Solved Problem on Dynamics

A car, considered a point, with mass m turns around a circular runway of radius R. The coefficient of friction between road and tires is μ. Assume g for the acceleration due to gravity. Determine the maximum speed that the car may have in the curve without slipping.

Problem data:

Mass of car: m;
Radius of the curve: R;
Coefficient of friction: μ;
Acceleration due to gravity: g.

Problem diagram:

Figure 1

The forces acting in the car are:

\( \vec{F}_{g} \):gravitational force;
\( \vec{N} \):normal reaction force;
\( {\vec F}_{f} \):force of friction.

Solution

Applying Newton's Second Law for circular motion

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{cp}=ma_{cp}} \tag{I} \end{gather} \]

In the vertical direction, the gravitational force \( \vec{F}_{g} \) and normal reaction force \( \vec{N} \) cancel out

\[ \begin{gather} N=F_{g} \tag{II} \end{gather} \]

The gravitational force is given by

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

substituting the expression (III) into (II)

\[ \begin{gather} N=mg \tag{IV} \end{gather} \]

In the radial direction, we have the force of friction \( {\vec F}_{f} \) given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{f}=\mu N} \tag{V} \end{gather} \]

substituting the expression (IV) into (V)

\[ \begin{gather} F_{f}=\mu mg \tag{VI} \end{gather} \]

the force of friction is the net force in the radial direction, substituting the expression (VI) into (V)

\[ \begin{gather} \mu \cancel{m}g=\cancel{m}a_{cp} \tag{VII} \end{gather} \]

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

substituting this value in the expression (VII)

\[ \begin{gather} \mu g=\frac{v^{2}}{R}\\ v^{2}=\mu Rg \end{gather} \]

the maximum speed with which the car can make the curve will be

\[ \bbox[#FFCCCC,10px] {v=\sqrt{\mu Rg\;}} \]

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