Forest UCM MiNF

Mechanics in Noninertial Reference Frames

Linearly accelerating reference frames

Let [math]\mathcal S_0[/math] represent an inertial reference frame and \mathcal S represent an noninertial reference frame with acceleration [math]\vec A[/math] relative to [math]\mathcal S_0[/math].

Ball thrown straight up

Consider the motion of a ball thrown straight up as viewed from [math]\mathcal S[/math].

Using a Galilean transformation (not a relativistic Lorentz transformation)

At some instant in time the velocities add like

[math]\dot {\vec {r}_0} = \dot {\vec r}+ \vec V[/math]

where

[math]\vec V[/math] = velocity of moving frame [math]\mathcal S[/math] with respect to [math]\mathcal S_0[/math] at some instant in time

[math]\Rightarrow \dot {\vec r} = \dot {\vec {r}_0} - \vec V[/math]

taking derivative with respect to time

[math]\Rightarrow \ddot {\vec r} = \ddot {\vec {r}_0} - \dot \vec V= \ddot {\vec {r}_0} - \vec A [/math]

[math]\Rightarrow m\ddot {\vec r} = m\ddot {\vec {r}_0} - m \vec A= \vec F - m\vec A= \vec F - \vec {F}_{\mbox {inertial}}[/math]

where

[math]\vec {F}_{\mbox {inertial}} = m \vec A \equiv[/math] inertial force

in your noninertial frame, the ball appears to have a force causing it to accelerate in the [math]\vec A[/math] 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 [math]\vec A[/math].

What is the pendulums equilibrium angle [math]\theta_0[/math]

In frame [math]\mathcal S_0[/math]

[math]\sum \vec F = m \ddot{\vec r_0}= \vec T + m \vec g[/math]

In frame [math]\mathcal S[/math]

[math]\sum \vec F = m \ddot{\vec r}= \vec T + m \vec g - m \vec A= \vec T + m \left ( \vec g -\vec A \right ) = \vec T + m \vec g_{eff} [/math]

If the pendulum is at rest and not oscillating then

[math]\sum \vec F = 0 = \vec T + m \vec g_{eff} [/math]

[math]g_{eff}[/math] is the vector sum of [math]g[/math] and [math]A[/math] which are orthogonal to each other in this problem thus

[math]\theta = tan^{-1}{\frac{g}{A}}[/math]

The pendulum oscillation frequency as seen in the accelerating car is

[math]\omega= \sqrt{\frac{g_{eff}}{l}} = \sqrt{\frac{\sqrt{g^2+A^2}}{l}}[/math]

Using lagrangian mechanics in the inertial frame

[math]\vec{r} = (l \sin \theta + \frac{1}{2} a t^2) \hat i + l \cos \theta \hat j[/math]

[math]\dot \vec{r} = (l \cos \theta \dot \theta + a t) \hat i + \sin \theta \dot \theta \hat j[/math]

[math] T = \frac{1}{2} mv^2 = \frac{1}{2} m \left ( (l \cos \theta \dot \theta + a t)^2 + \sin^2 \theta \dot \theta^2 \right )[/math]

[math]= \frac{1}{2} m \left ( l^2 \dot \theta^2 + 2 atl\dot \theta \cos \theta + a^2t^2\right )[/math]

[math]U =- mgy = -mgl \cos \theta[/math]

[math]\mathcal L = \frac{1}{2} m \left ( l^2 \dot \theta^2 + 2 atl\dot \theta \cos \theta + a^2t^2\right ) +mgl \cos \theta[/math]

[math] \left ( \frac{\partial \mathcal {L} }{\partial \theta} \right ) = \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot \theta} \right ) [/math]

[math] -matl \dot \theta \sin \theta -mgl\sin \theta = \frac{d}{dt} \left ( ml^2\dot \theta + matl \cos \theta \right ) [/math]

[math] -at \dot \theta\sin \theta -g\sin \theta = l\ddot \theta - at \sin \theta \dot \theta + a \cos \theta[/math]

[math] l\ddot \theta=-g\sin \theta + a \cos \theta [/math]

The tides

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

The moon of mass [math](M_m)[/math] pulls on the earth of mass [math](M_e)[/math] such that

[math]\vec F = G \frac{M_m M_e}{d_0^3}\vec d_0 = M_e \vec A[/math]

where [math]d_0[/math] is the earth-moon distance of separation.

