This rock looks like a steak...

or maybe a slice of a tree with visible tree rings.

But, okay, we will go with "this rock".

NOT. They are interacting and the interaction is stable against small spatial perturbations. I.e., it will oscillate around its equilibrium and rapidly radiate off the excess energy to settle back down to the equilibrium distance.

You can read about shearing forces in Chapter 5.3.

I.e., it can be shown using calculus and a ton of trig!

Hans Christian Ørsted in Denmark, and Michael Faraday in England made this discovery.

And, in fact, the electromagnetic forces produced by the muscle fibers in your quadriceps muscles DO overwhelm the gravitational force of the Earth, every time you go up a stairway!

Time does NOT slow down. However, time measurements by two different observers disagree if the two observers are timing the same events from different positions relative to the "very massive" body, like a black hole.

ALL forms of energy affect the curvature of spacetime

It is interesting to think about other large, nearby objects that pull gravitationally on us.

$$\text{You }\approx50 - 100\,kg \\ \updownarrow \\ \text{Earth }\approx 6\times 10^24\, kg,\, \\ @ r=6.371\times 10^6 \,m$$

vs.

$$\text{You on the sidewalk @ 350 Fifth Ave., NYC } \\ \approx 50 - 100\,kg\\ \updownarrow \\ \text{Empire State Building }\approx 3.31\times 10^8\, kg,\\ \text{w/ center of mass, elevation } h\approx 200\,m$$

How do those forces compare?

1. The cosmological abundance of elements in the universe is a huge topic in astrophysics.
2. The abundance or elements in solar system objects like planets, moons, comets, asteroids and meteorites that fall to Earth is a huge topic for planetary physicists, like UCF's Planetary Sciences Group.

This topic, action at a distance, has been a topic of scientific discussion for centuries.

Should be submicroscopic. VERY submicroscopic!

I.e., chemistry! The periods and groups of the periodic table are helpful because they encode the behavior of ELECTRONS in neutral atoms of each element. Hinrich's spiral "Programme of Atomechanics"

Instead of an average velocity relative to two points:

1. \(\left(x_1,\, y_1\right)\) @ time \(t_1\)
2. \(\left(x_2,\, y_2\right)\) @ time \(t_2\)

where the average speed is

$$\vec{\bar{v}}=\left(\frac{x_2-x_1}{t_2-t_1}\right)\hat{i}+\left(\frac{y_2-y_1}{t_2-t_1}\right)\hat{j}$$

When you take the instantaneous limit, \(\Delta t\longrightarrow 0\) , the other two Δs decrease, also, but the ratios \(\frac{\Delta x}{\Delta t}\) and \(\frac{\Delta y}{\Delta t}\) do not vanish; they converge to a finite value, the components \(v_x\) and \(v_y\)of the instantaneous velocity at time t, where \(t_1<t<t_2\),</p>

$$\vec{v}\left(t\right)=v_x \left(t\right)\hat{i}+v_y \left(t\right)\hat{j}$$

Compare to the \(v\left(t\right)\) formula with the instantaneous limit in the next paragraph.

Two different parameterizations of the path \(\Gamma\), one by increments of (approximately) equal distance, \(t\), and a second parameterization, \(\color{red}\theta\) , as it might be for an aggressive race car driver, with equal increments of time.

I.e., we parametrize the path along the x-axis.

We call this a path parameter. Cf., Parametrizations of Plane Curves in LibreText, for further reading.

Like a light-second

$$1.00\, LS= 3.00\times 10^8\, m$$

or the Earth-Sun distance, which we call the astronomical unit,

$$1.00\, AU = 499\, LS$$

etc.

For any \(\Delta\) the order of subtraction is always

$$\text{final}-\text{initial}$$

$$\text{later}-\text{earlier}$$

and if some kind of sequence exists, \(\left(u_1, u_2, u_3 \ldots\right) \) $$\Delta u=un - u{n-1}$$

Basic vocabulary

Tricky but important point.

...in motion relative to Earth or to any other object. For instance, when docking your capsule to a spacecraft on orbit, you are moving and your target, the spacecraft, is a moving target.

So your sense of steering includes a moving frame of reference. In fact, you are in a frame of reference that is in free fall!

In the usual sense, we can imagine three mutually perpendicular rigid meter sticks...

"starr-160503-4526-Triticum_aestivum-Glenn_seedheads_and_meter_stick_for_height-Hawea_Pl_Olinda-Maui" by Starr Environmental is licensed with CC BY 2.0. To view a copy of this license, visit https://creativecommons.org/licenses/by/2.0/

...but it could be a set of rigid curves, like on this sundial.

