A car traveling at constant speed safely negotiates a frictionless banked curved. Calculate the speed of the car when the road is banked at 20 degrees and the radius of its path is 50 meters.
HINT – Do not rotate your axes.
Answer v = 13.35 m/s
To solve this problem, we can use the concept of the centripetal force and the force of gravity that act on the car.
The centripetal force is the force that keeps an object moving in a circular path and is given by the equation Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the car, v is its velocity, and r is the radius of the circular path.
The force of gravity is given by the equation Fg = mg, where Fg is the force of gravity, m is the mass of the car, and g is the acceleration due to gravity.
When the road is banked, the component of the normal force perpendicular to the surface of the road counters the force of gravity, allowing the centripetal force to provide the necessary circular motion. The component of the normal force parallel to the surface of the road provides the necessary friction to prevent slipping.
Using trigonometry, we can calculate the normal force component perpendicular to the surface of the road. It is given by the equation N⊥ = mgcosθ, where N⊥ is the normal force component perpendicular to the surface of the road, m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of the banked road.
Since the normal force perpendicular to the surface of the road only opposes the force of gravity, we can set N⊥ = Fg and solve for the mass of the car:
mgcosθ = mg
cosθ = 1
This means that the angle θ is 20 degrees.
Now, we can equate the centripetal force to the normal force component parallel to the surface of the road. It is given by the equation Fc = N∥ = mgsinθ, where N∥ is the normal force component parallel to the surface of the road, m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of the banked road.
Substituting the equation for the centripetal force (Fc = mv^2/r) and the equation for the normal force component parallel to the surface of the road (N∥ = mgsinθ), we get:
mv^2/r = mgsinθ
Since the mass of the car appears on both sides of the equation, we can cancel it out. So, we're left with:
v^2/r = gsinθ
Now, we can solve for the speed of the car, v:
v^2 = rgsinθ
v = √(rgsinθ)
Plugging in the values for r (50m) and θ (20 degrees), we get:
v = √(50 * 9.8 * sin20°)
v ≈ √(980 * 0.342)
v ≈ √335.76
v ≈ 18.33 m/s
Therefore, the speed of the car when the road is banked at 20 degrees and the radius of its path is 50 meters is approximately 18.33 m/s.