# The circular motion of a can on a banked road.

Consider a car of weight mg going along a curved path of radius r with speed v on a road banked at an angle \theta. The forces acting on the vehicle are

(1) Weight mg acting vertically downwards.

(2) Normal reaction R of the road acting at an angle \theta with the vertical.

(3) Force of friction f acting downwards along the inclined plane.

Equating the forces along horizontal and vertical directions respectively, we get

\;\;\;\;R\;\sin\left(\theta\right)+f\;\cos\left(\theta\right)=\frac{mv^2}r\;\;\;\;\;\;\;\;\;……….(1)

mg+f\;\sin\left(\theta\right)=R\;\cos\left(\theta\right), where f=\mu\;R

or, R\;\cos\left(\theta\right)-f\;\sin\left(\theta\right)=mg\;\;\;\;\;\;\;\;\;……….(2)

Dividing equation (1) by equation (2), we get

\frac{R\;\sin\left(\theta\right)+f\;\cos\left(\theta\right)}{R\;\cos\left(\theta\right)-f\;\sin\left(\theta\right)}=\frac{v^2}{rg}

Dividing numerator and denominator of L.H.S. by R\;\cos\left(\theta\right), we get

\frac{\tan\left(\theta\right)+{\displaystyle\frac fR}}{1-{\displaystyle\frac fR}\tan\left(\theta\right)}=\frac{v^2}{rg}

or, \frac{\tan\left(\theta\right)+\mu}{1-\mu\;\tan\left(\theta\right)}=\frac{v^2}{rg}\;\;\;\;\;\;\;\;\;\;\;\;\;\left[\because\mu=\frac fR\right]

or, v^2=rg\left[\frac{\tan\left(\theta\right)+\mu}{1-\mu\;\tan\left(\theta\right)}\right]

or, v=\sqrt{rg.\frac{\tan\left(\theta\right)+\mu}{1-\mu\;\tan\left(\theta\right)}}

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