Answer
Verified
414.8k+ views
Hint:In this question, we need to find the maximum speed of the car at which car doesn’t skid while moving in a circular banked road. We basically balance all the factors of the forces exerted on a car in x-axis direction and y-axis direction.
To solve any balancing problems, whether motion is translational or rotational, we always balance forces acting along x-axis and y-axis.
In case of rotational motion, we balance torques acting on the body.
Complete step-by-step answer:
Several formulas used in our calculations are-
Gravitational attraction force – mg
Frictional force (f) – acting between car’s tire and road, and parallel to the slope of road
Normal force (N)- It is a reaction force of the gravitational force exerted by the road on car
Centripetal force $\left( \dfrac{m{{v}^{2}}}{r} \right)$ - acting towards the center of the circular path followed by the car.
Let’s take inclination angle A of road with respect to the x-axis as shown in the above diagram. Now we can balance all the x-axis and y-axis factors of forces. Refer to the figure.
(1) Balancing all the forces acting along the x-axis-
$f\cos (A)+N\sin (A)=\dfrac{m{{v}^{2}}}{r}$
Substituting the value $f=\mu N$ into equation, we get
$\begin{align}
& \mu N\cos (A)+N\sin (A)=\dfrac{m{{v}^{2}}}{r} \\
& \Rightarrow N\left( \mu \cos A+\sin A \right)=\dfrac{m{{v}^{2}}}{r}......(1) \\
\end{align}$
(2) Balancing all the forces acting along the y-axis-
$\begin{align}
& N\cos A=f\sin A+mg \\
& \Rightarrow N\cos A=\mu N\sin A+mg \\
& \Rightarrow N\left( \cos A-\mu \sin A \right)=mg........(2) \\
\end{align}$
(3) Frictional force-
$f=\mu N$
Now we will use these three equations to find the value of speed (v).
Now we can divide equation (1) and (2), we get
\[\begin{align}
& \dfrac{{{v}^{2}}}{rg}=\dfrac{\left( \mu +\tan A \right)}{\left( 1-\mu \tan A \right)} \\
& \Rightarrow v=\sqrt{\dfrac{rg\left( \mu +\tan A \right)}{\left( 1-\mu \tan A \right)}} \\
\end{align}\]
This is an expression for maximum speed of a car on a banked road in circular motion.
Note: (1) Roads are banked to prevent high speed cars from skidding.
(2) If a car is moving at a speed more than mentioned above, it will skid.
(3) If the force of friction is not strong enough, the vehicle will skid. Also, note that friction acts downwards along the road, as the tendency of the car is to skid up.
To solve any balancing problems, whether motion is translational or rotational, we always balance forces acting along x-axis and y-axis.
In case of rotational motion, we balance torques acting on the body.
Complete step-by-step answer:
Several formulas used in our calculations are-
Gravitational attraction force – mg
Frictional force (f) – acting between car’s tire and road, and parallel to the slope of road
Normal force (N)- It is a reaction force of the gravitational force exerted by the road on car
Centripetal force $\left( \dfrac{m{{v}^{2}}}{r} \right)$ - acting towards the center of the circular path followed by the car.
Let’s take inclination angle A of road with respect to the x-axis as shown in the above diagram. Now we can balance all the x-axis and y-axis factors of forces. Refer to the figure.
(1) Balancing all the forces acting along the x-axis-
$f\cos (A)+N\sin (A)=\dfrac{m{{v}^{2}}}{r}$
Substituting the value $f=\mu N$ into equation, we get
$\begin{align}
& \mu N\cos (A)+N\sin (A)=\dfrac{m{{v}^{2}}}{r} \\
& \Rightarrow N\left( \mu \cos A+\sin A \right)=\dfrac{m{{v}^{2}}}{r}......(1) \\
\end{align}$
(2) Balancing all the forces acting along the y-axis-
$\begin{align}
& N\cos A=f\sin A+mg \\
& \Rightarrow N\cos A=\mu N\sin A+mg \\
& \Rightarrow N\left( \cos A-\mu \sin A \right)=mg........(2) \\
\end{align}$
(3) Frictional force-
$f=\mu N$
Now we will use these three equations to find the value of speed (v).
Now we can divide equation (1) and (2), we get
\[\begin{align}
& \dfrac{{{v}^{2}}}{rg}=\dfrac{\left( \mu +\tan A \right)}{\left( 1-\mu \tan A \right)} \\
& \Rightarrow v=\sqrt{\dfrac{rg\left( \mu +\tan A \right)}{\left( 1-\mu \tan A \right)}} \\
\end{align}\]
This is an expression for maximum speed of a car on a banked road in circular motion.
Note: (1) Roads are banked to prevent high speed cars from skidding.
(2) If a car is moving at a speed more than mentioned above, it will skid.
(3) If the force of friction is not strong enough, the vehicle will skid. Also, note that friction acts downwards along the road, as the tendency of the car is to skid up.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
Other Pages
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The values of kinetic energy K and potential energy class 11 physics JEE_Main
In the reaction of KMnO4 with H2C204 20 mL of 02 M class 12 chemistry JEE_Main
Dependence of intensity of gravitational field E of class 11 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
BF3 reacts with NaH at 450 K to form NaF and X When class 11 chemistry JEE_Main