Defined a vehicle can be parked on a slope.
The velocity of the vehicle on a circular balanced road is given by,
$v_{\max }=\left[r g\left(\frac{\mu_{s}+\tan \theta}{1-\mu_{s} \tan \theta}\right)\right]^{\frac{1}{2}}$
Here we have to take $\mu_{s}=0$ as the surface is smooth.
$v_{\max }=\left[\operatorname{rg}\left(\frac{0+\tan \theta}{1-0}\right)\right]^{\frac{1}{2}}$
$\therefore v_{\max } =\left[r g\left(\frac{\tan \theta}{1}\right)\right]^{-}$
$\therefore v_{\max } =\sqrt{r g \tan \theta}$
At this speed, frictional force is not needed at all to provide the necessary centripetal force. Driving at this speed on a banked road will cause little wear and tear on the tyres. $v_{0}$ is called the optimum speed.
A modern grand-prix racing car of mass $m$ is travelling on a flat track in a circular arc of radius $R$ with a speed $v$. If the coefficient of static friction 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 $g =$ acceleration due to gravity)
A motor cyclist moving with a velocity of $72\, km/hour$ on a flat road takes a turn on the road at a point where the radius of curvature of the road is $20$ meters. The acceleration due to gravity is $10 m/sec^2$. In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than
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A particle is describing circular motion in a horizontal plane in contact with the smooth inside surface of a fixed right circular cone with its axis vertical and vertex down. The height of the plane of motion above the vertex is $h$ and the semivertical angle of the cone is $\alpha $ . The period of revolution of the particle
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