1,616 Matching Annotations
  1. Oct 2020
    1. This property allows us to define a different kind of energy for the system than its kinetic energy, which is called potential energy.
    2. Spring Potential Energy

      good workout

    1. Compare the work required to accelerate a car of mass 2000 kg from 30.0 to 40.0 km/h with that required for an acceleration from 50.0 to 60.0 km/h.

      Good workout

    2. Determining a Stopping Force

      We worked this problem out in Session 14

    3. By using the work-energy theorem, you did not have to solve a differential equation to determine the height.

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

    4. Loop-the-Loop

      A good example to work out

    5. We can also get this result from the work-energy theorem.

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

    6. The importance of the work-energy theorem, and the further generalizations to which it leads, is that it makes some types of calculations much simpler to accomplish than they would be by trying to solve Newton’s second law.

      A much more interesting way to view the universe.

  2. pressbooks.online.ucf.edu pressbooks.online.ucf.edu
    1. One very important and widely applicable variable force is the force exerted by a perfectly elastic spring

      Extremely important for all scientists

    2. Work Done by a Variable Force over a Curved Path

      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.

    3. paths

      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.

    1. For example, in the rectangular coordinate system in a plane, the scalar x-component of a vector is its dot product with the unit vector ^ii^\mathbf{\hat{i}}, and the scalar y-component of a vector is its dot product with the unit vector ^jj^\mathbf{\hat{j}}:

      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.

    2. We can use the commutative and distributive laws to derive various relations for vectors, such as expressing the dot product of two vectors in terms of their scalar components.

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

    1. NASA researchers test a model plane in a wind tunnel. (credit: NASA/Ames)

      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!

    1. A higher coefficient would also allow the car to negotiate the curve at a higher speed

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

    2. frictional force

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

    3. In Motion in Two and Three Dimensions, we examined the basic concepts of circular motion. An object undergoing circular motion, like one of the race cars shown at the beginning of this chapter, must be accelerating because it is changing the direction of its velocity. We proved that this centrally directed acceleration, called centripetal acceleration, is given by the formula ac=v2rac=v2r{a}_{\text{c}}=\frac{{v}^{2}}{r} where v is the velocity of the object, directed along a tangent line to the curve at any instant. If we know the angular velocity ωω\omega, then we can use ac=rω2.ac=rω2.{a}_{\text{c}}=r{\omega }^{2}. Angular velocity gives the rate at which the object is turning through the curve, in units of rad/s. This acceleration acts along the radius of the curved path and is thus also referred to as a radial acceleration.

      Very nice synopsis for uniform circular motion.

    4. particularly low pressures.

      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

    1. we can find the coefficient of kinetic friction if we can find the normal force on the skier

      Good

    1. Figure 5.26 The weight of a tightrope walker causes a wire to sag by 5.0∘5.0∘5.0^\circ.

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

    2. Once you have determined the tension in one location, you have determined the tension at all locations along the rope.

      Important

    3. outward

      .

  3. Sep 2020
    1. forces. Zooming in deeper, your body’s atoms are held together by electrical forces, and the core of the atom, called the nucleus, is held together by the strongest force we know—strong nuclear force

      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.
    2. Forces affect every moment

      ...of all objects in the universe.

    3. these situations

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

    1. skier

      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.

    2. Finding a Particle Acceleration

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

    1. The position of a particle

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

    2. It is remarkable that for each flash of the strobe

      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

    3. he velocity vector becomes tangent to the path of the particle.

      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.

    4. The instantaneous velocity vector is now

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

    5. Brownian Motion

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

    6. The displacement vector Δ→rΔr→\Delta \mathbf{\overset{\to }{r}} is found by subtractin

      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., \(x\left(t_2\right)-x\left(t_1\right)\)
    7. The position vector from the origin of the coordinate system

      Make sure to review this part. Vectors are essential.

    1. Thus, to describe the motion of any of the jets accurately, we must assign to each jet a unique position vector in three dimensions as well as a unique velocity and acceleration vector.

      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.
    1. A train is moving up a steep grade

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

    2. Height of free fall

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

    1. Kinematic Equations for Objects in Free Fall

      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.

