A racetrack has the shape of an inverted cone, as the drawing shows. On this surface the cars race in circles that are parallel to the ground, and the surface is at an angle = 20°. For a speed of 29 m/s, at what value of the distance d should a driver locate his car if he wishes to stay on a circular path without depending on friction?
To determine the required distance (d) for the car to stay on a circular path without depending on friction, we need to consider the relationship between the gravitational force and the centripetal force.
1. First, let's consider the forces acting on the car:
- Gravitational Force (Fg): This force acts vertically downwards and is given by the equation Fg = mg, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
- Centripetal Force (Fc): This force acts towards the center of the circular path and is given by the equation Fc = (mv^2) / r, where v is the speed of the car and r is the radius of the circular path.
2. In this case, the gravitational force provides the centripetal force, so Fg = Fc. Therefore, we can equate the two equations:
mg = (mv^2) / r
3. Since we are interested in finding the distance (d) from the center of the cone at which the driver should locate his car, we need to express r in terms of d. Looking at the given diagram, we can see that the radius of the circular path is the horizontal distance from the center to the edge of the track. This distance can be expressed as r = d * sin(θ).
4. Substituting the expression for r in the equation mg = (mv^2) / r, we get:
mg = (mv^2) / (d * sin(θ))
5. Now we can solve for d:
d = (v^2 * m * sin(θ)) / g
6. Plugging in the given values: v = 29 m/s, m = mass of the car (not provided), θ = 20°, and g = 9.8 m/s^2, we can calculate the value of d.
To answer this question, we need to understand the concept of centripetal force and the forces acting on a car moving in a circular path.
In this scenario, the car has two types of forces acting on it: the normal force (N) and the gravitational force (mg). The normal force acts perpendicular to the surface, while the gravitational force acts vertically downward.
To stay on a circular path without depending on friction, the centripetal force must provide the necessary inward force to keep the car moving in a circle. In this case, the centripetal force is provided solely by the component of the gravitational force perpendicular to the surface.
Let's find the value of this component. We'll start by considering the forces acting on the car.
The gravitational force (mg) can be separated into two components: one parallel to the inclined surface and one perpendicular to it. The perpendicular component is given by mg * cos(θ), where θ is the angle of inclination.
Since we want the car to remain on a circular path without friction, the centripetal force must be equal to this perpendicular component of the gravitational force.
The formula for centripetal force is Fc = m * (v^2 / r), where Fc is the centripetal force, m is the mass of the car, v is the speed, and r is the radius of the circular path.
We know the speed of the car (v = 29 m/s), and we want to find the value of the distance (d) from the center of the circular path. The radius of the path is given by r = d * sin(θ).
Setting the centripetal force equal to the perpendicular component of the gravitational force, we have:
m * (v^2 / r) = mg * cos(θ)
Simplifying the equation, we get:
m * v^2 = mg * cos(θ) * r
Substituting the values we know:
m * (29^2) = m * g * cos(20°) * (d * sin(20°))
The mass of the car (m) cancels out from both sides of the equation:
29^2 = g * cos(20°) * (d * sin(20°))
Simplifying further:
d = (29^2) / (g * cos(20°) * sin(20°))
Now, we can calculate the value of d using the acceleration due to gravity, which is approximately 9.8 m/s^2:
d = (29^2) / (9.8 * cos(20°) * sin(20°))
Calculating this expression gives us the value of d, which is approximately 45.3 meters.
So, to stay on a circular path without depending on friction, the driver should locate his car at a distance of approximately 45.3 meters from the center of the circular path.