Question

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a \(100{\rm{ m}}\) radius curve banked at \(15.0^\circ \). (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at \(20.0{\rm{ km}}/{\rm{h}}\)?

Short answer

1. The ideal speed to take the turn is \(16.2{\rm{ m}}/{\rm{s}}\).
2. Minimum coefficient of friction is \(0.27\).

Step-by-step solution

Definition of Coefficient of friction

The coefficient of friction is defined as the ratio of frictional force to the normal reaction force. It is a scalar and dimensionless quantity with no units.

Calculating ideal speed for radius curve

The ideal speed to take a safe turn is,

\(v = \sqrt {rg\tan \theta } \)

Here, \(r\)is the radius of the curved road, \(g\) is the acceleration due to gravity, and \(\theta \) is the ideal banking angle.

Substitute \(100{\rm{ m}}\) for \(r\), \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\), and \(15.0^\circ \) for \(\theta \),

\(\begin{aligned}{}v = \sqrt {\left( {100{\rm{ m}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \tan \left( {15.0^\circ } \right)} \\ = 16.2{\rm{ m}}/{\rm{s}}\end{aligned}\)

Hence, the ideal speed to take the turn is \(16.2{\rm{ m}}/{\rm{s}}\).

Calculating the minimum coefficient of friction

The coefficient of friction is,

\(\mu = \frac{{{v^2}}}{{rg}}\)

Substitute \(16.2{\rm{ m}}/{\rm{s}}\) for \(v\), \(100{\rm{ m}}\) for \(r\), \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\),

\(\begin{aligned}{}\mu = \frac{{{{\left( {16.2{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{\left( {100{\rm{ m}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 0.27\end{aligned}\)

Hence, the minimum coefficient of friction is \(0.27\).