    1. The equation –v=v0+v2v–=v0+v2\overset{\text{–}}{v}=\frac{{v}_{0}+v}{2} reflects the fact that when acceleration is constant, v is just the simple average of the initial and final velocities.

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

    2. We now make the important assumption that acceleration is constant.

      Just like Galileo did.

  4. Aug 2020
    1. In the control tower at a regional airport,

      Also a good exercise

    2. A barge is pulled by the two tugboats

      good exercise

    1. Ask yourself: Does this make any sense?

      ALWAYS!!!

    2. If all else fails, bound it. For physical quantities for which you do not have a lot of intuition, sometimes the best you can do is think something like: Well, it must be bigger than this and smaller than that. For example, suppose you need to estimate the mass of a moose. Maybe you have a lot of experience with moose and know their average mass offhand. If so, great. But for most people, the best they can do is to think something like: It must be bigger than a person (of order 102 kg) and less than a car (of order 103 kg). If you need a single number for a subsequent calculation, you can take the geometric mean of the upper and lower bound—that is, you multiply them together and then take the square root. For the moose mass example, this would be

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

    1. This simple fact can be used to check for typos or algebra mistakes, to help remember the various laws of physics, and even to suggest the form that new laws of physics might take.

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

    1. yotta

      Based on another familiar term. ; )

    2. Check out a comprehensive source of information on SI units at the National Institute of Standards and Technology (NIST) Reference on Constants, Units, and Uncertainty.

      Well worth checking out.

    3. base units. All other physical quantities can then be expressed as algebraic combinations of the base quantities. Each of these physical quantities is then known as a derived quantity and each unit is called a derived unit.

      The distinction between base units and derived units is important.

    4. the units for some physical quantities must be defined through a measurement process

      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.
    5. Amount of substance mole (mol)
    6. The metric system is also the standard system agreed on by scientists and mathematicians.

      We will mostly use metric units in PHY2048.

    1. Mathematical Formulas

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

  5. May 2020
    1. For a star 500 light-years away, the light we detect tonight left 500 years ago and is carrying 500-year-old news.

      Electromagnetic waves carry energy, momentum and information!

  6. Apr 2020
    1. Given that maximum air displacements are possible at the open end and none at the closed end, other shorter wavelengths can resonate in the tube

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

    1. ear. This wave function models the displacement of the medium of the resulting wave at each position along the x-axis. If two linear waves occupy the same medium, they are said to interfere.

      Interference!!!!

    2. to a linear restoring force of the medium

      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.

    3. linear wave equation

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

    4. partial derivative

      Have you learned about partial derivatives yet?

    5. the wave speed is constant and represents the speed of the wave as it propagates through the medium, not the speed of the particles that make up the medium

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

    6. A wave function is any function such that

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

    7. Whether the temporal term ωtωt\omega t is negative or positive depends on the direction of the wave.

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

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

      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.

    1. Waves may be transverse, longitudinal, or a combination of the two.

      Read this paragraph

    2. waves on stringed instruments
    3. In a longitudinal wave

      like sound waves

    4. This fundamental relationship holds for all types of 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, \(v=\lambda f\)

    5. the equilibrium position is indicated by the dotted line

      I.e., sea level!

    6. The amplitude of the wave (A) is a measure of the maximum displacement of the medium from its equilibrium position.

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

    7. Matter waves are a central part of the branch of physics known as quantum mechanics.

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

    8. Electromagnetic waves are associated with oscillations in electric and magnetic fields and do not require a medium.

      You will study these in PHY2049.

    9. vibrating tuning fork

      Check this video in YouTube with two tuning forks.

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

    10. For sound waves, the disturbance is a change in air pressure

      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, \(p\left(x,\, t\right)\)

    11. Mechanical waves transfer energy and momentum

      Transport of energy and momentum across spacetime.

    12. elastic restoring forc

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

    1. 34. One type of BB gun uses a spring-driven plunger to blow the BB from its barrel. (a) Calculate the force constant of its plunger’s spring if you must compress it 0.150 m to drive the 0.0500-kg plunger to a top speed of 20.0 m/s. (b) What force must be exerted to compress the spring?

      good workout involving max speed

    2. 32. Fish are hung on a spring scale to determine their mass. (a) What is the force constant of the spring in such a scale if it the spring stretches 8.00 cm for a 10.0 kg load? (b) What is the mass of a fish that stretches the spring 5.50 cm? (c) How far apart are the half-kilogram marks on the scale?

