Question

Part of a racing track is to be designed
for a curvature of 72 m. We are not
recommending the vehicles to drive
faster than 216 kmph. With what angle
should the road be tilted? By what height
will its outer edge be, with respect to the
inner edge if the track is 10 m wide?

Answer 1

Answer:

Ө =  tan inverse( 5 )

 Ө = 78.69 degree

 h = 9.805 meter      

Explanation:

Given data :-  

radius of curvature r = 72m

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

width of tack = w = 10 m

What we have to find out :-

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

 2) height of outer edge h = ?

 

Solution :-

1)The angle of banking is given by,  

 

Ө =  tan inverse ( v square / rg )

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

     =  (3600 /720 )

 Ө =  ( 5 )

 Ө = 78.69 degree  

2) now,

 h = w (sin Ө)

 h = 10 (sin 78.69 degree  )  

 h = 10 ( 0.9805 )

 h = 9.805 meter