A car of mass M = 800kg traveling at 40.0km/hour enters a banked turn covered with ice. The road is banked at an angle θ, and there is no friction between the road and the car's tires. (Figure 1) . Use g = 9.80m/s2 throughout this problem.
Part A
What is the radius r of the turn if θ = 20.0∘ (assuming the car continues in uniform circular motion around the turn)?
wrong^^
To find the radius of the turn, we can start by considering the forces acting on the car while it is in the banked turn. Since the car is in uniform circular motion, there must be a net inward force acting towards the center of the turn.
Let's analyze the forces acting on the car:
1. Gravitational force (weight): The weight of the car acts vertically downwards. The gravitational force is given by Fg = Mg, where M is the mass of the car and g is the acceleration due to gravity.
2. Centripetal force: This force is responsible for keeping the car moving in a circular path. In the absence of friction, the only force providing the necessary centripetal force is the component of the normal force acting towards the center of the turn.
3. Normal force: The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force acts perpendicular to the road surface and has both vertical and horizontal components.
To find the radius of the turn, we need to determine the relationship between these forces. In this problem, we are given the angle of banking (θ), the mass of the car (M), and the speed of the car (v). We need to find the radius (r) of the turn.
We can start by determining the forces acting in the vertical direction:
Vertical forces:
- Gravitational force (Fg) = Mg
- Vertical component of the normal force (Nv) = N * cos(θ), where N is the total normal force and θ is the angle of banking.
Since the gravitational force and the vertical component of the normal force must balance each other (since there is no vertical acceleration), we have:
Mg = N * cos(θ) ----(1)
Now, let's consider the forces acting in the horizontal direction:
Horizontal forces:
- Horizontal component of the normal force (Nh) = N * sin(θ)
- Centripetal force (Fc) = M * v^2 / r, where v is the speed of the car and r is the radius of the turn.
The centripetal force and the horizontal component of the normal force must balance each other (since the car is in uniform circular motion), so we have:
Fc = M * v^2 / r = N * sin(θ) ----(2)
We can rewrite equation (2) in terms of the gravitational force using equation (1):
M * v^2 / r = Mg * sin(θ) / cos(θ)
Simplifying:
v^2 / r = g * tan(θ)
Finally, we can solve for the radius (r) of the turn:
r = v^2 / (g * tan(θ))
Substituting the given values (v = 40.0 km/hour, g = 9.80 m/s², and θ = 20.0∘) into the equation, we can calculate the radius (r) of the turn.
the bank angle provides the necessary centripetal force
m * v^2 / r = m * g * sin(θ) * cos(θ)
convert v to m/s, then solve for r