      Good workout

    3. A suspension bridge oscillates with an effective force constant of 1.00×108N/m1.00×108N/m. (a) How much energy is needed to make it oscillate with an amplitude of 0.100 m? (b) If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart 1.00×104J1.00×104J of energy each second, how long does it take for the bridge’s oscillations to go from 0.100 m to 0.500 m amplitude.

      Nice workout

    1. 15.6 Forced Oscillations

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

    1. 15.5 Damped Oscillations

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

    1. if you are familiar with

      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.

    2. energy proportional to the amplitude squared

      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 \(c\) :

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

    3. can be approximated

      after a ton of algebra

    4. the atoms can be separated

      molecule broken

    5. when displaced a small amount from the equilibrium position.

      the famous case of "small" displacements.

    6. A practical application of the concept of stable equilibrium points is the force between two neutral atoms in a molecule.

      Entree to current research areas!

    7. The potential energy curve in Figure resembles a bowl.

      good

    8. is proportional to the square of the amplitude

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

    9. Energy in Simple Harmonic Motion

      The heart of the matter

    1. Using the small angle approximation gives an approximate solution for small angles,

      "small"

    1. new equilibrium position

      $$y_1$$

    2. The spring is hung from the ceiling and the equilibrium position is marked as yoyo{y}_{o}.

      with only the mass of the spring

    3. Vertical Motion and a Horizontal Spring
    4. Cancelling out like terms

      Uh oh. The Math Dept. would not approve!

    5. equations of motion and Newton’s second law

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

    6. The position, velocity, and acceleration can be found for any time.

      ...as usual with equations of motion!

    7. can be modeled with the following 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. $$a=A\cos\left(\varphi\right) \ b=-A\sin\left(\varphi\right)$$

    8. pulling it out to x=+0.02m.x=+0.02m.x=+0.02\,\text{m}\text{.} The block is released from rest

      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.

    9. The velocity of the mass on a spring, oscillating in SHM, can be found by taking the derivative of the position equation

      We know this already

    10. This is the generalized equation for SHM

      Good.

    11. Figure 15.6

      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 $$x\left(0\right)=A\ \dot{x}\left(0\right)=0$$

    12. Figure 15.5

      This figure is a nice synopsis

    13. repeats every integer multiple of the period.

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

    14. from rest

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

    15. analyze the equations of motion

      ✔︎

    16. For one thing, the period T and frequency f of a simple harmonic oscillator are independent of amplitude.

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

    17. The stiffer the spring is, the smaller the period T. The greater the mass of the object is, the greater the period T.

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

    18. The mass now begins to accelerate in the positive x-direction

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

    19. maximum negative velocity at x=0x=0x=0.

      also max KE

    20. The mass is displaced to a position x=Ax=Ax=A and released from rest.

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

    21. In simple harmonic motion, the acceleration of the system, and therefore the net force, is proportional to the displacement and acts in the opposite direction of the displacement.

      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 \(\dot{q}\).

    22. much highe

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

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

      This also means the wavelengths are much smaller for ultrasound.

    23. Period and Frequency in Oscillations

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

    1. This is the weight of a typical grain of pollen.

      Cavendish, however, could measure quantities like this!

    2. Cavendish determined it to be

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

    1. for objects within a few kilometers of Earth’s surface, we can take

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

    2. the radius of Earth

      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:

    1. will be cancelled

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

    2. the magnitude of the angular momentum of this mass segment is

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

    3. small mass segments

      pixellation

    4. What is the torque

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

    5. Angular Momentum and Torque on a Meteor

      good example

    6. lever arm

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

    7. Angular Momentum of a Particle

      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.

  7. Mar 2020
    1. Example

      Oscillator potential hides in this example!!

    2. A pendulum in the shape of a rod

      like a "grandfather clock"

    3. a little bit trickier

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

    4. A uniform thin rod with axis at the end

      We kind of did this integral in Session 21.

