An airplane is flying above the Earth's surface at a height of 10.9 km. The centripetal acceleration of the plane is 16.7 m/s2. What is the angular velocity of the plane moving in uniform circular motion? Take the radius of the earth to be 6400 km.
Please help, I've tried everything
I changed the height to angle by dividing the height by the radius
I then plugged that into the formula
wf^2 = wi^2 +2(acceleration)(angle)
but I'm not getting the right answer
centripetal acc = v^2 / r
... v is the tangential velocity
16.7 = v^2 / 6.41E6
v / r = angular velocity (in radians)
if r is the radius above earth center the speed is totally impossible. I suspect a typo
not sure about impossible...
this is around escape velocity
To find the angular velocity of the plane, we can start by finding the angle at which the plane is located above the Earth's surface.
Given that the plane is flying at a height of 10.9 km and the radius of the Earth is 6400 km, we can calculate the angle using trigonometry.
angle = arctan(height / radius)
angle = arctan(10.9 km / 6400 km)
Next, we need to convert the angle from degrees to radians since angular velocity is typically measured in radians per second.
angle_radians = angle * (π / 180)
Now, we can use the centripetal acceleration formula to find the angular velocity of the plane:
ω^2 = ωi^2 + 2αθ
Given that the centripetal acceleration is 16.7 m/s^2 and the angle is in radians, we have:
ω^2 = ωi^2 + 2(16.7 m/s^2)(angle_radians)
Finally, we can solve for ω by taking the square root of both sides:
ω = sqrt(ωi^2 + 2(16.7 m/s^2)(angle_radians))
Make sure to substitute the correct values for ωi^2 (initial angular velocity, if given) and solve the equation using a calculator to find the final angular velocity of the plane.