A modern grand - prix racing car of mass $m$ is traveling on a flat track in a circular arc of radius R with a speed u . If the coefficient of static between the tyres and the track is $\mu_{s}$, then the magnitude of negative lift $F_{L}$ acting downwards on the car is : (Assume forces on the four tyres are identical and $\mathrm{g}=$ acceleration due to gravity)

(A) $m\left(\frac{v^{2}}{\mu_{s} R}-g\right)$

(B) $-m\left(g+\frac{v^{2}}{\mu_{s} R}\right)$

(C) $m\left(g-\frac{v^{2}}{\mu_{s} R}\right)$

(D) $m\left(\frac{v^{2}}{\mu_{s} R}+g\right)$

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Laws of Motion

A modern grand - prix racing car of mass

m

is traveling on a flat track in a circular arc of radius R with a speed u . If the coefficient of static between the tyres and the track is

μ_{s}

, then the magnitude of negative lift

F_{L}

acting downwards on the car is :
(Assume forces on the four tyres are identical and

g =

acceleration due to gravity)

Answer: (A)

The track is flat, so the normal reaction supports the weight together with the downward aerodynamic lift:

N = m g + F_{L}

Friction supplies the centripetal force, and at the limiting speed it takes its largest value

μ_{s} N

μ_{s} (m g + F_{L}) = \frac{m v ^{2}}{R}

Solving for the lift,

F_{L} = m (\frac{v ^{2}}{μ _{s} R} - g)

The downward lift has magnitude

m (\frac{v ^{2}}{μ _{s} R} - g)

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Single correct option