"Israel-05367 - Sundial" by archer10 (Dennis) is licensed with CC BY-SA 2.0. To view a copy of this license, visit https://creativecommons.org/licenses/by-sa/2.0/

This is a set of basic questions that apply to a LOT of the scientific enterprise.

Joule experiment, free expansion.

A little bit like a wavelet, in that the ring is the source of the reflected wave, on the right hand end of the string.

Boundary conditions, so important!! Just like initial conditions being important.

Bypass this chapter section. It is better for a more advanced class, but you may enjoy working through it.

Photons: $$\vec{p}=\hbar\, \vec{k}$$ $$E=\hbar\, \omega$$

Bypass this chapter section. It is better for a more advanced class, but you may enjoy working through it.

SHAPES OF ORBITS determined by energy and angular momentum states:

1. Center of mass, again.
2. The effective potential energy which we know from applying basic planetary angular momentum concepts. $$V_{eff}\left(r\right)=\frac{L^2}{2mr^2}+-\frac{GMm}{r}$$

You will use this technology, Gauss' Law, many times in PHY2049

the field concept is an abstraction useful here and in PHY2049

The Ice Skater Effect. Classic example of conservation of angular momentum.

This example is not even a trick shot in a circus -- i.e., why would anybody want to do this. But it is still illustrative

An interesting exercise. The gymnast slows down his spin just in time to land extended to his full height. Requires a lot of practice, probably.

We discussed this in Session 25

This is an important distinction to make for each and every moment of inertia calculation. The rotation axis

1. might or might not go through the center of mass,
2. it might be aligned along the length of a rod-shaped object or perpendicular to the long axis of the rod shaped axis,
3. it might be along the edge or a rectangular box-shaped object,
4. or it might be any weird axis you need to study,
5. and you might have an irregular object without even any symmetry. Many symmetries in these objects

Asteroid Eros, very irregular

This will be abstracted in PHY2049, when studying the topic of electrostatic potential, which is defined as a property of a charged object or an array of charged objects. Electrostatic potential is not measured in Joules. It is measured in Volts.

Good to keep in mind, even in PHY2049

Recall our discussion in Session 14.1, where we studied

• the shelving example of Ch. 7.1 (pressbooks.online.ucf.edu link.

• the shelving example (Webcourses internal link)

good workout

Good study problem

Good basic workout

Good workout

We worked this problem out in Session 14

True here, but not always the case on a complicated problem.

A good example to work out

This is the primary reason why I refrain from teaching this kinematic equation earlier in a semester

A much more interesting way to view the universe.

Extremely important for all scientists

This example is a little bit unphysical, in that keeping an object on a parabola requires some F = ma. There must be some kind of constraint, like a parabolic track that keeps an object on the track.The external \(\color{red}\vec{F}\) here would be in addition to the constraint forces. So analyzing the forces would be tricky. E.g., you'd also need to know a temporal parameter for the path, which would give you the magnitude of the tangent velocity vector at any time in the path. SO... we will ignore this example!

It is a nice workout, however.

This diagram is slightly blooped up. Path (b) is just the hypotenuse. Path (a), we may assume, is the drop of 1 m folowed by a traverse right of 3 m.

One can use this idea to compute components of a vector in a new coordinate system (p, q). Just take the dot product of the vector with each new unit vector,

$$\vec{A}\cdot \hat{p}=A_p$$

etc.

This is a helpful skill to use from time to time: getting products of two vectors in terms of each vector's components.

Mary Jackson, the engineer in Hidden Figures, worked at a wind tunnel just like this one, at NASA/Langley in Virginia. Long may she reign!

Racing tires are designed to have much larger coefficients of friction, even \(\mu_s >1\).

E.g., friction from the road surface on the tires.

Very nice synopsis for uniform circular motion.

Incorrect. The pressure gradient from the center of a hurricane (maybe 900 millibars) outward to a region of fair weather (1023 millibars), a distance of a few hundred miles, is about 10%, not particularly big, yet it has enormous effects over that distance

Good

We worked on an asymmetric version of this in Session 11, 9/29/20

Important

.

There are four fundamental forces in nature:

1. Universal gravitation
2. the electromagnetic interaction
3. the weak nuclear force (which is now unified with electromagnetism -- we now speak of the electroweak interaction)
4. the strong nuclear force, which binds protons and neutrons into a nucleus.

...of all objects in the universe.

THE most boring example ever. However, it is valid.

