Question

A circular turn on racing track has radius of 150m what must be banking angle if optimum for taking the turn is to be 180km/

Answer 1

This attachment may help u all

Answer 2

Concept Introduction: Banking are done in highways and circular tracks.

Given:

We Have been Given: radius is

$150m$

Speed is

$180km {hr}^{ - 1} = 180 \times \frac{5}{18} = 50m {s}^{ - 1}$

To Find:

We have to Find: Banking Angle.

Solution:

According to the problem, Banking angle is

$\tan( \alpha ) = \frac{ {v}^{2} }{rg}$

therefore putting the values

$\tan( \alpha ) = \frac{50 \times 50}{150 \times 10} \\ \tan( \alpha ) = \frac{25}{15} \\ \tan( \alpha ) = \frac{5}{3} \\ \alpha = { \tan }^{ - 1} (\frac{5}{3} )$

Final Answer: The Banking angle is

$\alpha = { \tan }^{ - 1} ( \frac{5}{3})$

#SPJ2