---------------------------- Original Message ----------------------------
Subject: Re: http://aethro-kinematics.com/
From: 'Akira Kanda' <ka...@cs.toronto.edu>
Date: Thu, July 12, 2018 9:51 am
To: 'Robert Neil Boyd Ph D' <rnbo...@comcast.net>
--------------------------------------------------------------------------
Neil,
This kind of anti-sceince attitude coming from insecurity of physicists
which will make physics a laughing stock to even middle school students.
Only an absolute anti-science thinkers will say that everything is OK with
the claim that the kinetic energy of mv is (mv^2)/2 even though the work
needed to do acceleration from m0 to mv needs not be (mv^2)/2. If
physicists are to regain public trust and respect they have to come clear
on this fatal issue. I have shown you that physics is nothing but a cult!
Main stream or dissident, this problem is shared among all of physicists.
Scientific honesty is in question.
We are determined to nail this fake science which mislead the entire
humanity for centuries. As I told you, Newton rejected energy. I presume
he knew this problem. We will make sure that this lie will stop. If
e=(mv^2)/2 does not bother you, what you are doing is not science, it is
cult! More and more people are now realizing that the so called physics
is more corrupted than the middle age Vatican cosmology. Vatican
theologians were honest. When they discovered that all of their 'proof'
for the existence of God was false as they assumed the existence of God to
prove the existence of God, they abandoned predicative reasoning and went
into modal logic from which today's information science benefits a lot.
I am pointing out the truth that the trouble of physics did not start with
relativity theory and quantum mechanics. They are also victims of the Post
Newtonian classical mechanics and em theory. When they introduced the
concept of energy without noticing that this concept is illegitimate,
physics stepped into a deep dung heap. After that physics became a cult a
most destructive cult.
I do not put cloth on my mouth when it comes to scientific truth. So,
sometimes I may be too direct. But it is what has to be done. What we need
is the truth and it does not come from indoctrination or social skill. We
have to discuss! I do not tolerate any bull shitting! I will clearly say,
something is wrong when it is wrong. This is the way of science. I also
make it clear that I do not understand if I do not understand. It is
called scientific honesty which the 20th century physics completely lost.
So, I will remind you that if you really think you are OK with what you
said to me about e=(mv^2)/2, you must make sure that you never use the
concept of energy in your work. What is Poingting vector about, b.t.w.?
Aether theory does not deal with Pointing vector?
Akira
>
> I wish you would bother to read the messages I take the time and
> trouble to compose and to send to you.
>
> It is obvious you did not read the entire of my most recent
> message.
>
>
> I am not a theoretician. I am an experimentalist.
>
> Your complaints are best directed elsewhere.
>
> I am not going to get involved with your mv ^2 arguments which
> have no bearing on any my experiments or instrumented
> observations, especially when you demonstrate that you do not
> listen to anything I say.
>
> Have a nice day.
