# Mechanics in Noninertial Reference Frames

## Linearly accelerating reference frames

Let represent an inertial reference frame and represent an noninertial reference frame with acceleration relative to .

### Ball thrown straight up

Consider the motion of a ball thrown straight up as viewed from .

Using a Galilean transformation (not a relativistic Lorentz transformation)

At some instant in time the velocities add like

where

= velocity of moving frame with respect to at some instant in time

taking derivative with respect to time

where

inertial force
in your noninertial frame, the ball appears to have a force causing it to accelerate in the direction.

The inertial force may also be referred to as a fictional force

an example is the "fictional" centrifugal force for rotational acceleration.

The observer in a noninertial reference frame will feel these frictional forces as if they are real but they are really a consequence of your accelerating reference frame

example

A force pushes you back into your seat when your Jet airplane takes off
you slam on the brakes and hit your head on the car's dashboard

### Pedulum in an accelerating car

Consider a pendulum mounted inside a car that is accelerating to the right with a constant acceleration .

What is the pendulums equilibrium angle

In frame

In frame

If the pendulum is at rest and not oscillating then

is the vector sum of and which are orthogonal to each other in this problem thus

The pendulum oscillation frequency as seen in the accelerating car is

Using lagrangian mechanics in the inertial frame

### The tides view by a ref frame accelerating towards the moon

The gravitational force between the moon and the earth accelerates the earth towards the moon.

This makes the earth an accelerated reference frame

The moon of mass pulls on the earth of mass such that

where is the earth-moon distance of separation.

Earth's acceleration towards the moon that makes the earth a non-inertial reference frame

The moon of mass pulls on a test mass of water on the surface of the earth's ocean such that

As seen in the Earth non-inertial reference frame

where

a net non-graviational force hold the mass M_o in top of the ocean (Bouyant force)= M(displaced water)*g = constant force independent of position

Let's consider two cases, one where M_o is directly between the moon and the earth and the other when the mass is directly on the opposite side of the earth from the moon.

Case 1
The mass is directly between the moon and the earth
In this case and making pull towards the moon

Case 2
The mass is directly on the other side of the earth with respect to the moon
In this case BUT making pull away from the moon

## Magnitude of the Tides

Consider a drop of water of mass on the surface of the ocean.

The three forces influencing this drop from the previous sections are

Archimedes Principle
An object in a fluid is buoyed up with a force equal to the weight of the water displaced by the object.

All of the above forces act normal to the surface of the water.

are the result of a gravitational force which is conservative.

A potential may be defined for the two forces above such that

: is NOT always pointed along the x-axis towards the moon

Since is at fixed distance
Since is parallel to x

Since

are forces that are exerted such that the are always perpendicular to the surface of the ocean ( they are normal to the surface) then the sum of the two potential's, , is constant on the surface

The ocean's surface is an equipotential surface ( all points on the surface of the ocean are have the same gravitational potential energy

If you consider two points on the surface of the earth where one point is high tide (HT) and the other point is low tide (LT) then

The change in the gravitational potential energy between the earth and the ocean for high tide and low tide is

but

At HT

At LT

: geometric series when

### Height of Tides due to Moon and Sun

kg = mass of moon
kg = mass of earth
kg = mass of sun

m
m = earth-moon distance
height of tides due to the moon
m = earth-sun distance
height of tides due to sun
spring tides
The case when the moon, earth and sun are aligned to give maximum height tide (moon can be either full or new doesn't matter because the bulge is on both sides of the earth)
neap tides
This is the case then the sun, earth, and sun are aligned to form a right triangle. The tide effect should cancel so the height is only 54-25=29 cm.

## Euler's Theorem of rotation

Original form:

Any displacement of a rigid body in three dimensional space such that a point on the rigid body, say O, remains fixed, is equivalent to a rotation about a fixed axis through the point O.

Expressed using Modern mathematical terms

Euler's theorem on the axis of a three dimensional rotation
If is a 3x3 orthogonal matrix and is proper , then there is a non-zero vector such that

The matrix R above represents a spatial rotation that is a linear one-to-one mapping that transforms the coordinate vector P into p.

The above theorem can be proven by treating it as a matrix algebra problem where you want to prove. ie

First let's find the eigenvalues and eigenvectors of the matrix

A consequence of Euler's theorem is that a single rotation may be represented as a combination of rotation.

