Answer:[tex]\theta =\tan ^{-1}(\frac{v^2}{gR})[/tex]
Explanation:
let \theta be the inclination at which curve is tilted
v is the speed of car and R is the radius of curve
m is the mass of car
Suppose R is the reaction offered by road to car
Resolving R in x and y direction we get
[tex] R\cos \theta [/tex] will balance weight and [tex]R\sin \theta [/tex] will provide the necessary centripetal Force
thus [tex]R\sin \theta =\frac{mv^2}{R}[/tex] Â ------------1
[tex]R\cos \theta =m g[/tex] Â ----------------2
Divide 1 & 2 we get
[tex]\frac{R\sin \theta }{R\cos \theta }=\frac{mv^2}{mgR}[/tex]
[tex]\tan \theta =\frac{v^2}{gR}[/tex]
[tex]\theta =\tan ^{-1}(\frac{v^2}{gR})[/tex] Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â