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)

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

Preparation Products

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions