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You are in an airplane circling an airport at a speed of 600 km/hour on a circular path of radius 3 km. A stewardess accidentally drops a cup , of mass .1 kg, on the floor. She releases it from rest at a perpendicular distance of 1.5 m from the floor. (That is to say, the shortest distance from the cup to a spot on the floor is 1.5 m.)

a) Show that the cup hits the closest point on the floor when it falls to good approximation.

b) What is its speed, relative to the airplane, when it hits the floor? (That is what is the magnitude of the velocity of the cup, measured in the reference frame of the airplane?) Show that it is somewhat higher than it would be if the cup dropped the same distance in an inertial reference frame.

User JulienVan
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1 Answer

6 votes

Answer:

a) (0.0004·
\mathbf{\hat i} - 1.55·
\mathbf{\hat j}) m

b) 0.31 km/h·
\mathbf{\hat i} + 0.00008 km/h·
\mathbf{\hat j}

Step-by-step explanation:

The given parameters are;

The circular motion of the airplane = 600 km/hour

The radius of the circular path = 3 km

The time it takes the cup to reach the floor of the airplane is given as follows;

s = u·t + 1/2·g·t²

1.5 = 1/2 × 9.81 × t²

t² = 1.5/(1/2 × 9.81)

t = √(1.5/(1/2 × 9.81)) ≈ 0.553 s

The location of the plane in 0.533 s is 600 km/hour × 0.533 s ≈ 83.83 meters

The angular velocity, ω = v/r = 600 km/hour/3 km = 200°/hour

Therefore, the angle covered in 0.533 s is 0.533 s × 200°/hour ≈ 0.0296°

tan(0.0296)*3000 ≈ 1.55 m vertical

3000 - cos(0.0296)*3000 ≈ 0.0004 m horizontal

The location of the cup relative to the plane will be

0 - (-0.0004·i + 1.55·j) = 0.0004·i - 1.55·j m

b) The speed of the cup relative to the plane when it hits the floor is given by the

The speed of the plane = 600 km/hour × cos(90 - 0.0296)j + 600 km/hour × sin(90 - 0.0296)j = -0.31 km/hour·i + 599.99 km/hour·j

The speed of the cup = 600 km/hour·j

The relative speed = 600 km/hour·j - ( -0.31 km/hour·i + 599.99 km/hour·j) = 0.31 km/hour·i + 0.00008 km/hour·j

The relative speed of the cup to the plane = 0.31 km/h·i + 0.00008 km/h·j

Therefore, given that there will be no relative speed in an inertia frame, the speed of the cup relative to the plane will be higher than when dropped in an inertia frame

User Siavash Alp
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