For students from Florida, skiing is a fun physical activity in what we call "mountains" (large piles of rock, thousnads of feet tall), on a cold white slippery substance called "snow," (the crystalline form of \(H_2O\) water). Cf., this skiing area in Montana.

A good practice problem. You can answer part c with a spreadsheet.

A nice mini-workout, and you can check your work.

Strobe photo, straight drop https://youtu.be/SmEevo_PL7k

This one is not a strobe photo but in slow motion, it ain't bad. https://youtu.be/cxvsHNRXLjw

Enormously important. This is how we define any vector, a tangent vector, so that vectors can be used in curved spacetime. For math fiends, here is nice blurb on manifold tangent vectors.

We will actually go to this definition when we study aerodynamic drag forces and terminal velocity.

A nice mini-workout, and you can check your work.

This is not the same as the three dogs + 1 example from the Ch. 2.3. That resultant vector was a SUM of the first three dogs' pull forces.

This displacement is a DIFFERENCE of two vectors,

$$\Delta\vec{r}=\vec{r}\left(t_2\right)-\vec{r}\left(t_1\right)$$

using vector subtraction, which can be done by

1. graphically drawing from the tip of the first vector to the tip of the second vector, as in this diagram, or
2. subtracting component by component, e.g., \(x\left(t_2\right)-x\left(t_1\right)\)

Make sure to review this part. Vectors are essential.

We have already seen this in our study of free fall. To integrate the equations of motion (i.e., involving some acceleration, \(\ddot{s}\left(t\right)\)), one needs initial conditions -- initial position and initial velocity.

Our entire program of integration means

• we take each moment at which we know velocity and position
• and use that information as initial conditions
• to solve the equations of motion for position and velocity at the next moment of time.

We will review this in class on 9/8/20.

Note the minus sign \(-\frac{1}{2}gt^2\) in the author's version of this formula.

I hardly ever use the third equation. It is related to conservation of total mechanical energy, which is good, but difficult to use properly, in my opinion.

This is why it works, but only for constant acceleration systems

Just like Galileo did.

Session 1.2 on August 25, 2020, breakout rooms.

Another scale model is the "grapefruit model," illustrated nicely in this Youtube:

https://youtu.be/TAQGmYWwZFk

Also a good exercise

good exercise

ALWAYS!!!

Helpful when doing mental arithmetic. You get skilled with this, you can beat people on calculators!

"Checking the units" is a simple check on your derivations and solutions. Very important

Based on another familiar term. ; )

Well worth checking out.

The distinction between base units and derived units is important.

This is significant.

• The original methods of measuring the second involved the length of a day, a specific fraction, \(\frac{1}{86400}\) of a day.
• The meter similarly was a specific fraction \(\frac{1}{10000000}\) of the distance on the line of longitude between the Equator and the North Pole that passes through Paris.
• The current measurement method for meter and second use the frequency and wavelength of light emitted by krypton and cesium. E.g., a second is the amount of time taken by 9192631770 cycles of a specific microwave wavelength in the X-band uniquely emitted by a cesium atom.
• These methods also mean that the speed of light is defined exactly, \(c=299 792 458\frac{m}{s}\). Very nice.

Derived from Latin mole, meaning bulk in English.<br> Cf., definition of molecule.

We will mostly use metric units in PHY2048.

We will be using a ton of these formulas, even the series expansions.

Check your numbers: [h5p id="1"]

Electromagnetic waves carry energy, momentum and information!

Boundary conditions are determined by the physical factors like length etc. You will find this all over physics!!

Interference!!!!

Linear restoring force means porportional to displacement \(x\), like the spring force \(\vec{F}=-kx\left(\hat{i}\right)\).

However, because the Taylor series exists, we can cobble together a linear wave equation from a nonlinear system, like the fluid we call air, for "small" displacements away from the equilibrium at ambient atmospheric pressure.

the relation \(\frac{\omega^2}{k^2}=v^2\) is known as a dispersion relation.

Have you learned about partial derivatives yet?

Important distinction. The medium moves about equilibrium, but the wave energy and momentum move through the medium.

This is why there is a famous differential equation, the "one-way wave equation," viz. $$\frac{df}{dt}-v\frac{df}{dt}=0$$

$$\frac{d}{dt}\varphi=\frac{d}{dt}\left(kx-\omega t\right)=k\frac{dx}{dt}-\omega$$

$$kv-\omega=0\text{ since }\varphi\text{ is a constant}$$

This means that the velocity of the phase \(\varphi\) is \(v=\frac{\omega}{k}=\lambda f\) since \(k=\frac{2\pi}{\lambda}\).