[math]\Rightarrow \vec A = G \frac{M_m }{d_0^3}\vec d_0 =[/math] Earth's acceleration towards the moon that makes the earth a non-inertial reference frame

The moon of mass [math](M_m)[/math] pulls on a test mass of water [math](M_o)[/math] on the surface of the earth's ocean such that

[math]\vec F = G \frac{M_m M_o}{d^3}\vec d[/math]

As seen in the Earth non-inertial reference frame

[math]M_o \ddot \vec r = \sum \vec F - M_o \vec A[/math]

[math]= \left ( M_o \vec g + \vec F_N + G \frac{M_m M_o}{d^3}\vec d \right ) - G \frac{M_m }{d_0^3}\vec d_0 [/math]

where

[math]\vec F_N =[/math] a net non-graviational force hold the mass M_o in top of the ocean (Bouyant force)

[math]M_o \ddot \vec r = \left ( M_o \vec g + \vec F_N + G \frac{M_m M_o}{d^3}\vec d \right ) - G \frac{M_m }{d_0^3}\vec d_0 [/math]

[math]= M_o \vec g + \vec F_N + \vec{F}_{\mbox{tide}}[/math]

[math]\vec{F}_{\mbox{tide}} = G \frac{M_m M_o}{d^3}\vec d - G \frac{M_m }{d_0^3}\vec d_0 [/math]

[math] = G M_m M_o \left ( \frac{\vec d}{d^3} - \frac{\vec d_0}{d_0^3}\right ) [/math]

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 [math]\vec d || \vec d_0 [/math] and [math]d_0 \lt d[/math] making [math]\vec{F}_{\mbox{tide}}[/math] pull [math]M_o[/math] towards the moon

Case 2

The mass is directly on the other side of the earth with respect to the moon

In this case [math]\vec d || \vec d_0 [/math] BUT [math]d_0 \gt d[/math] making [math]\vec{F}_{\mbox{tide}}[/math] pull [math]M_o[/math] away from the moon

Magnitude of the Tides

Consider a drop of water of mass [math]m[/math] on the surface of the ocean.

The three forces influencing this drop from the previous sections are

[math]\vec F_g =M_o \vec g \;\;\;\; \vec{F}_{\mbox{tide}} \;\;\;\;\; \vec F_N[/math]

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.

[math]\vec F_g = M_o \vec g \;\;\;\; \vec{F}_{\mbox{tide}} [/math]

are the result of a gravitational force which is conservative.

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

[math]\vec F_g = M_o \vec g = -\nabla U_{eg} \;\;\;\; \vec{F}_{\mbox{tide}} = - \nabla U_{\mbox{tide}} [/math]

[math]\vec{F}_{\mbox{tide}} = G \frac{M_m M_o}{d^3}\vec d - G \frac{M_m }{d_0^3}\vec d_0 \equiv - \nabla U_{\mbox{tide-1}}- \nabla U_{\mbox{tide-2}}[/math]

[math]G \frac{M_m M_o}{d^3}\vec d \equiv - \nabla U_{\mbox{tide-1}}[/math]

[math]\Rightarrow U_{\mbox{tide-1}}=G M_m M_o\frac{1}{d} \;\;\;\;\;[/math]:[math] \;\;\;\;\vec d[/math] is NOT always pointed along the x-axis towards the moon

[math] G \frac{M_m M_o}{d_0^3}\vec d_0 \equiv \nabla U_{\mbox{tide-2}}[/math]

[math] U_{\mbox{tide-2}} = - \int \vec F_o \cdot d \hat x= \int G \frac{M_m M_o }{d_0^2}\hat {d}_0 \cdot d \hat x[/math]

[math] = G \frac{M_m M_o}{d_0^2}\int\hat {d}_0 \cdot d \hat x[/math] Since [math]d_0[/math] is at fixed distance

[math] = G \frac{M_mM_o }{d_0^2}\int\ d x[/math] Since [math]d_0[/math] is parallel to x

[math] = G M_m M_o\frac{x}{d_0^2}[/math]

[math] U_{\mbox{tide}}=-G M_m M_o \left ( \frac{1}{d} + \frac{x}{d_0^2} \right ) [/math]

Since

[math]\vec F_g = M_o \vec g = -\nabla U_{eg} \;\;\;\; \vec{F}_{\mbox{tide}} = - \nabla U_{\mbox{tide}} [/math]

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, [math]U_{eg} + U_{\mbox{tide}}[/math], is constant on the surface

[math]\Rightarrow[/math] 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

[math]U_{eg}(HT) + U_{\mbox{tide}}(HT)=U_{eg}(LT) + U_{\mbox{tide}}(LT)[/math]

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

[math]M_o gh=U_{eg}(HT)-U_{eg}(LT)[/math]

but

[math]U_{eg}(HT) -U_{eg}(LT)= U_{\mbox{tide}}(LT) - U_{\mbox{tide}}(HT)[/math]

[math]M_ogh= -G M_m M_o \left ( \frac{1}{\sqrt{d_0^2+r^2}} + \frac{0}{d_0^2} \right )- (-G M_m M_o) \left ( \frac{1}{d_0-R_e} + \frac{-R_e}{d_0^2} \right ) [/math]

At HT

[math]d = d_0 - R_e \;\;\;\; x = -R_e[/math]

At LT

[math]d = \sqrt{d_0^2 + r^2} \;\;\;\; x = 0[/math]

