Consider a tall building located on the Earth's equator. As the Earth rotates, a person on the top floor of the building moves faster than someone on the ground with respect to an inertial reference frame because the person on the ground is closer to the Earth's axis. Consequently, if an object is dropped from the top floor to the ground a distance h below, it lands east of the point vertically below where it was dropped.
(a)
How far to the east will the object land? Express your answer in terms of h, g, and the angular speed
ω of Earth's rotation.
Let's consider the Earth's rotation ω and the height h from which the object is dropped. For this problem, we'll ignore air resistance.
We can start by writing the equation for the centripetal acceleration of a point on the Earth's surface, which is given by:
ac = ω²R
where ac is centripetal acceleration and R is the Earth's radius.
At the top of the building, the centripetal acceleration is:
ac_top = ω²(R + h)
Since we are considering an inertial reference frame, we can calculate the difference in centripetal acceleration between the top floor and the ground:
Δac = ac_top - ac = ω²(R + h) - ω²R = ω²h
Now, let's analyze the motion of the falling object. When an object falls from a height h, it takes a time t to reach the ground, which can be calculated from:
h = 1/2 g t²
t = √(2h/g)
During this time, the object is also experiencing the horizontal acceleration due to the difference in centripetal acceleration (Δac). Thus, we can find its horizontal displacement by applying the equation:
Δx = 1/2 * Δac * t²
Substitute the values for Δac and t:
Δx = 1/2 * ω²h * (2h/g)
Simplify the expression:
Δx = ω²h²/g
Therefore, the object will land ω²h²/g meters to the east of the point vertically below where it was dropped.