Solved Problem on Dynamics

A car runs on a banked curve with radius R and slope θ. What will be the maximum speed to negotiate the curve without friction?

Problem data:

Radius of the curve: R;
Curve slope: θ.

Problem diagram:

Figure 1 - Banked curve in the old AVUS autodrome in Berlin, Germany.

The forces that act in the car are the gravitational force \({\vec F}_{g} \) pointing vertically down, the normal reaction force \( \vec{N} \) perpendicularly to the track (Figure 1).

Solution

The bank angle of the track is θ, the gravitational force \({\vec F}_{g} \) is perpendicular to the horizontal plane, makes an angle of 90°, as the sum of the interior angles of a triangle equals to 180°, the angle between the gravitational force \({\vec F}_{g} \) and the banked lane (Figure 2-A)

\[ \alpha +\theta +90°=180° \Rightarrow \alpha =180°-90°-\theta \Rightarrow \alpha =90°-\theta \]

Figure 2

The angle between the banked curve and horizontal plane is θ, and the angle between the centripetal force \({\vec F}_{cp} \) and the horizontal is also θ, are alternate angles (Figure 2-B).
The angle between the normal reaction force \(\vec{N} \) and the centripetal force \({\vec F}_{cp} \) is α (Figure 2-C), and the angle between the normal force and the vertical axis is θ, \( \alpha +\theta =90° \) (Figure 2-D). This angle will be used to decompose the normal force (Figure 3).

Drawing the forces in a coordinate system xy, we can apply Newton's Second Law for circular motion

\[ \begin{gather} \bbox[#99CCFF,10px] {{\vec F}_{cp}=m{\vec a}_{cp}} \tag{I} \end{gather} \]

In the x direction, the centripetal force \( {\vec F}_{cp} \) is given by the component of normal force in the x direction, \( {\vec N}_{x} \) (Figure 3)

\[ \begin{gather} F_{cp}=N_{x} \tag{II} \end{gather} \]

Figure 3

The N_x component is given by

\[ \begin{gather} N_{x}=N\sin \theta \tag{III} \end{gather} \]

The centripetal acceleration is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {a_{cp}=\frac{v^{2}}{R}} \tag{IV} \end{gather} \]

Substituting expressions (III) and (IV) into expression (I)

\[ \begin{gather} N\sin \theta =m\frac{v^{2}}{R} \tag{V} \end{gather} \]

In the y direction, there is no motion, the gravitational force and the component of the normal force in y direction cancel out

\[ \begin{gather} F_{g}=N_{y} \tag{VI} \end{gather} \]

The gravitational force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{VII} \end{gather} \]

The component N_y is given by

\[ \begin{gather} N_{y}=N\cos \theta \tag{VIII} \end{gather} \]

Substituting the expressions (VII) and (VIII) into expression (VI)

\[ \begin{gather} N\cos \theta =mg \tag{IX} \end{gather} \]

Equations (V) and (IX) can be written as a system of two equations.

\[ \left\{ \begin{array}{l} N\sin \theta =m\dfrac{v^{2}}{R}\\ N\cos \theta=mg \end{array} \right. \]

dividing the first equation by the second equation

\[ \begin{gather} \frac{\cancel{N}\sin \theta }{\cancel{N}\cos \theta }=\frac{\cancel{m}\dfrac{v^{2}}{R}}{\cancel{m}g} \end{gather} \]

From the Trigonometry \( \dfrac{\sin \theta }{\cos \theta} =\tan \theta \)

\[ \begin{gather} \tan\theta =\frac{v^{2}}{Rg}\\[5pt] v^{2}=Rg\tan\theta \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v=\sqrt{\;Rg\tan\theta \;}} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .