
A vehicle of mass m is driven along an un-banked curved path of radius of curvature r with a speed v. if μ is necessary minimum coefficient of friction between tyres of vehicle and road so that the vehicle does not skid, then:
$\begin{align}
& A.\mu \propto {{v}^{2}} \\
& B.\mu \propto \dfrac{1}{{{v}^{2}}} \\
& C.\mu \propto r \\
& D.\mu \propto \dfrac{1}{r} \\
\end{align}$
Answer
569.1k+ views
Hint: When the vehicle drives along a curved road it performs circular motion. To do circular motion, centripetal force is necessary for a safe turn; centripetal force must be equal to force of friction. Use this condition and calculate required conditions by keeping speed and radius of curvature separately.
Complete answer:
In this question it is given that a vehicle having mass m is driven along a straight i.e. un-banked curved path which has a radius of curvature r. A vehicle is driving with a speed v. μ is necessary minimum coefficient of friction between tyres of vehicle and the road.
Now we need to find out the condition so that the vehicle does not skid. When a car is moving along a curved road, it is performing a circular motion since the road is curved which is an arc of a Circle centripetal force is a necessary force for circular motion. In case of a vehicle moving along a curved horizontal road, the necessary centripetal force is provided by force of friction between the tyres of the car and the road surface.
Then, for a safe turn,
Centripetal force = force of friction
$\begin{align}
& \dfrac{m{{v}^{2}}}{r}=\mu N \\
& \dfrac{m{{v}^{2}}}{r}=\mu mg \\
\end{align}$
Where, μ is the coefficient of friction between tyres of car and road surface i.e. for not skidding.
$\begin{align}
& \mu mg=\dfrac{m{{v}^{2}}}{r} \\
& \mu =\dfrac{{{v}^{2}}}{rg} \\
\end{align}$
So when we keep road curvature constant i.e. r is constant then we can say that
$\mu \propto {{v}^{2}}$
I.e. μ is directly proportional to square of speed of vehicle now; if we keep speed v constant then $\mu $ is inversely proportional to radius of curvature r.
i.e. $\mu \propto \dfrac{1}{r}$
Therefore the correct option is option A and option D.
So, the correct answer is “Option A and D”.
Note:
Centripetal force is necessary for circular motion and in absence of centripetal force, the vehicle would move along a tangent and circular motion is not possible. This process of raising the outer edge of the road over its inner edge through a certain angle is known as banking of the road.
Complete answer:
In this question it is given that a vehicle having mass m is driven along a straight i.e. un-banked curved path which has a radius of curvature r. A vehicle is driving with a speed v. μ is necessary minimum coefficient of friction between tyres of vehicle and the road.
Now we need to find out the condition so that the vehicle does not skid. When a car is moving along a curved road, it is performing a circular motion since the road is curved which is an arc of a Circle centripetal force is a necessary force for circular motion. In case of a vehicle moving along a curved horizontal road, the necessary centripetal force is provided by force of friction between the tyres of the car and the road surface.
Then, for a safe turn,
Centripetal force = force of friction
$\begin{align}
& \dfrac{m{{v}^{2}}}{r}=\mu N \\
& \dfrac{m{{v}^{2}}}{r}=\mu mg \\
\end{align}$
Where, μ is the coefficient of friction between tyres of car and road surface i.e. for not skidding.
$\begin{align}
& \mu mg=\dfrac{m{{v}^{2}}}{r} \\
& \mu =\dfrac{{{v}^{2}}}{rg} \\
\end{align}$
So when we keep road curvature constant i.e. r is constant then we can say that
$\mu \propto {{v}^{2}}$
I.e. μ is directly proportional to square of speed of vehicle now; if we keep speed v constant then $\mu $ is inversely proportional to radius of curvature r.
i.e. $\mu \propto \dfrac{1}{r}$
Therefore the correct option is option A and option D.
So, the correct answer is “Option A and D”.
Note:
Centripetal force is necessary for circular motion and in absence of centripetal force, the vehicle would move along a tangent and circular motion is not possible. This process of raising the outer edge of the road over its inner edge through a certain angle is known as banking of the road.
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