Then consider \(\varphi=kx+\omega t\) instead. That would correspond to a leftward propagating wave, as the text mentions.

Check out some of the resonance and standing wave demonstrations in my Demonstrations playlist in YouTube.

https://youtu.be/ImLbZjF_gaI

Read this paragraph

https://youtu.be/i3ZMoymgqbU

like sound waves

Even the nastiest physical system one could imagine, like the plasma states of the atmosphere of the Sun, when you comb through the calculus and extreme geometry, you can still make out wave equations like this one, \(v=\lambda f\)

I.e., sea level!

Or, in the case of sound waves, from its equilibrium pressure state -- the ambient atmospheric pressure.

You might get to quantum waves in PHY2049 or PHY3101 or higher courses.

You will study these in PHY2049.

Check this video in YouTube with two tuning forks.

https://youtu.be/Hmo-1sMsaNQ

Very complex time evolution equations for pressure, but they boil down under certain conditions to linear wave equations for the pressure as a function of position and time, \(p\left(x,\, t\right)\)

Transport of energy and momentum across spacetime.

Why we study oscillation in the simple spring-mass system, \(\vec{F}\left(x\right)=-kx\left(\hat{i}\right)\)

good workout involving max speed

Good workout

Nice workout

Bypass for now. Physics and engineering majors will see plenty of instruction on forced oscillation in upper level courses.

Bypass for now. Physics and engineering majors will see plenty of instruction on damped oscillation in upper level courses.

Another way to think about this approximation is the Taylor series for a function \(f\left(x\right)\) near its minimum at \(x = a\).

$$f\left(x\right)\approx f\left(a\right)+f^{\prime}\left(a\right) \left(x-a\right)+\frac{1}{2}f^{\prime\prime}\left(a\right)\left(x-a\right)^2$$

At the minimum, the derivative is zero, so \(f^{\prime}\left(a\right)=0\). Therefore, the first order term drops out; the zeroth order term, \(f\left( a\right)\) and the second order term \(\frac{1}{2}f^{\prime \prime}\left(a\right)\left(x-a\right)^2\) remain, i.e.,

$$f\left(x\right)\approx f\left(a\right)+\frac{1}{2}f^{\prime\prime}\left(a\right)\left(x-a\right)^2$$

If the function \(f\) is your potential, then you now have an oscillator potential for this small neighborhood of \(x=a\). We know the solutions for \(x\left(t\right)\) in this neighborhood, as we discussed in remote session 26.

Many physical systems can be studied profitably from this standpoint, the Taylor series expansion of the potential near one of its minima.

In the electromagnetic field of electric field \(\vec{E}\) and magnetic field \(\vec{B}\), the energy density of the field, \(u\), is also proportional to the squares of the fields: $$u=\frac{1}{2}\epsilon_0E^2+\frac{1}{2}\frac{1}{\mu_0}B^2$$

The constants \(\epsilon_0\) and \(\mu_0\) are related to the speed of light \(c\) :

$$c^2=\frac{1}{\epsilon_0 \mu_0}$$

after a ton of algebra

molecule broken

the famous case of "small" displacements.

Entree to current research areas!

good

Because, when \(x=\pm A\), the kinetic energy is zero, no matter what your initial conditions were, and all the energy \(E\) is potential.

The heart of the matter

"small"

$$y_1$$

with only the mass of the spring

Cf., simulations of hanging oscillator

Uh oh. The Math Dept. would not approve!

I.e., use a differential equation based on \(F=m\ddot{x}\).

...as usual with equations of motion!

Alternate model uses the generic solution but with the phase \(\phi\) converted into a sine term. E.g.,

$$x\left(t\right)=a\, \cos\left(\omega t\right)+b\, \sin\left(\omega t\right)$$

The constants \(a\) and \(b\) look like components of a vector whose magnitude is \(A\), viz. $$a=A\cos\left(\varphi\right) \ b=-A\sin\left(\varphi\right)$$

Initial conditions easiest to use: $$x\left(0\right)=0.02\, m \ \dot{x}\left(0\right)=0.0\frac{m}{s}$$

Result: \(\varphi = 0\). Nice.

We know this already

Good.

The velocity is leftward, derivative is negative, $$\dot{x}<0$$ during the first half of the motion, then rightward, positive derivative, $$\dot{x}>0$$ in the second half of the cycle... from THESE initial conditions $$x\left(0\right)=A\ \dot{x}\left(0\right)=0$$

This figure is a nice synopsis

That is, when \(t=T,\,2T,\,3T\dots\)

so the work done is not in KE, it is in the potential energy.

