Question

Part of racing track is to be desinged for a curvature of 72m.we are not recommending the vehicle to drive faster than 216kmph . With what angle should the road be tilled?

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Answer 1

Answer:

 Ө =  tan inverse( 5 )

 Ө = 78.69 degree

 

Explanation:

Given data :-  

radius of curvature r = 72m

vmax = 216 km/hr = 216 x (5/18 ) = 60 m/s

What we have to find out :-

 1) angle should the road be tilted i.e angle of banking = Ө= ?

 

Solution :-

1)The angle of banking is given by,  

 

Ө =  (  v square/ rg )

    = $tan^{-1}$ ( $60^{2}$  / 72 x10 )              g = 10 m/$s^{2}$  

     =  (3600 /720 )

 Ө =  ( 5 )

 Ө = 78.69 degree