## Angular velocity vector

If we denote the angular velocity vector describing the angular velocity of a rotating body as the the magnitude and direction of this vector may be expressed as

magnitude of the rate of rotation (angular speed)

direction of the rotation axis which is given by the right-hand rule and may is expressed using the components of a right handed coordinate system ( ie or )

the angular velocity vector describing the angular velocity of a rotating body
The angular velocity vector describing the angular velocity of a rotating non-inertial reference frame.

### Linear velocity from angular velocity

= linear velocity of a point located at the position on a body rotating about the origin of the coordinate system used to specify .

Consider two coordinate systems where represent an inertial reference frame and represent a frame with a velocity relative to .

Using a Galilean transformation (not a relativistic Lorentz transformation)

the velocity in the moving frame may be expressed in terms of the velocity in the rest frame and the velocity of the moving frame as.

Now consider the situation when an object is rotating. The relationship between angular and linear motion should still satisfy the general express

A point on this rotating body located by the position vector from the inertial reference frame will have the linear velocity

If we fix a moving reference frame so its origin coincides with then the position of the same point on the object is given by the same position vector . The angular velocity of this point is given by

If the moving reference frame is rotating with an angular velocity then the point on the object located by the vector is moving with a linear velocity as viewed by of

Both the linear and the angular velocities add

## Time derivative in Rotating Frame

### Nomenclature

The following variables are used to describe the kinematics of a rotating frame

linear velocity of the rotating frame
linear acceleration of the rotating frame
angular velocity of the rotating frame

For example; a reference frame fixed to the earth with its origin at the center of the earth would be rotating with respect to an inertial frame with its origin at the same location with an angular velocity of one revolution every 24 hours.

### Rate of change of a general vector Q

let

an arbitrary vector representing position

Let's write the vector in terms of three orthonormal unit vectors representing an arbitrary right-handed coordinate system

= rate of change of as described by (relative to ) the rotating frame

Using unit vectors that are attached to the rotating frame

In a cartesian coordinate system

From the inertial reference frame the above unit vectors used by are changing their direction with time

The expansion coefficients (vector components) of the vector are the same in both frames as the moving frames is not moving fast enough to be relativistic

as seen earlier

The time derivative of a vector as measured in the inertial frame is related to the derivative in the non-inertial frame by

## Newton's second law in a rotating frame

Newton's 2nd law in an inertial reference fram

To determine the second derivative in a non-inertial frame we can take the derivative of the velocity relationship

let

then

returning to Newton's second law

: changing order of cross products introduces negative sign

### Alternative derivation using Lagrangian

Hamilton's principle is defined for an inertial reference frame, therefor the lgrangian is written from that perspective.

kinetic energy in the inertial reference frame

As seen before

where

= velocity of moving frame with respect to inertial frame at some instant in time

If we consider a non-inertial reference frame that is only rotating with an angular velocity with an origin fixed to the origin of the inertial reference frame.

Then

substituting this Gallilean transformation into the expression for the kinetic energy in the inertial reference fram

: scalar triple product is invariant under cyclic permutation

this is a scaler so ignore the unit vector time derivative

### psuedo Force names

Coriolis Force
Centrifugal Force

where

velocity of object in the rotating frame
velocity of the rotating frame

For the rotating earth

m/s
mi/hr

For objects with velocity v < 500 mph you can neglect the coriolis force.
Also note that the coriolis force depends on the velocity of the object while the centrifugal does not.

## Centrifugal Force

The direction of this fictional force on the earth's non-inertial reference frame is radially outward from the axis of rotation.

Insert picture

### Free fall example

The force of gravity between two objects with mass M and m is

In a non-inertial reference frame attached to the rotating earth the effective gravitational force is

If we assume the objects velocity v < 500 mph then you can neglect the coriolis force.

The coriolis force decreases the pull of gravity and is a function of the lattitude () The larger the stronger the effective gravitational force.

Launch rockets from Florida not Alaska

### Free fall tangential component

The tangential component of the centrifugal force is directed along the surface of the sphere such that

This has a maximum at = 45 degrees

## Coriolis Force

### Coriolis Force in Earth's reference frame

The Earth rotates from west to east.