✔︎

I think we demonstrated this with variation of the swing angle back when we used pendula in the classroom.

We already know this from earlier in the semester: $$\omega^2=\frac{k}{m}$$

Incorrect. Its acceleration is rightward as soon as it crosses equilibrium, \(x<0\), i.e., in (c).

also max KE

Initial conditions: $$\vec{F}\left(0\right)=kA\left(-\hat{i}\right)$$ $$x\left(0\right)=A$$ $$\vec{v}\left(0\right)=0$$

These words imply, as we have already seen, that the total energy \(E\) is proportional to the square of the displacement, \(q\), using a generic coordinate \(q\) and to the square of the coordinate velocity, \(\dot{q}\). Or, as the math department might say, the function \(E\) is "quadratic" in \(q\) and \(\dot{q}\).

$$\frac{\text{human}}{\text{ultra}}=\left(2\times 10^4\right)/ \left(2.5\times 10^6\right)\approx 0.01$$

So that is "small" in the sense of calculus shortcuts.

This also means the wavelengths are much smaller for ultrasound.

Basic terms, must be familiar with them and where they fit into the equations of motion.

Dissipates energy.

Cavendish, however, could measure quantities like this!

actually in Cavendish's day, it was $$G=6.754\times 10^{-11}N\,m^2/kg^2$$

"a few kilometers"

I.e., from the top of Mt. Everest (elevation \(\approx 9\, km\)) to the bottom of the Mariana Trench (depth \(\approx 11\, km\). Cf., NASA planetary fact sheet.

E.g., $$\text{Everest} \longrightarrow\frac{1}{6380^2}=\frac{1}{40704400}\approx 2.457\times 10^{-8}$$ $$\text{sea level}\longrightarrow\frac{1}{6371^2}=\frac{1}{40589641}\approx 2.464\times 10^{-8}$$ $$\text{Challenger Deep}\longrightarrow\frac{1}{6360^2}=\frac{1}{40449600}\approx 2.472\times 10^{-8}$$

Not true for the Earth-moon system, as discussed, but good for most terrestrial objects like the oceans and for human objects like stadia and great pyramids.

Earth-moon system:

Earth and the average depth of the ocean (blue line):

Spectrum stadium, a blip next to Earth itself:

due to the symmetry of the circle of radius \(R_i\)

i.e., the orbital angular momentum of a single pixel

pixellation

Not so important in this context, but a good exercise.

good example

Be sure to work this out with some careful sketching. You can even define orbital angular momentum \(\vec{l}\) for a straight line trajectory.

Orbital angular momentum as opposed to spin angular momentum. It is amazing to think that, on a complex gravitational trajectory like an ellipse, parabola or hyperbola, the vector product of \(\vec{r}\) and \(\vec{p}\) is conserved. The magnitudes of momentum and radial distance and their geometric relationship are controlled by gravity to lock the object into its elliptical, parabolic or hyperbolic orbit.

Bypass

Oscillator potential hides in this example!!

like a "grandfather clock"

an area integral, but not as hard as \(dx\, dy\)

We kind of did this integral in Session 21.

Session 21 intro

pixellation

We made a study of this several weeks ago

bypass

We will not dwell too much on angular acc.

not sure this is correct, but magnitude \(r\) is constant in uniform circular motion The real reason that \(\frac{d\vec{r}}{dt}=0\) is that in uniform circular motion, the position vector \(\vec{r}\) only changes in the \(\hat{\theta}\) or \(-\hat{\theta}\) direction, which are parallel or antiparallel to \(\vec{\theta}\). The cross product of parallel or antiparallel vectors is zero!

A good diagram

Vector product aka cross product, using the right hand rule.

...as in the motorcyclist problem.

Use linear mass densities 8.0g/m and 2.7 g/m.

Use this integral and its associated example to tackle the problem of the one meter rod, using linear mass densities \(\lambda_{iron}=8.0\frac{g}{cm}\) and \(\lambda_{al}=2.7\frac{g}{cm}\)

Try this with \(m_1=1.000\, kg\) and \(m_2=1.500\, kg\). A good study problem.

brain burner!

good study problem

unlike the motorcyclist making a jump

nice

I like this example

We discussed this

yup, this is it.

This is why an astronomer can focus on the trajectory of a star or planet or galaxy as if all of its mass is concentrated at a point located at the center of mass.

bypass this section

Good approximation for an ideal gas, for instance.

Like a set of boxcars in the freight yard hooking together and moving off at a new speed.

Skateboarder interaction