>
>
> On 7/11/2018 2:56 PM, Akira Kanda wrote:
>
> Neil, let us make it very simple and clear. Just straight
> into the issues. Everything started with Newton's theology
> of the universe. Everything else is its derivative. I am not
> discussing the derivatives. I am asking about the most
> important foundation. I do not see why you are taalking about
> the frill issue of terminal sciences to consider this crisis
> at the deepest level of physical science. It is just a matter
> of very basic logical reasoning. We need no fancy frills to
> cover up the real issues. I am asking physicists a very
> simple question on the concept of energy and its conservation
> law. Interestingly Newton did not work on this issue thought
> he considered momentum. As I keep saying Newton was a great
> improvement to all of his successors who tried to cash in the
> glory of his work through twisting his original work. The
> work needed to accelerate from m0 to mv is depending upon the
> way we accelerate. It is not correct to say that it is
> (mv^2)/2, which is the case only for very limited cases such
> as constant acceleration. Then why is the 'kinetic energy' of
> mv (mv^2)/2? As I said, this is a middle school level
> physics question. Why is it that nobody at the top of the
> hierarchy of the tens of billions of dollars public funding
> guzzling community can not come up with any satisfactory
> answer to this middle school level question. Their 'answer'
> is always diversion from the real issues. I do not need any
> reference to other field of science in which they use theory
> of physics which is based upon this highly questionable
> concept of energy and their conservation law. The answer must
> come from something as basic as the theory of physics if not
> more. This is called normal reasoning, or logic. Anything
> which uses this questionable assumption of kinetic energy can
> not answer to this question at the pain of vicious
> circularity. It is this totally corrupted culture of physics
> which lead theoretical physic to this totally laughable and
> well deserved end. Now young generation of scientists
> consider physicists liars or totally confused people, a most
> corrupted authoritarian minds in human history. I think they
> have some kind of intellectual identity crisis emanating from
> their superiority complex. So, you are saying that to
> understand why the kinetic energy of mv is (mv^2)/2 one has
> to understand the most advanced cosmology which is based upon
> this assumption about kinetic energy and the conservation law
> of energy in general. I feel dizzy now Neil. We
> mathematicians are normal thinkers. We make our human
> reasoning a little more precise. This is our world. The world
> of physics is so advanced and esoteric that we need to think
> differently form the way we normal people do? Akira
> Have a look at this:
> http://aethro-kinematics.com/ On 7/11/2018 3:51 AM,
> Akira Kanda wrote: Neil, Here are some
> more to ask. [1] As I said the concept of energy makes no
> sense at all. It is a myth. Then why are people talking about
> dark energy? I insist that those who discuss dark energy
> provide coherent acceptable concept of energy. First explain
> why the kinetic energy of mv is (mv^2)/2. Whenever, I ask
> this uncomfortable question, there is a deadly science.
> Physicists just ignore it and move on. What is going on? Come
> on, it is the 'most important and fundamental' concept of
> physics which curiously Newton did not talk about.
> It seems to me you need to investigate what is energy a
> little bit more. .
> I had questions about what is energy in the 7th grade.
> I was given a definition of work and a definition of
> power, both of which require energy. I was also
> given the definitions of kinetic energy and potential
> energy. https://en.wikipedia.org/wiki/Kinetic_energy
> https://en.wikipedia.org/wiki/Potential_energy Type of
> energy Description Mechanical
> the sum of macroscopic translational
> and rotational kinetic and potential energies
> Electric potential energy due to or stored in electric
> fields Magnetic
> potential energy due to or stored in magnetic fields
> Gravitational potential energy due to
> or stored in gravitational fields
> Chemical potential energy due to chemical
> bonds Ionization
> potential energy that binds an electron to its atom
> or molecule Nuclear
> potential energy that binds nucleons to form
> the atomic nucleus (and nuclear reactions) Chromodynamic
> potential energy that binds quarks
> to form hadrons Elastic
> potential energy due to the deformation of a material
> (or its container) exhibiting a restorative force
> Mechanical wave kinetic and
> potential energy in an elastic material due to a
> propagated deformational wave Sound
> wave kinetic and potential energy in a fluid due to a
> sound propagated wave (a particular form of
> mechanical wave) Radiant potential
> energy stored in the fields of propagated by
> electromagnetic radiation, including light
> Rest
> potential energy due to an object's rest mass Thermal
> kinetic energy of the microscopic
> motion of particles, a form of disordered equivalent of
> mechanical energy In the International System of
> Units (SI), the unit of energy is the joule, named after
> James Prescott Joule. It is a derived unit. It is equal to
> the energy expended (or work done) in applying a force
> of one newton through a distance of one metre. However
> energy is also expressed in many other units not part
> of the SI, such as ergs, calories, British Thermal
> Units, kilowatt-hours and kilocalories, which
> require a conversion factor when expressed in SI units. The
> SI unit of energy rate (energy per unit time) is the watt,
