(D) The necessary centripetal force for circular motion is provided by the static friction between the road and the tyres.The condition for the car no

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(D) The necessary centripetal force for circular motion is provided by the static friction between the road and the tyres.The condition for the car not to skid is $f \leq \mu_s N$.Since the track is level,the normal force $N = mg$.The centripetal force required is $F_c = \frac{mv^2}{R}$.Equating the maximum friction to the centripetal force: $\mu_s mg = \frac{mv_{max}^2}{R}$.Solving for $v_{max}$:$v_{max}^2 = \mu_s Rg$$v_{max} = \sqrt{\mu_s Rg}$.

Class 11 Physics Laws of Motion Circular motion with Friction

Medium

$A$ car of mass $m$ is moving on a level circular track of radius $R.$ If $\mu_s$ represents the static friction between the road and tyres of the car,the maximum speed of the car in circular motion is given by

AIPMT 2012 All AIPMT

A
$\mu_s Rg$
B
$Rg\sqrt{\mu_s}$
C
$\mu_s\sqrt{Rg}$
D
$\sqrt{\mu_s Rg}$

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$A$ car of mass $m$ is moving on a level circular track of radius $R.$ If $\mu_s$ represents the static friction between the road and tyres of the car,the maximum speed of the car in circular motion is given by

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$A$ motorcyclist of mass $m$ is to negotiate a curve of radius $r$ with a speed $v$. The minimum value of the coefficient of friction so that the negotiation may take place safely is

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