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Q] Obtain an expression for maximum safety speed with which a vehicle can be safely driven along a curved banked roadAndQ] Show that the angle of banking is independent of mass of vehicle.​

Answer»

Refer to the attachment first for better understanding!

(1)

Net force acting on the vehicle,

\to\sf{N cos \theta = mg + f sin \theta } \\

\to\sf{N cos \theta- f sin \theta  = mg ....(1)} \\

\to\sf{N sin \theta + f cos \theta =\dfrac{ mv^<klux>2</klux> }{ r} ....(2)} \\ \\

Dividing (2) and (1) we get,

\to\sf{\dfrac{N sin \theta + f cos \theta }{N cos \theta  - f sin \theta  } = \dfrac{v^2}{ rg}}\\ \\

Now LET's divide the numerator and denominator of the L.H.S by N cos θ,

\to\sf{\dfrac{\dfrac{N sin \theta }{N cos \theta  } + \dfrac{f cos \theta }{N cos \theta  }}{\dfrac{N cos \theta }{N cos \theta  } - \dfrac{f sin \theta  }{N cos \theta  }} = \dfrac{v^2}{ rg}} \\ \\

\to\sf{\dfrac{{tan \theta  + \dfrac{f}{N} }} {1- \dfrac{f}{N} tan \theta  }= \dfrac{v^2}{ rg}} \\ \\

We know that,

\to\sf{ f = \mu N }\\ \\

\to \sf{\mu = f / N } \\ \\

Hence,

\to\sf{\dfrac{{tan \theta  + \mu }} {1-\mu tan \theta  }= \dfrac{v^2}{ rg}}\\ \\

\to\boxed{\sf{v = \sqrt{rg\bigg(\dfrac{{tan \theta  + \mu }} {1- \mu tan \theta  } \bigg)}}} \\ \\

When there is no FRICTION acting between the road and the tire the coefficient of friction (μ) will be zero.

\to\sf{v = \sqrt{rg\bigg(\dfrac{tan \theta  +0 } {1-0  }\bigg) }}\\ \\

Hence the maximum safety speed with which a vehicle can be safely driven along a curved banked road is,

\to\boxed{\sf{v = \sqrt{rg tan \theta }}}

(2)

On rearranging the above equation we get,

\to\sf{\theta = tan^{-1}\bigg(\dfrac{v^2}{rg}\bigg)}

From the above equation, we can conclude that the angle of BANKING doesn't DEPEND upon the mass of the vehicle.



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