
The maximum speed that can be achieved without skidding by a car on circular unbanked road of radius R and coefficient of static friction \[\mu\], is
A. $\mu Rg$
B. $Rg\sqrt{\mu }$
C. $\mu \sqrt{Rg}$
D. $\sqrt{\mu Rg}$
Answer
162.9k+ views
Hint: In order to answer this question, we need to know the maximum speed at which an automobile can go on a round, banked road without skidding. Basically, we balance every aspect of the forces acting in both x- and y-axes on a car.
We constantly balance the forces acting along the x-axis and the y-axis to address any balancing problems, whether the motion is translational or rotational.
We balance the torques acting on the body in the case of rotational motion.
Formula used:
1. Gravitational force= mg
2. Frictional force, $f=\mu N$,
Where \[\mu \] is the coefficient of friction and N is the normal force.
3. Centripetal force= $\dfrac{m{{v}^{2}}}{R}$
Where m is the mass of car, v is the velocity and R is the radius of the road
Complete step by step solution:
We’ll start by equating all the forces and by trying to keep the body in equilibrium as shown in the diagram below:

Let’s balance all the forces along the x axis
$f\cos \theta +N\sin \theta =\dfrac{m{{v}^{2}}}{R}$
We will now substitute $f=\mu N$ in the above equation we get,
$\mu N\cos \theta +N\sin \theta =\dfrac{m{{v}^{2}}}{R}$
$\mu (\cos \theta +\sin \theta )=\dfrac{m{{v}^{2}}}{R}$…….. (i)
Now similarly let’s balance all the forces along y axis
$N\cos \theta =f\sin \theta +mg$
$\Rightarrow N\cos \theta =\mu N\sin \theta +mg$
$\Rightarrow N(\cos \theta -\mu \sin \theta )=mg$…… (ii)
And frictional force Frictional force, $f=\mu N$….. (iii)
Now we will divide the equation (i) by (ii), we get
$\dfrac{{{v}^{2}}}{Rg}=\dfrac{(\mu +\tan \theta )}{(1-\mu \tan \theta )}$
$\Rightarrow v=\sqrt{\dfrac{Rg(\mu +\tan \theta )}{(1-\mu \tan \theta )}}$
For maximum velocity the road is unbanked,$\theta =0{}^\circ $
Therefore, ${{v}_{\max }}=\sqrt{\mu Rg}$
Hence, the correct option is D. $\sqrt{\mu Rg}$
Note: Roads are banked to stop fast cars from skidding. If a car is moving faster than the aforementioned speed, it will slide. If there is inadequate frictional force, the car will slide. Additionally, keep in mind that while the car tries to slide along the road, friction acts downwardly along the surface.
We constantly balance the forces acting along the x-axis and the y-axis to address any balancing problems, whether the motion is translational or rotational.
We balance the torques acting on the body in the case of rotational motion.
Formula used:
1. Gravitational force= mg
2. Frictional force, $f=\mu N$,
Where \[\mu \] is the coefficient of friction and N is the normal force.
3. Centripetal force= $\dfrac{m{{v}^{2}}}{R}$
Where m is the mass of car, v is the velocity and R is the radius of the road
Complete step by step solution:
We’ll start by equating all the forces and by trying to keep the body in equilibrium as shown in the diagram below:

Let’s balance all the forces along the x axis
$f\cos \theta +N\sin \theta =\dfrac{m{{v}^{2}}}{R}$
We will now substitute $f=\mu N$ in the above equation we get,
$\mu N\cos \theta +N\sin \theta =\dfrac{m{{v}^{2}}}{R}$
$\mu (\cos \theta +\sin \theta )=\dfrac{m{{v}^{2}}}{R}$…….. (i)
Now similarly let’s balance all the forces along y axis
$N\cos \theta =f\sin \theta +mg$
$\Rightarrow N\cos \theta =\mu N\sin \theta +mg$
$\Rightarrow N(\cos \theta -\mu \sin \theta )=mg$…… (ii)
And frictional force Frictional force, $f=\mu N$….. (iii)
Now we will divide the equation (i) by (ii), we get
$\dfrac{{{v}^{2}}}{Rg}=\dfrac{(\mu +\tan \theta )}{(1-\mu \tan \theta )}$
$\Rightarrow v=\sqrt{\dfrac{Rg(\mu +\tan \theta )}{(1-\mu \tan \theta )}}$
For maximum velocity the road is unbanked,$\theta =0{}^\circ $
Therefore, ${{v}_{\max }}=\sqrt{\mu Rg}$
Hence, the correct option is D. $\sqrt{\mu Rg}$
Note: Roads are banked to stop fast cars from skidding. If a car is moving faster than the aforementioned speed, it will slide. If there is inadequate frictional force, the car will slide. Additionally, keep in mind that while the car tries to slide along the road, friction acts downwardly along the surface.
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