A car is moving on a circular track banked at | vedclass

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

Question

(D) The maximum speed $V_{\max}$ on a banked road with friction is given by $V_{\max} = \sqrt{\frac{rg(\mu + 	an 	heta)}{1 - \mu 	an 	heta}}$.The

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(D) The maximum speed $V_{\max}$ on a banked road with friction is given by $V_{\max} = \sqrt{\frac{rg(\mu + 	an 	heta)}{1 - \mu 	an 	heta}}$.The optimum speed $V_o$ (where no friction is required) is given by $V_o = \sqrt{rg 	an 	heta}$.Given $	heta = 45^{\circ}$ and $V_{\max} = 2V_o$.Substituting the expressions: $\sqrt{\frac{rg(\mu + 	an 	heta)}{1 - \mu 	an 	heta}} = 2 \sqrt{rg 	an 	heta}$.Squaring both sides: $\frac{\mu + 	an 	heta}{1 - \mu 	an 	heta} = 4 	an 	heta$.Since $	an 45^{\circ} = 1$,we have $\frac{\mu + 1}{1 - \mu} = 4(1)$.$\mu + 1 = 4 - 4\mu$.$5\mu = 3$.$\mu = \frac{3}{5} = 0.6$.

$A$ car is moving on a circular track banked at an angle of $45^{\circ}$. If the maximum permissible speed of the car to avoid slipping is twice the optimum speed of the car to avoid the wear and tear of the tyres,then the coefficient of static friction between the wheels of the car and the road is

Similar Questions

$A$ car when passing through a convex bridge exerts a force on it which is equal to:

$A$ car is driven on a banked road of radius of curvature $20 \ m$ with maximum safe speed. In order to increase its safe speed by $10 \%$,the increase in the radius of curvature will be (Angle of banking and friction is unchanged in both the cases.) (in $m$)

How is centripetal force provided while taking a turn on a level circular track?

The normal reaction $N$ for a vehicle of $800 \, kg$ mass,negotiating a turn on a $30^{\circ}$ banked road at maximum possible speed without skidding is $... \times 10^{3} \, kg \cdot m/s^{2}$. [Given $\cos 30^{\circ} = 0.87, \mu_{s} = 0.2$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module