    5. The need to use an infinitesimally small piece of mass dm suggests that we can write the moment of inertia by evaluating an integral over infinitesimal masses

      Session 21 intro

    1. Angular Acceleration

      We will not dwell too much on angular acc.

    2. Since →rr→\mathbf{\overset{\to }{r}} is constant

      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!

    3. Figure 10.5 For counterclockwise rotation in the coordinate system shown

      A good diagram

    4. These three vectors are related to each other

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

    5. Now we expand our description of motion to rotation

      ...as in the motorcyclist problem.

    1. Find the center of mass

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

    2. the summation becomes an integral:

      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 \(\lambda_{al}=2.7\frac{g}{cm}\)

    3. Two particles of masses m1m1{m}_{1} and m2m2{m}_{2} move uniformly

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

    4. Two particles of masses m1m1{m}_{1} and m2m2{m}_{2} separated by a horizontal distance D are let go from the same height h at different times

      brain burner!

    5. Three point masses

      good study problem

    6. no net external force acting on it

      unlike the motorcyclist making a jump

    7. Therefore, there is no meaningful application of the center of mass of a unit cell beyond as an exercise.

      nice

    8. Suppose we included the sun in the system

      I like this example

    9. The radius of Earth is 6.37×106m6.37×106m6.37\times {10}^{6}\,\text{m}, so the center of mass of the Earth-moon system is (6.37 − 4.64) ×106m=1.73×106m=1730km×106m=1.73×106m=1730km×\,{10}^{6}\,\text{m}=1.73\times {10}^{6}\,\text{m}=1730\,\text{km} (roughly 1080 miles) below the surface of Earth.

      We discussed this

    10. position of the center of mass has units of meters; that suggests a definition

      yup, this is it.

    11. This is Newton’s second law

      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.

    1. the total kinetic energy of the system is conserved. Such an interaction is called elastic.

      Good approximation for an ideal gas, for instance.

    2. Multiple objects can collide and stick together, forming a single object

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

    3. A single object can explode into multiple objects (one-to-many).

      Skateboarder interaction

    1. but those changes are identical in magnitude, though opposite in sign.

      Newton's third law is best understood in this way. The "equal but opposite reaction" is encoded quantitatively by the changes of momentum, or exchanges of momenta.

    1. when a force is applied for an infinitesimal time interval dt, it causes an infinitesimal impulse d→JdJ→d\mathbf{\overset{\to }{J}},

      So, for instance, the moon exchanges momentum with Earth as it travels around its orbit, but second by second, we can compute the amount of impulse, e.g.,

      \(\vec{J}=\Delta \vec{p}=\int_{0.0\,s}^{1.0\,s} \left[ \frac{Gm_{\oplus}m_{moon}}{\left[r\left(t\right)\right]^2}\right]dt\)

    1. An elephant and a hunter are having a confrontation.

      good example

    2. The momentum p of an object is the product of its mass and its velocity:

      Sir Isaac Newton's invention, "quantity of motion."

    1. the positions where its kinetic energy is zero

      turning points in the motion

    2. Finding the force from the potential energy is mathematically easier than finding the potential energy from the force, because differentiating a function is generally easier than integrating one.

      Good strategy note

    3. Force due to a Quartic Potential Energy

      I like this Example!

    1. In Relativity, we will see that conservation of energy still applies to a non-classical particle, but for that to happen, we have to make a slight adjustment to the definition of energy.

      Correctamente! Slight adjustment, indeed, plus E becomes a component of the total relativistic momentum in four dimensions, \(\vec{p}=\left(p_x,\, p_y,\, p_z\right)\longrightarrow \vec{p}=\left(E,\, p_x,\, p_y,\, p_z\right) \).

    1. we can rearrange the differentials dt, etc., as algebraic quantities in this expression

      This would not be allowed in the math department, however.

  8. Feb 2020
    1. The string vibrates around an equilibrium position, and one oscillation is completed when the string starts from the initial position, travels to one of the extreme positions, then to the other extreme position, and returns to its initial position.

