt{\frac{h\times r\times g}{AB}}=\sqrt{\frac{0.5\times 800\times 9.8}{39.2}}$
or, $v=10\text{m/s}$