> which is a joule per second. Thus, one joule is one
> watt-second, and 3600 joules equal one watt-hour. The
> CGS energy unit is the erg and the imperial and
> US customary unit is the foot pound. Other energy units such
> as the electronvolt, food calorie or thermodynamic kcal
> (based on the temperature change of water in a heating
> process), and BTU are used in specific areas of science
> and commerce. You also seem to want to do away with
> fields entirely. That is not going to work out too well for
> you. And what do you propose to replace fields with? I am
> interested in physical systems and in being able to predict
> their behaviors so I can build new technologies that function
> perfectly. More later, Neil [2] To
> those who say that wave is the foundation of physics, I have
> the following question: What does it mean to apply force to a
> part of wave media? More generally what does it mean to apply
> force to a physical body which has geometric shape and
> dimension? Newton knew the difficulty and so he idealized the
> situation only to consider force upon a point mass. As we
> know force is a vector which has a pointed end. How can this
> pointed end exert force upon a cube for example? What does it
> mean precisely by a sphere moving with speed v. If the
> sphere is spinning too, what does it really mean? I do not
> know how to articulate these concepts mathematically. I
> understand Newton's idea of a force vector pushing a point
> mass OK. No problem. I do not know what contemporary
> physicists are doing and what they are saying at all. It
> sounds all obscure, sorry for harsh words but I think the
> kings must be told now. When I asked them these uneasy
> uncomfortable questions, most of them just walked way or
> changed the topics. Some of them became visibly less than
> cordial and called me 'crank'. I met some theoretical
> physicists who told me 'I hate people like you.' [3] I can
> not understand what does it mean by force exerted on a
> non-point shape physical object changes the shape of the
> object etc. I just do not understand on what ground these
> people are making their argument. We are having serious
> problems with figuring out what is the kinetic energy of a
> point mass m moving with speed v. This is not as simple as
> theoretical physicists think. Give me a mathematical
> definition of a ball moving with speed v while spinning. So
> far I met no theoretical physicist who came up with any
> decent answer. same switch the precise mathematical
> explanation of what happens when we kick a ball on the ground
> hat they do and what they say makes practical sense and they
> never claim that they can explain the nature of things at
> all. [4] Going back to aether theory, I do not understand
> how is it possible that such a complex theory of aether can
> explicate the most basic physics of electromagnetic charges.
> It makes good sense that Gauss and Weber did not use either
> aether nor em field. Neil, I have a question. Maxwell and his
> field theory failed because force field violates action
> reaction law. We can not represent local phenomena globally
> using force field. This is the lesson we have to learn from
> newton who ignored the concept of force field for very good
> reason. My question is that aether is a global entity which
> covers the entire universe. How such global continuum entity
> gives birth to totally local theory of electromagnetism of
> Gauss-Weber. For me the direction should e opposite. We have
> to start with Gauss-Weber and then move on to electrical
> engineering. Electrical engineering will not give us the
> physics of electromagnetism. I just do not understand how is
> it possible to give birth to particle based theory of
> electromagnetism from continuum theory of em fields or
> aether. It makes no mathematical sense. In mathematics, it
> took a long time to reach the world of real continuum from
> discrete world of natural numbers. It did not go the other
> way around. The price we paid for the wrong approach of em
> field theory of Maxwell was that we ended up wit a
> contradiction: the so called Lawrence force. Of course no
> theoretical physicists see that this is a problem. They never
> ever thought that this a problem. The truth of the matter is
> that this is in complete contradiction with Newton's second
> law. Sorry electrical engineering may be OK but the physical
> 'theory' of electromagnetic field is nonsense. So, Niel, I am
> asking you how this contradiction is resolved in aether
> theory. I asked this question to some aether theorists and I
> received no answer. They were as rude as mainstream. Akira
>
- Arthrokinematics Of Thoracic Spine
- Arthrokinematics Spin
- Arthrokinematics Def
- Arthrokinematics
- Arthrokinematics Wiki
- Arthrokinematics Of Talocrural Joint
Learning Objectives
By the end of this section, you will be able to:
Theoretical physics Aethro-Dynamics Textbook Perfect Paperback Steven Rado New. Free shipping DIGITAL BOOK Theoretical Basis for Nursing (5th Edition) - INSTANT DELIVERY. Introduction to Theoretical Physics, Leigh Page, 2d Ed. Hardcover 5th Prtg 1944. Posted by kanda, Jul 12, 2018 6:54 AM. Introduction 'Arthrokinematics' refers to the movement of joint surfaces. The angular movement of bones in the human body occurs as a result of a combination of rolls, spins, and slides. A roll is a rotary movement, one bone rolling on another. Aethro-Kinematics renders an alternate mechanical solution for the polarization of light. Thus, it reinstates Faraday and Maxwell's gaseous model of the Aether and resumes the original task of exploring all 'action at a distance forces' as fluid dynamical behavior of the all-pervading Aether.