      Very good description of the periodic motion of a string on a guitar. The frequency of oscillation is in the audible range, e.g., A above middle C is 440 Hz.

    1. If you have observed the classroom demonstration with the spinning bicycle wheel

      I hope to do this demonstration during Lecture 16, on Feb. 27.

    1. Atwood Machine

      Bypass

    2. Two Attached Blocks

      Bypass

    3. What Does the Bathroom Scale Read in an Elevator?

      Bypass

    4. First consider the horizontal or x-axis:

      I can now see why a few of you, on HW 4, used this notation, \( F_{net\ x}=T_{2x} - T_{1x}=0\) rather than the way I usually write it, \(\sum F_x = T_{2x}-T_{1x}=0\).

    5. This implies

      Notice the subtle change of notation here, using the magnitude w as opposed to the vector notation in Fig. 6.3.

    6. Different Tensions at Different Angles

      Spider-man problem

    1. Inertial Forces and Noninertial (Accelerated) Frames: The Coriolis Force

      Bypass. This section requires Calculus 3 concepts to be done properly, although the weather diagrams, Fig. 6.28 are instructive to view.

    2. Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug’s xy-position, velocity, and acceleration using vectors or graphs.

      Not sure why this paragraph is here.

    3. the horizontal component of lift

      It also accelerates a tide of circulating blood into the pilot -- blackouts occur

    4. due to the force of the air on the wing

      "force of the air on the wing" ↔︎ Many scientists describe aerodynamic lift as the pressure differential \(\Delta p\) between the top and bottom of the wing, effecting every square inch of the wing.

  9. pressbooks.online.ucf.edu pressbooks.online.ucf.edu
    1. the increment of work dW

      Here is a mini-differential equation

    1. Often, an analytic solution is intractable or impossible, requiring lengthy numerical solutions or simulations to get approximate results.

      MANY scientific jobs require this skill, numerical solution of differential equations.

    1. This means a skydiver with a mass of 75 kg achieves a terminal velocity of about 350 km/h while traveling in a pike (head first) position, minimizing the area and his drag.

      about the same as a small 2" sphere

    2. Ferrari Testarossa 0.37 Dodge Ram Pickup 0.43

      Amazing comparison.

    3. wind tunnel

      Cf., also the NASA Langley Research Center's wind tunnel fact sheet.

    1. but →ww→\mathbf{\overset{\to }{w}} has components along both axes,

      The free body diagram on the right is a bit different than the sketches I used in Session 12.<br> The author's \(\vec{w}_y\) vector is attached at its tail to the mass point, whereas in my sketch, it is slid over and attaches at the tip of \(\vec{w}\). But both are acceptable and useful. As I have mentioned several times, how you diagram forces etc. can be different from the author's and from my way of diagramming, and still be fine.

    2. Consider the skier on the slope

      A good study problem

    1. Newton’s Third Law of Motion

      Our demonstration in Session 10, with the two skateboarders out in the hallway, is good to remember here.

    1. changes in motion

      By "changes in motion," Newton meant \(\Delta \vec{v}\) and therefore \(\vec{a}=\frac{\Delta \vec{v}}{\Delta t}\)

  10. pressbooks.online.ucf.edu pressbooks.online.ucf.edu
    1. without the necessity of physical contact between objects

      Or, once could redefine contact to mean an interaction "at a distance," as the author mentions here.

    2. A quantitative definition of force can be based on some standard force

      This is because force is a derived quantity, based on time, space and mass measurements.

    3. Isaac Newton

      Galileo's famous student!

    1. b

      Be careful checking your answers here, too.

    2. a

      It is not clear to me what this expression is supposed to be. Be careful when checking your answer for h. For example,

      $$\frac{24\frac{m}{s}}{9.8 \frac{m}{s^2}}=2.49\, s$$

    3. Olympus Mons

      Another planet? No problem. A walk in the park. : )

    4. solid volcanic rocks

      also known as "bombs" "Accretionary lava ball comes to rest on the grass after rolling off the top of an ‘a‘a flow in Royal Gardens subdivision. Accretionary lava balls form as viscous lava is molded around a core of already solidified lava (photo by J.D. Griggs, 7/2/83, JG2562) (picture #012)."