Consider a reference frame fixed to the earth in the Northern Hemisphere at the latitude angle 90 - . In a standard spherical coordinate system the "z-axis" is directed towards the north pole and is the polar angle measures from this fixed zenith direction, is the azimuthal angle corresponding to longitudinal (east-west). Lines of latitude represent the angle measured from the equator to one of the Earth's poles (90 - )

The non-inertial frame shown above has the axis pointing to the west and the axis pointing upward along the radial direction from the Earth's center.

The rotation vector as described in the non-inertial frame is

I you launch a projectile in the Northern Hemisphere so its velocity tangential to the Earth's surface and moving from the equator towards the north pole, the projectile will be deflected towards the east (right).

East

If you launch the projectile towards the equator from the Northern hemisphere.

West

In both cases the projectile is deflected to the right of its velocity direction.

How about a projectile launched to the West (East)

A projectile moving West (East) will deflected down(up) and to the right of its trajectory

What happens in the Southern Hemisphere?

The rotation vector as described in the non-inertial frame in the Southern Hemisphere is

Now if

Projectile directed from South to North

west

In the Northern hemisphere a projectile would swerve East if you directed it to the North

Projectile directed from South to North

East

How about a projectile launched to the West (East)

A projectile moving West (East) in the Souther Hemisphere will deflected down(up) and to the left of its trajectory

#### Coriolis effect on storm systems

The Earth (and a non-inertial reference frame attached to it) has a rotational velocity

Weather systems involve a large mass of air moving at speeds up to 100 miles per hour in the case of a tropical storms.

Over large distances this acceleration can have an effect on the course of a storm but not the short distances involved in tornados.

Tropical storms (cyclones) are examples of how the coriolis force determine the rotational direction of an air mass that is quickly moving towards a low pressure area. As the air mass moves along the pressure gradient is is deflected to the right. This results in a clockwise (counter-clockwise) rotation in teh Northern (Southern) hemisphere. The radius of the cyclone can be estimated as

( or of the latitude angle)

### Free fall

Consider an object of mass m close to the surface of the earth and falling in a vacuum.

As viewed from a non-inertial reference frame attached to the rotating earth Newton's second law is

The equation of motion may be written as

where

the acceleration due to a non-rotating earth
the acceleration observed
Note
by expressing the equation of motion in terms of \vec g you remove the dependence of leaving only a dependence on and

If I use a coordinate system where the x-axis if pointing East then the z-axis will be pointing radially upward such that

left

writing out the individual components

#### 0th order Approx

let

Then

x-directions

: integrate and the constant is the initial velocity of the projectile
: integrate again
:

similarly in the y-direction

z-direction

#### 1st order Approx

substitute the 0th order solutions and integrate again

x-direction

y-direction

: ignoring higher order terms

z-direction

: ignoring higher order terms
: ignoring

General equations of motion

If the object is dropped from a height h

If I drop an object down an evacuated 100 m deep hole at the equator

### Foucault Pendulum

The equation of motion may be written as

## Interpretation of rotating reference frame

### Polar

Vector Notation convention:

Position:

Because points in a unique direction given by we can write the position vector as

: does not have the units of length

The unit vectors ( and ) are changing in time. You could express the position vector in terms of the cartesian unit vectors in order to avoid this

The dependence of position with can be seen if you look at how the position changes with time.

#### Velocity in Polar Coordinates

Consider the motion of a particle in a circle. At time the particle is at and at time the particle is at

If we take the limit ( or ) then we can write the velocity of this particle traveling in a circle as

or

Thus for circular motion at a constraint radius we get the familiar expression

If the particle is not constrained to circular motion ( i.e.: can change with time) then the velocity vector in polar coordinates is

=
or in more compact form

linear velocity Angular velocity

Finding the derivative directly

Cast the unit vector in terms of cartesian coordinates and take the derivative.

#### Acceleration in Polar Coordinates

Taking the derivative of velocity with time gives the acceleration

We need to find the derivative of the unit vector with time.

Consider the position change below in terms of only the unit vector

Using the same arguments used to calculate the rate of change in :

If we take the limit ( or ) then we can write the velocity of this particle traveling in a circle as

or

Finding the derivative of directly

Cast the unit vector in terms of cartesian coordinates and take the derivative.

Substuting the above into our calculation for acceleration:

F_r = m \ddot r \hat r = m \left ( \ddot{r} -r\dot{\phi}^2 \right) \hat{r} =