An airplane is flying in a loop, following a circular path with a radius of 2.5 km in the vertical plane. What is the minimum speed that the airplane can maintain without being forced out of the circular path?
To make it over the top the mass times centripetal acceleration must be equal to m g
m g = m v^2/R
9.81 = v^2 / 2500 meters
v^2 = 9.81 * 2500 = 24525
v = 156.6 meters/ second
156.6m/s * (1 km / 1000 meters) * (3600 seconds / hour) = 564 km/hr
Well, it seems like this airplane is caught in a bit of a loop-de-loop situation! To find the minimum speed it needs to maintain, we can turn to physics.
Keeping in mind that the centripetal force required to keep an object moving in a circular path is given by F = mv^2/r, where F is the force, m is the mass, v is the velocity, and r is the radius.
If the airplane wants to stay on track without being flung out of the loop, the centripetal force provided by its weight must be greater than or equal to the centripetal force required to maintain the circular path.
So, we have:
mg ≥ (m*v^2)/r,
where g is the acceleration due to gravity.
By canceling out the mass, we find that the minimum speed v can be calculated by:
v^2 ≥ g * r.
Plugging in the given values, with g being approximately 9.8 m/s^2 and r being 2.5 km (or 2,500 m), we find:
v^2 ≥ 9.8 m/s^2 * 2,500 m.
Simplifying further, we get:
v^2 ≥ 24,500 m^2/s^2.
So, if we take the square root of both sides, we find:
v ≥ √24,500 m/s.
Therefore, the minimum speed the airplane needs to maintain without being forced out of the circular path is at least √24,500 m/s.
Now, I must say it's highly unlikely for an airplane to reach such speeds, unless it suddenly turns into a supersonic jet or superhero invincible jet! So, while physics gives us an answer, it's good to keep in mind that real-life airplanes are designed for much more realistic speeds. Safety first, after all!
To find the minimum speed that the airplane can maintain without being forced out of the circular path, we need to consider the force acting on the airplane.
In this case, the force acting on the airplane is the centripetal force, which is given by the equation:
F = (m * v^2) / r
Where:
F = Centripetal force
m = Mass of the airplane
v = Velocity of the airplane
r = Radius of the circular path
Since we want to find the minimum speed, we need to find the maximum force that the airplane can withstand without being forced out of the path. This occurs when the force of gravity is equal to the centripetal force.
The force of gravity can be calculated using the equation:
F_gravity = m * g
Where:
g = Acceleration due to gravity (approximately 9.8 m/s^2)
Setting the force of gravity equal to the centripetal force, we have:
m * g = (m * v^2) / r
Simplifying the equation, we get:
v^2 = g * r
Taking the square root of both sides, we get:
v = √(g * r)
Plugging in the values, we have:
v = √(9.8 * 2.5)
Calculating the value, we find:
v ≈ 7.91 m/s
Therefore, the minimum speed that the airplane can maintain without being forced out of the circular path is approximately 7.91 m/s.
To find the minimum speed that the airplane can maintain without being forced out of the circular path, we need to consider the forces acting on the airplane.
The main force at play is centrifugal force, which acts outward and tries to pull the airplane away from the circular path. It is given by the equation:
Fc = mv^2 / r
Where Fc is the centrifugal force, m is the mass of the airplane, v is its velocity, and r is the radius of the circular path.
For the airplane to remain in the circular path, the inward centripetal force (provided by gravity in this case) must be equal to the outward centrifugal force. The centripetal force is given by:
Fcp = mg
Where Fcp is the centripetal force and g is the acceleration due to gravity.
Setting these two forces equal to each other, we can solve for the minimum velocity (v) of the airplane:
mv^2 / r = mg
Cancel out mass (m) from both sides:
v^2 / r = g
Multiply both sides by r:
v^2 = gr
Take the square root of both sides:
v = sqrt(gr)
Now we can substitute in the given values:
r = 2.5 km = 2500 m
g = 9.8 m/s^2
v = sqrt(9.8 * 2500) = sqrt(24500) = 156.45 m/s
Therefore, the minimum speed that the airplane can maintain without being forced out of the circular path is approximately 156.45 m/s.