Question

For traffic moving at $60\mspace{6mu} km/hr$ along a circular track of radius $0.1\mspace{6mu} km$, the correct angle of banking is

• A. $\frac{(60)^{2}}{0.1}$
• B. $\tan^{- 1}\left\lbrack \frac{(50/3)^{2}}{100 \times 9.8} \right\rbrack$
• C. $\tan^{- 1}\left\lbrack \frac{100 \times 9.8}{(50/3)^{2}} \right\rbrack$
• D. $\tan^{- 1}\sqrt{60 \times 0.1 \times 9.8}$

Answer 1

Answer: (B)

$\tan^{- 1}\left\lbrack \frac{(50/3)^{2}}{100 \times 9.8} \right\rbrack$

Answer 2

b)

Sol. $v = 60km/hr = \frac{50}{3}m/s$, $r = 0.1km = 100m,$

$g = 9.8m/s^{2}$ (given)

Angle of banking $\tan\theta = \frac{v^{2}}{rg}$or$\theta = \tan^{- 1}\left( \frac{v^{2}}{rg} \right)$

$= \tan^{- 1}\left\lbrack \frac{(50/3)^{2}}{100 \times 9.8} \right\rbrack$