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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:

  • Option 1)

    \sqrt {{{\text{m}\,\text{Rg}} \mathord{\left/ {\vphantom {{\text{m}\,\text{Rg}} {\mu _\text{s} }}} \right. \kern-\nulldelimiterspace} {\mu _\text{s} }}}

  • Option 2)

     

    \sqrt {{{\text{Rg}} \mathord{\left/ {\vphantom {{\text{Rg}} {\mu _\text{s} }}} \right. \kern-\nulldelimiterspace} {\mu _\text{s} }}}

  • Option 3)

    \sqrt{\mu_{s}Rg}

  • Option 4)

     

     

    \sqrt {\mu _\text{s} \,\text{Rg}}

 

Answers (1)

best_answer

As we learnt in

Skidding of object on a Rotating Platform -

Centripetal force leq Force of friction

momega^{2}rleqmu mg

herefore omega_{max}=sqrt{(mu g/r)}

omega=Angular velocity

r = radius

mu= coefficient of friction

- wherein

It is the maximum velocity of rotation of the platform, so that object will not skid on it.

 

 

 

 

 

In given condition

Static frictional force =Fc

\mu _{s}mg=\frac{MV^{2}}{R}

V_{max }=\sqrt{\mu _{s}Rg}


Option 1)

\sqrt {{{\text{m}\,\text{Rg}} \mathord{\left/ {\vphantom {{\text{m}\,\text{Rg}} {\mu _\text{s} }}} \right. \kern-\nulldelimiterspace} {\mu _\text{s} }}}

Incorrect

Option 2)

 

\sqrt {{{\text{Rg}} \mathord{\left/ {\vphantom {{\text{Rg}} {\mu _\text{s} }}} \right. \kern-\nulldelimiterspace} {\mu _\text{s} }}}

Incorrect

Option 3)

\sqrt{\mu_{s}Rg}

Correct

Option 4)

 

 

\sqrt {\mu _\text{s} \,\text{Rg}}

Correct

Posted by

Aadil

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