- Observe the kinematics of rotational motion.
- Derive rotational kinematic equations.
- Evaluate problem solving strategies for rotational kinematics.
Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating ω, α, and t. To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion:
[latex]v={v}_{0}+{at}[/latex] (constant a)
Note that in rotational motion a = at, and we shall use the symbol a for tangential or linear acceleration from now on. As in linear kinematics, we assume a is constant, which means that angular acceleration α is also a constant, because a = rα. Now, let us substitute v = rω and a = rα into the linear equation above:
rω =rω0 + rat.
The radius r cancels in the equation, yielding
ω =ω0 + at. (constant a)
where ω0 is the initial angular velocity. This last equation is a kinematic relationship among ω, α, and t —that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart.
Making Connections
Starting with the four kinematic equations we developed in One-Dimensional Kinematics, we can derive the following four rotational kinematic equations (presented together with their translational counterparts):
| Rotational | Translational | |
|---|---|---|
| [latex]theta =bar{omega }t[/latex] | [latex]x=bar{v}t[/latex] | |
| ω = ω0 + αt | v = vo + at | (constant α, a) |
| [latex]theta ={omega }_{0}t+frac{1}{2}{alpha t}^{2}[/latex] | [latex]x={v}_{0}t+frac{1}{2}{text{at}}^{2}[/latex] | (constant α, a) |
| ω2 = ω02+ 2αθ | v2 = vo2 + 2ax | (constant α, a) |
In these equations, the subscript 0 denotes initial values (θ0, x0, and t0 are initial values), and the average angular velocity [latex]bar{omega}[/latex] and average velocity [latex]bar{v}[/latex] are defined as follows:
[latex]bar{omega}=frac{{omega }_{0}+omega }{2}text{ and }overline{v}=frac{{v}_{0}+v}{2}[/latex].
The equations given above in Table 1 can be used to solve any rotational or translational kinematics problem in which a and α are constant.
Problem-Solving Strategy for Rotational Kinematics
- Examine the situation to determine that rotational kinematics (rotational motion) is involved. Rotation must be involved, but without the need to consider forces or masses that affect the motion.
- Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful.
- Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
- Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion.
- Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles.
- Check your answer to see if it is reasonable: Does your answer make sense?
Example 1. Calculating the Acceleration of a Fishing Reel
A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110 rad/s2 for 2.00 s as seen in Figure 1. (a) What is the final angular velocity of the reel? (b) At what speed is fishing line leaving the reel after 2.00 s elapses? (c) How many revolutions does the reel make? (d) How many meters of fishing line come off the reel in this time?
Strategy
In each part of this example, the strategy is the same as it was for solving problems in linear kinematics. In particular, known values are identified and a relationship is then sought that can be used to solve for the unknown.
Solution for (a)
Here α and t are given and ω needs to be determined. The most straightforward equation to use is ω = ω0+αt because the unknown is already on one side and all other terms are known. That equation states that
We are also given that ω0 = 0 (it starts from rest), so that
ω = 0 + (110 rad/s2)(2.00 s) = 220 rad/s
Solution for (b)
Now that ω is known, the speed v can most easily be found using the relationship
where the radius r of the reel is given to be 4.50 cm; thus,
Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have m × rad = m .