[math]\frac{1}{\sqrt{d_0^2+r^2}} \approx \frac{1}{\sqrt{d_0^2+R_e^2}} = \frac{1}{d_0 \sqrt{1 + \left ( \frac{R_e}{d_0} \right )^2}} \approx\frac{1}{d_0} \left ( 1 - \frac{1}{2} \left ( \frac{R_e}{d_0} \right )^2 \right ) [/math]

[math]M_ogh= -G M_m M_o \left [ \left ( \frac{1}{\sqrt{d_0^2+r^2}} \right )- \left ( \frac{1}{d_0-R_e} + \frac{-R_e}{d_0^2} \right ) \right ][/math]

[math]= -G M_m M_o \left [ \frac{1}{d_0} \left ( 1 - \frac{1}{2} \left ( \frac{R_e}{d_0} \right )^2 \right ) - \left ( \frac{1}{d_0-R_e} + \frac{-R_e}{d_0^2} \right ) \right ][/math]

[math]= -G M_m M_o \left [ \frac{1}{d_0} \left ( 1 - \frac{1}{2} \left ( \frac{R_e}{d_0} \right )^2 \right ) - \frac{1}{d_0}\left ( \frac{1}{1-\frac{R_e}{d_0}} + \frac{-R_e}{d_0^2} \right ) \right ][/math]

[math]\frac{1}{1-x} = 1 + x + x^2 = \sum_n^{\infty} x^n[/math] : geometric series when[math] x \lt 1[/math]

[math]M_o gh= -G M_m M_o \left [ \frac{1}{d_0} \left ( 1 - \frac{1}{2} \left ( \frac{R_e}{d_0} \right )^2 \right ) - \frac{1}{d_0}\left (1 + \frac{R_e}{d_0} + \left( \frac{R_e}{d_0} \right)^2 - \frac{R_e}{d_0^2} \right ) \right ][/math]

[math] -G M_m M_o \left [ \frac{1}{d_0} \left ( 1 - \frac{1}{2} \left ( \frac{R_e}{d_0} \right )^2 \right ) - \frac{1}{d_0}\left (1 + \left( \frac{R_e}{d_0} \right)^2 \right ) \right ][/math]

[math] -G M_m M_o \left [ \frac{1}{d_0} \left ( - \frac{3}{2} \left ( \frac{R_e}{d_0} \right )^2 \right ) \right ][/math]

[math] \frac{G M_m M_o}{d_o} \left [ \frac{3}{2} \left ( \frac{R_e}{d_0} \right )^2 \right ][/math]

[math]g = G\frac{M_e}{R_e^2}[/math]

[math]M_o gh=\frac{G M_m M_o}{d_o} \left [ \frac{3}{2} \left ( \frac{R_e}{d_0} \right )^2 \right ][/math]

[math]M_o \left ( G\frac{M_e}{R_e^2} \right) h=\frac{G M_m M_o}{d_o} \left [ \frac{3}{2} \left ( \frac{R_e}{d_0} \right )^2 \right ][/math]

[math] h= \frac{3}{2} \frac{ M_m R_e^4}{M_ed_o^3}[/math]

Height of Tides due to Moon and Sun

[math] h= \frac{3}{2} \frac{ M_m R_e^4}{M_ed_o^3}[/math]

[math] M_m = 7.35 \times 10^{22}[/math] kg = mass of moon

[math] M_e = 5.98 \times 10^{24}[/math] kg = mass of earth

[math] M_s = 1.99 \times 10^{30}[/math] kg = mass of sun

[math] R_e = 6.37 \times 10^{6}[/math] m

[math]d_m = 3.84 \times 10^{8}[/math] m = earth-moon distance

[math]h= 54 cm =[/math] height of tides due to the moon

[math] d_s = 1.495 \times 10^{11}[/math] m = earth-sun distance

[math]h = 25 cm =[/math] 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 [math]\mathbf{R}[/math] is a 3x3 orthogonal matrix [math](\mathbf{R}^T\mathbf{R} = \mathbf{RR}^T = 1)[/math] and [math]\mathbf R[/math] is proper [math]( \left | \mathbf{R} \right | = +1) [/math], then there is a non-zero vector [math]r[/math] such that [math]\mathbf {R}r = r[/math]

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

[math]\mathbf {R} \vec P = \lambda \vec p[/math]

First let's find the eigenvalues and eigenvectors of the matrix [math]\mathbf R[/math]

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 as [math]\omega[/math] the the magnitude and direction of this vector may be expressed as

[math]\vec \omega = \left | \vec \omega \right | \hat \omega[/math]

[math]\left | \vec \omega \right | \equiv\lt /math magnitude of the rate of rotation (angular speed) : \lt math\gt \hat \omega \equiv[/math] 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 [math]\hat i,\hat j ,\hat k[/math] or [math]\hat r,\hat \theta,\hat \phi \cdots[/math])

Forest_Ugrad_ClassicalMechanics