A plane flies horizontally at constant ground speed v=720km/h. The pilot is informed by an air traffic controller that he must immediately change his direction of flight while keeping the same altitude. The pilot performs a turning maneuver by rolling the plane to a banked position. Then she increases the airspeed in Δv to keep the plane at constant altitude. As a result, the plane describes an arc of radius R=8km. Find the increase in airspeed Δv in km/h. Assume that the lift force is proportional to the square of the speed and that it is always perpendicular to the plane of the wings.
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it's about use the Ek = Ep
and V is using the root mean square velocity
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it's about 95.5
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If it was going straight,it's weight would equal the lift force,since it is banked at an angle,the force will be split into components,
horizontal lift force equals Lsin=mv^2/r,the centripetal force hence describing a radius,
vertical component must equal the weight,lcostheta=mg,
then it is done,and please if u have done the lattice point question tell me,
To find the increase in airspeed Δv, we can start by analyzing the forces acting on the plane during the turning maneuver.
When the plane is in level flight, the lift force is equal to the weight of the plane. However, when the plane is banking during the turning maneuver, the lift force will have two components: one perpendicular to the wings (lifting force) and another component toward the center of the turn (centripetal force).
Let's denote:
- m as the mass of the plane
- g as the acceleration due to gravity (approximately 9.8 m/s²)
- L as the lifting force perpendicular to the wings
- Fc as the centripetal force
The centripetal force is given by Fc = m * ac, where ac is the centripetal acceleration.
The centripetal acceleration can be calculated using ac = v² / R, where v is the ground speed and R is the radius of the turn.
Since the lift force is perpendicular to the wings, it can be written as L = k * v², where k is a constant of proportionality.
Now let's analyze the forces acting on the plane:
1. Weight force (mg): Always directed downward and equal to m * g.
2. Lifting force (L): Perpendicular to the wings and equals k * v².
3. Centripetal force (Fc): Directed towards the center of the turn and equals m * (v² / R).
Since the plane is at a constant altitude, the lifting force and the weight force must balance each other, resulting in L = mg.
Setting these two equal, we have:
k * v² = m * g
Now let's find the increase in airspeed Δv:
During the turning maneuver, the pilot increases the airspeed by Δv while maintaining the same altitude. Therefore, the new speed of the plane will be v + Δv.
Substituting the new speed into the equation:
k * (v + Δv)² = m * g
Expanding the equation:
k * (v² + 2vΔv + Δv²) = m * g
Since Δv is assumed to be small, we can neglect the Δv² term:
k * v² + 2k * vΔv = m * g
Substituting k * v² = m * g from earlier:
m * g + 2k * vΔv = m * g
Canceling out the m * g terms:
2k * vΔv = 0
Dividing both sides by 2k * v:
Δv = 0
Surprisingly, the increase in airspeed Δv is 0 km/h.
This result implies that the pilot can perform the turning maneuver and change the direction of flight while keeping the same altitude without needing to increase the airspeed. This is possible because the balance between the lifting force and the weight force stays the same during the maneuver.
Please note that this solution assumes idealized conditions and simplifications, such as neglecting drag forces and the effects of changing altitude. In reality, there may be other factors to consider.