Question

In coordinates fixed on the Earth the Coriolis force \(\boldsymbol{F}=2 m[\boldsymbol{v}, \boldsymbol{\Omega}]\), where \(\Omega\) is the angular velocity vector of the Earth, acts on a material point of mass \(m\) moving with velocity \(v\) with respect to the Earth. A stone is thrown (with initial velocity 0 ) into a mine shaft of depth \(10 \mathrm{~m}\) at the latitude of Leningrad \(\left(\lambda=60^{\circ}\right)\). By how much will the Coriolis force cause it to deviate from the vertical?

Short answer

Question: Determine the deviation caused by the Coriolis force on a stone thrown into a mine shaft with a depth of 10 meters, at the latitude of Leningrad (\(\lambda = 60^{\circ}\)).

Step-by-step solution

1. Determine the time it takes for the stone to reach the bottom of the mine shaft

Using the kinematic equation: \[d = v_0 t + \frac{1}{2} a t^2\] where \(d\) is the distance traveled, \(v_0\) is the initial velocity, \(a\) is the acceleration due to gravity (\(9.81 m/s^2\)), and \(t\) is the time. Since the initial velocity is 0, the equation simplifies to: \[10 = \frac{1}{2}(9.81) t^2\] Now, solve for \(t\): \[t^2 = \frac{20}{9.81}\] \[t = \sqrt{\frac{20}{9.81}}\]

2. Calculate the velocity of the stone at the bottom of the mine shaft

Using another kinematic equation: \[v = v_0 + at\] Since the stone is thrown with an initial velocity of 0, this equation simplifies to: \[v = (9.81)(t)\]

3. Determine the angular velocity vector of the Earth at the given latitude

The Earth's angular velocity, \(\boldsymbol{\Omega}\), at a latitude \(\lambda\) is given by: \[\boldsymbol{\Omega} = \begin{pmatrix}0\\ \Omega \cos \lambda \\ \Omega \sin \lambda \end{pmatrix}\] where \(\Omega = 7.27 \times 10^{-5} \,\text{rad/s}\) is the Earth's angular velocity. Substituting \( \lambda = 60^{\circ} \), we get: \[\boldsymbol{\Omega} = \begin{pmatrix}0\\ 3.64 \times 10^{-5} \\ 6.29 \times 10^{-5} \end{pmatrix} \, \text{rad/s}\]

4. Calculate the Coriolis force vector acting on the stone

Using the given formula for the Coriolis force: \[\boldsymbol{F} = 2m[\boldsymbol{v}, \boldsymbol{\Omega}]\] The mass \(m\) of the stone cancels out in the final deviation calculation as we'll see later, so we can ignore \(m\) for now. We find the cross product of \(\boldsymbol{v}\) and \(\boldsymbol{\Omega}\): \[\boldsymbol{F} = 2[\boldsymbol{v}, \boldsymbol{\Omega}]\]

5. Determine the deviation caused by the Coriolis force

Ultimately, we want to find the horizontal distance the stone deviates due to the Coriolis force during its fall. The deviation from the vertical can be calculated as: \[\text{Deviation} = \frac{F}{g} \times t\] Substituting the previously calculated values for \(t\), \(F\), and \(g = 9.81\) m/s², we can find the deviation of the stone caused by the Coriolis force. By following the steps above and carefully analyzing the expressions for the various parameters, you'll be able to determine the deviation caused by the Coriolis force on the stone thrown into the mine shaft at the given latitude.

Key concepts

Angular Velocity

Angular velocity describes how fast an object rotates around a specific point or axis. In many cases, it is represented by the Greek letter \( \Omega \). For the Earth, this vector points along its axis of rotation.

The Coriolis force in our exercise depends on Earth's angular velocity. This is because the force acts due to Earth's rotation, affecting moving objects. This problem requires using the angular velocity at the latitude of 60 degrees, where Leningrad is located.

The Earth's angular velocity is approximately \( 7.27 \times 10^{-5} \, ext{rad/s} \). However, the effect at any point depends on the latitude, which we factor in by using trigonometric functions. Calculate it as this vector: \[ \boldsymbol{\Omega} = \begin{pmatrix}0\ \Omega \cos \lambda \ \Omega \sin \lambda \end{pmatrix} \] with the values adjusted for the specified latitude.

Kinematics

Kinematics is a branch of physics dealing with motion without considering the forces that cause it. A key part of this exercise involves using kinematic equations to determine the time it takes for a stone to fall 10 meters in a mine shaft.

Starting from rest, the equation used is a well-known formula: \[ d = v_0 t + \frac{1}{2} a t^2 \] Here, \( v_0 \) is the initial velocity, and \( a = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. This simplifies to finding \( t \) when \( d = 10 \, \text{m} \).

Solving gives us the time duration \( t \) that influences the Coriolis force's effect later in the problem.

Cross Product

The cross product is a mathematical operation used on two vectors to produce another vector. This operation is crucial in calculating the Coriolis force. The Coriolis effect occurs because the velocity of the stone and Earth's angular velocity interact through a cross product.

The cross product of vectors \( \boldsymbol{v} \) and \( \boldsymbol{\Omega} \), represented as \[ [\boldsymbol{v}, \boldsymbol{\Omega}] \] results in a vector that is perpendicular to both \( \boldsymbol{v} \) and \( \boldsymbol{\Omega} \). This resultant vector helps determine the subtler forces affecting the stone's path.

It's worth noting that the mass factor \( m \) in the Coriolis force formula cancels in deviation calculations, simplifying our focus to just the geometric interaction between the vectors.

Deviation Calculation

Determining the deviation from the vertical path due to the Coriolis force involves calculating the horizontal shift during the stone's fall.

Once the Coriolis force \( \boldsymbol{F} \) is evaluated using the cross product, we can find the stone's deviation using:

• Time of fall \( t \) from kinematics
• Gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \)

The formula for deviation is: \[ \text{Deviation} = \frac{F}{g} \times t \] This relation quantifies the Coriolis effect, indicating how rotation alters the trajectory path of vertically moving objects, even if only slightly.