A modern grand-prix racing car of mass m is t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a car moving in a circular path on a flat track,the centripetal force is provided by the static friction force $f_{s}$.The maximum static fric

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For a car moving in a circular path on a flat track,the centripetal force is provided by the static friction force $f_{s}$.The maximum static friction force is given by $f_{s,max} = \mu_{s} N$,where $N$ is the normal force.The centripetal force required is $F_{c} = \frac{mv^{2}}{R}$.Thus,$\mu_{s} N = \frac{mv^{2}}{R}$,which gives $N = \frac{mv^{2}}{\mu_{s} R}$.The normal force $N$ on the car is the sum of its weight $mg$ and the downward negative lift force $F_{L}$ (aerodynamic downforce).Therefore,$N = mg + F_{L}$.Substituting the expression for $N$,we get $\frac{mv^{2}}{\mu_{s} R} = mg + F_{L}$.Solving for $F_{L}$,we get $F_{L} = \frac{mv^{2}}{\mu_{s} R} - mg = m \left(\frac{v^{2}}{\mu_{s} R} - g\right)$.

Similar Questions

The maximum tension which an inextensible ring of mass $0.1\, kg/m$ can bear is $10\,N$. The maximum velocity in $m/s$ with which it can be rotated is ........ $m/s.$

$A$ car sometimes overturns while taking a turn. When it overturns,which wheel leaves the ground first?

$A$ cyclist riding a bicycle at a speed of $14 \sqrt{3} \, m/s$ takes a turn around a circular road of radius $20 \sqrt{3} \, m$ without skidding. What is his inclination to the vertical (in $^{\circ}$)?

$A$ small object placed on a rotating horizontal turntable just slips when it is placed at a distance $4\, cm$ from the axis of rotation. If the angular velocity of the turntable is doubled,the object slips when its distance from the axis of rotation is (in $, cm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module