Solution for (c)
Here, we are asked to find the number of revolutions. Because 1 rev=2π rad, we can find the number of revolutions by finding θ in radians. We are given α and t, and we know ω0 is zero, so that θ can be obtained using [latex]theta = omega_{0}t+frac{1}{2}{{alpha t}}^{2}[/latex].
[latex]begin{array}{lll}theta &=& omega_{0}t+frac{1}{2}{{alpha t}}^{2} &=& 0+left(text{0.500}right)left(text{110}{text{rad/s}}^{2}right){left(text{2.00 s}right)}^{2}=text{220 rad}.end{array}[/latex]
Converting radians to revolutions gives
[latex]theta =(220 text{ rad})frac{1 text{ rev}}{2pi text{rad}}=35.0 text{ rev}[/latex]
Solution for (d)
The number of meters of fishing line is x, which can be obtained through its relationship with θ:
Discussion
This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.) The amount of fishing line played out is 9.90 m, about right for when the big fish bites.
Figure 1. Fishing line coming off a rotating reel moves linearly. Example 10.3 and Example 10.4 consider relationships between rotational and linear quantities associated with a fishing reel.
Example 2. Calculating the Duration When the Fishing Reel Slows Down and Stops
Arthrokinematics Of Thoracic Spine
Now let us consider what happens if the fisherman applies a brake to the spinning reel, achieving an angular acceleration of -300 rad/s2. How long does it take the reel to come to a stop?
Strategy
We are asked to find the time t for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is ω0 = 220 rad/s and the final angular velocity ω is zero. The angular acceleration is given to be α = -300 rad/s2. Examining the available equations, we see all quantities but t are known in ω = ω0+αt, making it easiest to use this equation.
Solution
The equation states
ω = ω0 + αt.
We solve the equation algebraically for t, and then substitute the known values as usual, yielding
[latex]t=frac{omega -{omega }_{0}}{alpha }=frac{0-text{220 rad/s}}{-text{300}{text{rad/s}}^{2}}=0text{.}text{733 s}[/latex].
Discussion
Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration.
Example 3. Calculating the Slow Acceleration of Trains and Their Wheels
Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350-m-radius wheels an angular acceleration of 0.250 rad/s2. After the wheels have made 200 revolutions (assume no slippage): (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train?
Strategy
Arthrokinematics Spin
In part (a), we are asked to find x, and in (b) we are asked to find ω and v. We are given the number of revolutions θ, the radius of the wheels r, and the angular acceleration α.
Solution for (a)
The distance x is very easily found from the relationship between distance and rotation angle:
Solving this equation for x yields
x = rθ.
Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities:
[latex]theta =left(text{200}text{rev}right)frac{2pi text{rad}}{text{1 rev}}=text{1257}text{rad}[/latex].
Now we can substitute the known values into x=rθ to find the distance the train moved down the track:
x = rθ = (0.350 m)(1257 rad) = 440 m
Solution for (b)
Arthrokinematics Def
We cannot use any equation that incorporates t to find ω, because the equation would have at least two unknown values. The equation [latex]{omega }^{2}={{omega }_{0}}^{2}+2alphatheta [/latex] will work, because we know the values for all variables except ω:
[latex]{omega }^{2}={{omega }_{0}}^{2}+2alphatheta[/latex]
Taking the square root of this equation and entering the known values gives
[latex]begin{array}{lll}omega & =& {left[0+2left(0text{.}text{250}{text{ rad/s}}^{2}right)left(text{1257}text{ rad}right)right]}^{1/2} & =& text{25.1 rad/s.}end{array}[/latex]
We can find the linear velocity of the train, v, through its relationship to ω:
Discussion
Arthrokinematics
The distance traveled is fairly large and the final velocity is fairly slow (just under 32 km/h).
There is translational motion even for something spinning in place, as the following example illustrates. Figure 2 shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels.
