Question

A velodrome is built for use in the Olympics. The radius of curvature of the surface is \(20.0 \mathrm{m}\). At what angle should the surface be banked for cyclists moving at \(18 \mathrm{m} / \mathrm{s} ?\) (Choose an angle so that no frictional force is needed to keep the cyclists in their circular path. Large banking angles are used in velodromes.)

Short answer

Answer: The surface of the velodrome should be banked at an angle of approximately \(9.47°\).

Step-by-step solution

List the given values

We are given the following values: - Radius of curvature (R) = 20.0 m - Speed of the cyclists (v) = 18 m/s

Use the formula for centripetal force

The centripetal force (F_c) acting on the cyclists can be found using the following formula: \(F_c = \frac{mv^2}{R}\) where m is the mass of the cyclists and v is their speed. Since we want no frictional force, the only force acting on the cyclists that provides the centripetal force is the horizontal component of the gravitational force (mg) acting on them.

Apply horizontal and vertical force balance

There are two forces acting on the cyclists: 1. The gravitational force (mg) acting vertically downwards. 2. The normal force (N) acting perpendicular to the surface. Let the angle of banking be θ. The horizontal component of the gravitational force (mg) will provide the centripetal force and the vertical component will be balanced by the normal force. So, we can write the following equations: \(F_c = mg \sin{\theta} = \frac{mv^2}{R}\) \(N = mg \cos{\theta}\)

Solve for the angle

We can divide the two equations to eliminate the mass (m): \(\frac{F_c}{N} = \frac{mg \sin{\theta}}{mg \cos{\theta}}\) By simplifying, we can directly calculate the tangent of the angle (θ) = \(tan{\theta}\) : \(tan{\theta} = \frac{\frac{mv^2}{R}}{mg} = \frac{v^2}{Rg}\) Now, we can plug in the given values for R and v, and the acceleration due to gravity (g = 9.81 m/s²): \(tan{\theta} = \frac{(18\,m/s)^2}{(20\,m)(9.81\,m\,/s^2)} = 0.1663\) To find the angle, we take the inverse tangent: θ = \(tan^{-1}(0.1663) = 9.47°\)

Conclusion

The surface of the velodrome should be banked at an angle of approximately \(9.47°\) for cyclists moving at \(18\,\frac{m}{s}\) so that no frictional force is needed to keep them in their circular path.