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Laws Of Motion

Question
CBSEENPH11017026

A road is 5m wide. Its radius of curvature is 50m. The outer edge is above the inner edge of road by a distance 0-5m. If the coefficient of friction of road is 0.3, find the upper limit of velocity on the road.

Solution

Given that, 
Radius of the road, r equals 50 m
Limiting space friction comma space straight mu space equals space 0.3
Tangent of the angle of banking of road is,

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Now the upper limit of the velocity is,
           space space space space v italic equals square root of r g fraction numerator mu italic plus t a n theta over denominator italic 1 italic minus mu t a n theta end fraction end root 
rightwards double arrow           nu equals square root of 50 cross times 10 fraction numerator 0.3 plus 0.1 over denominator 1 minus 0.3 cross times 0.1 end fraction end root 
So, upper limit of the velocity of road is, v = 14.36 m/s