A car moves along a circular track of radius | vedclass

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

Question

(D) For a car moving on a banked circular track with friction,the maximum speed $v_{max}$ is given by the formula:$v_{max} = \sqrt{gR \left( \frac{\mu

Vedclass · Accepted Answer

None

Vedclass · Answer

(D) For a car moving on a banked circular track with friction,the maximum speed $v_{max}$ is given by the formula:$v_{max} = \sqrt{gR \left( \frac{\mu + 	an 	heta}{1 - \mu 	an 	heta} ight)}$Given the banking angle $	heta = 30^o$,we know that $	an 30^o = \frac{1}{\sqrt{3}}$.Substituting this into the formula:$v_{max} = \sqrt{gR \left( \frac{\mu + \frac{1}{\sqrt{3}}}{1 - \mu \frac{1}{\sqrt{3}}} ight)}$Multiplying the numerator and denominator inside the square root by $\sqrt{3}$:$v_{max} = \sqrt{gR \left( \frac{\mu \sqrt{3} + 1}{\sqrt{3} - \mu} ight)}$Comparing this with the given options,none of the provided expressions match this result. Therefore,the correct choice is $D$.

$A$ car moves along a circular track of radius $R$ banked at an angle of $30^o$ to the horizontal. The coefficient of static friction between the wheels and the track is $\mu$. The maximum speed with which the car can move without skidding out is

Similar Questions

If the coefficient of friction between the tires and the road is $\mu$,the maximum safe speed is $10\;m/s$. If the coefficient of friction becomes $\mu' = \frac{\mu}{2}$,what will be the new maximum safe speed?

$A$ cyclist on a level road takes a sharp circular turn of radius $3 \; m$ $(g = 10 \; m \cdot s^{-2})$. If the coefficient of static friction between the cycle tyres and the road is $0.2$,at which of the following speeds will the cyclist not skid while taking the turn?

$A$ cyclist is travelling with velocity $v$ on a curved road of radius $R$. The angle $\theta$ through which the cyclist leans inwards is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module