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 =  (  / 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                                      

Answer 2

Given :

Radius of curvature of the track = 72 m

Maximum speed = 216 kmph = 60 m/s

Width of the track = 10 m

To Find :

The angle that the road should be tilted

The height of the outer edge of the track

Solution :

• The angle of banking of the road is given by

             $Tan\theta =\frac{ v^{2} }{rg}$

             $\theta=Tan^{-1}(\frac{v^{2}}{rg})$

             $\theta=Tan^{-1}(\frac{60^{2}}{72\times10})$

             $\theta =78.69^{0}$

• The height of the outer edge of the road is given by

            $Sin\theta = \frac{h}{w}$

            $h=wSin\theta$

            $h=10\times Sin(78.69)$

            $h = 9.805 m$  

 The angle that the road should be tilted is 78.69°.

 The height of the outer edge is 9.805 m.