Figure 2. The image shows a microwave plate. The fly makes revolutions while the food is heated (along with the fly).
Example 4. Calculating the Distance Traveled by a Fly on the Edge of a Microwave Oven Plate
A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times.)
Strategy
First, find the total number of revolutions θ, and then the linear distance x traveled.[latex]theta =bar{omega}t[/latex] can be used to find θ because [latex]bar{omega}[/latex] is given to be 6.0 rpm.
Solution
Entering known values into [latex]theta =bar{omega} t[/latex] gives
[latex]theta =bar{omega }t=left(text{6.0 rpm}right)left(text{2.0 min}right)=text{12 rev}[/latex].
As always, it is necessary to convert revolutions to radians before calculating a linear quantity like x from an angular quantity like θ:
[latex]theta =left(text{12 rev}right)left(frac{2pi text{rad}}{text{1 rev}}right)=75.4 text{ rad}[/latex].
Now, using the relationship between x and θ, we can determine the distance traveled:
Discussion
Quite a trip (if it survives)! Note that this distance is the total distance traveled by the fly. Displacement is actually zero for complete revolutions because they bring the fly back to its original position. The distinction between total distance traveled and displacement was first noted in One-Dimensional Kinematics.
Check Your Understanding
Solution
Rotational kinematics (just like linear kinematics) is descriptive and does not represent laws of nature. With kinematics, we can describe many things to great precision but kinematics does not consider causes. For example, a large angular acceleration describes a very rapid change in angular velocity without any consideration of its cause.
Section Summary
- Kinematics is the description of motion.
- The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time.
- Starting with the four kinematic equations we developed in the One-Dimensional Kinematics, we can derive the four rotational kinematic equations (presented together with their translational counterparts) seen in Table 1.
- In these equations, the subscript 0 denotes initial value ([latex]{x}_{0}[/latex] and [latex]{t}_{0}[/latex] are initial values), and the average angular velocity [latex]bar{omega}[/latex] and average velocity [latex]bar{v}[/latex] are defined as follows:[latex]bar{omega }=frac{{omega }_{0}+omega }{2}text{ and }bar{v}=frac{{v}_{0}+v}{2}[/latex].
Problems & Exercises
1. With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s. (a) What is its angular acceleration in rad/s2? (b) How many revolutions does it go through in the process?
2. Suppose a piece of dust finds itself on a CD. If the spin rate of the CD is 500 rpm, and the piece of dust is 4.3 cm from the center, what is the total distance traveled by the dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.)
3. A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2. (a) How long does it take to come to rest? (b) How many revolutions does it make before stopping?
4. During a very quick stop, a car decelerates at 700 m/s2.
(a) What is the angular acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement?
(b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 95.0 rad/s?
(c) How long does the car take to stop completely?
(d) What distance does the car travel in this time?
(e) What was the car’s initial velocity?
(f) Do the values obtained seem reasonable, considering that this stop happens very quickly?
Figure 3. Yo-yos are amusing toys that display significant physics and are engineered to enhance performance based on physical laws. (credit: Beyond Neon, Flickr)
5. Everyday application: Suppose a yo-yo has a center shaft that has a 0.250 cm radius and that its string is being pulled.
(a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.50 m/s2, what is the angular acceleration of the yo-yo?
(b) What is the angular velocity after 0.750 s if it starts from rest?
(c) The outside radius of the yo-yo is 3.50 cm. What is the tangential acceleration of a point on its edge?
Glossary
- kinematics of rotational motion:
- describes the relationships among rotation angle, angular velocity, angular acceleration, and time
Selected Solutions to Problems & Exercises
1. (a) [latex]80{text{rad/s}}^{2}[/latex] (b) 1.0 rev
Arthrokinematics Wiki
3. (a) 45.7 s (b) 116 rev
Arthrokinematics Of Talocrural Joint
5. (a) 600 rad/s2 (b) 450 rad/s (c) 21.0 m/s