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Laws Of Motion

Question
CBSEENPH11020513

A car is negotiating a curved road of radius R. The road is banked at angle straight theta. The coefficient of friction between the tyres of the car and the road is straight mu subscript straight s. The maximum safe velocity on this road is,

  • square root of gR open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tan space straight theta end fraction close parentheses end root
  • square root of straight g over straight R open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tanθ end fraction close parentheses end root
  • square root of straight g over straight R squared open parentheses fraction numerator mu subscript s plus tan theta over denominator 1 minus mu subscript s tan theta end fraction close parentheses end root
  • square root of g R squared open parentheses fraction numerator straight mu subscript straight s plus tanθ over denominator 1 minus straight mu subscript straight s tanθ end fraction close parentheses end root

Solution

A.

square root of gR open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tan space straight theta end fraction close parentheses end root

A car is negotiating a curved road of radius R. The road is banked at angle straight theta and the coefficient of friction between the tyres of car and the road is straight mu subscript straight s.
The given situation is illustrated as:

In the case of vertical equilibrium,
N cos straight theta = mg + f1 sin straight theta
rightwards double arrowmg = N cos straight theta space minus space straight f subscript 1 space sin space straight theta    ... (i)
In the case of horizontal equilibrium,
straight N space sin space straight theta space plus space straight f subscript 1 space cosθ space equals space mv squared over straight R  ... (ii)
Dividing Eqns. (i) and (ii), we get
straight v squared over Rg equals fraction numerator sin space theta space plus space mu subscript s space cos theta over denominator cos space theta space minus space mu subscript s space sin space theta end fraction space open square brackets f subscript 1 proportional to mu subscript s close square brackets

rightwards double arrow v space equals square root of R g open parentheses fraction numerator tan space theta space plus mu subscript s over denominator cos space theta space minus space mu subscript s space sin space theta space end fraction close parentheses end root
rightwards double arrow space v space equals space square root of R g open parentheses fraction numerator tan space theta space plus space mu subscript s over denominator 1 minus mu subscript s space tan space theta end fraction close parentheses end root