A velodrome is built for use in the Olympics. The radius of curvature of the surface is 19.6 m. At what angle should the surface be banked for cyclists moving at 11.5 m/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.)
Let the x-axis point toward the center of curvature and the y-axis point upward. Use Newton’s second law
ΣFy = N• cos θ = mg,
ΣFx = N• sin θ = m•v²/R,
Dividing the 2nd equation by the 1st equation, we obtain
tan θ = v² /R•g =11.5²/19.6•9.8 = 0.67
θ = 34.5 º
To find the angle at which the surface should be banked, we can make use of the concept of centripetal force.
The centripetal force required to keep the cyclists moving in a circular path is provided by the vertical component of the cyclist's weight. This vertical component is given by:
N = mg
where N is the normal force, m is the mass of the cyclist, and g is the acceleration due to gravity.
Since no frictional force is needed, we can consider the forces acting on the cyclist along the vertical direction, which are the cyclist's weight (mg) and the normal force (N).
From the free body diagram, we can find the vertical component of the cyclist's weight:
mg * sin(θ) = N
Where θ is the angle of banking.
Now, let's consider the forces acting on the cyclist in the horizontal direction. The horizontal force is responsible for providing the centripetal force required to keep the cyclist moving in a circular path.
The centripetal force is given by:
Fc = m * v^2 / r
Where Fc is the centripetal force, m is the mass of the cyclist, v is the velocity of the cyclist, and r is the radius of curvature of the velodrome.
This centripetal force is provided by the component of the gravitational force that acts along the horizontal direction. The horizontal component of the cyclist's weight is given by:
mg * cos(θ) = Fc
Substituting the value of Fc in terms of m, v, and r, we get:
mg * cos(θ) = m * v^2 / r
Simplifying the equation, we can find the value of the angle θ:
cos(θ) = v^2 / (rg)
θ = arccos(v^2 / (rg))
Plugging in the given values:
v = 11.5 m/s
r = 19.6 m
g = 9.8 m/s^2
θ = arccos((11.5 m/s)^2 / ((19.6 m)*(9.8 m/s^2)))
Calculating this expression, we find:
θ ≈ 36.7 degrees
Therefore, the surface of the velodrome should be banked at an angle of approximately 36.7 degrees.
To determine the angle at which the surface of the velodrome should be banked, we need to consider the forces acting on the cyclist. In order to keep the cyclist moving in a circular path without the need for friction, the gravitational force and the normal force must provide the necessary centripetal force.
First, let's start by determining the centripetal force required for the cyclist moving at 11.5 m/s. The centripetal force is given by:
Fc = (m * v^2) / r
where Fc is the centripetal force, m is the mass of the cyclist, v is the velocity, and r is the radius of curvature.
Now, since the cyclist doesn't require the frictional force to provide the centripetal force, we can assume that the normal force (N) provides the necessary force. The normal force has two components: one perpendicular to the surface (N vertical) and the other parallel to the surface (N horizontal).
The vertical component of the normal force cancels out the weight force (mg), so N vertical = mg.
The horizontal component of the normal force (N horizontal) provides the centripetal force required to keep the cyclist moving in a circular path. The angle at which the surface should be banked can be determined using this equation:
tan(θ) = N horizontal / N vertical
Now, let's substitute the values we have:
Fc = (m * v^2) / r
N vertical = mg
tan(θ) = N horizontal / N vertical
Plugging in the given values:
v = 11.5 m/s
r = 19.6 m
Since mass (m) is not given, it cancels out in the equation when we form the tangent.
tan(θ) = [(m * v^2) / r] / (mg)
= v^2 / (g * r)
To find the angle θ, we can use the inverse tangent function:
θ = arctan(v^2 / (g * r))
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, let's calculate the angle:
θ = arctan((11.5 m/s)^2 / (9.8 m/s^2 * 19.6 m))
Using a calculator:
θ ≈ arctan(1.377)
Therefore, the angle at which the surface of the velodrome should be banked is approximately 53.8 degrees.