Object dropped from tower with deflection problem

[telekpsi]

Homework Statement

a) An object with mass, m, is dropped from height, h, in a tower located at latitude 30 degrees to the north from the equator. How far will it be deflected? Please derive the deflection in terms of Earth's rotation angular velocity omega, h, g, and m. This part needs complete derivation.

b) Calculate the deflection if h=100m and mass=1kg

 

[heth]

There's the rest of the template to fill in too...

Homework Equations

The Attempt at a Solution

 

[thomate1]

a) The deflection of the object can be calculated using the formula: d = (2 * m * h * omega^2 * sin(30))/g

Where:
d = deflection
m = mass of the object
h = height of the tower
omega = Earth's rotation angular velocity
g = acceleration due to gravity

To derive this formula, we first need to understand the forces acting on the object. When the object is dropped, it experiences a downward force due to gravity, which is countered by an upward force called the normal force. However, the object also experiences a horizontal force due to the Earth's rotation. This force is known as the Coriolis force and it is directed perpendicular to the object's velocity.

To calculate the deflection, we need to find the horizontal component of the Coriolis force. This can be done by using the formula: F_c = 2 * m * v * omega * sin(theta)

Where:
F_c = Coriolis force
m = mass of the object
v = velocity of the object
omega = Earth's rotation angular velocity
theta = angle between the velocity vector and the direction of Earth's rotation (in this case, theta = 30 degrees)

Since the object is dropped from rest, its initial velocity is zero. Therefore, the horizontal component of the Coriolis force can be simplified to: F_c = 2 * m * omega * sin(30)

This horizontal force causes the object to deflect from its vertical path. The deflection can be calculated using the formula: d = (F_c * h)/g

Substituting the value of F_c, we get: d = (2 * m * h * omega^2 * sin(30))/g

b) Substituting the given values of h=100m and m=1kg, and using the value of Earth's rotation angular velocity (omega = 7.27 x 10^-5 rad/s), we get:

d = (2 * 1kg * 100m * (7.27 x 10^-5 rad/s)^2 * sin(30))/9.8 m/s^2 = 0.0000938 m

Therefore, the deflection of the object dropped from a height of 100m is approximately 0.0000938 meters.