Post by: Jaaanosik on 21/06/2018 16:12:24
(https://theelectromagneticnatureofthings.com/img/i9/pendulum01.png)
The equation (5.124) appears to me as incorrect.
The correct equation should look like this:
... because the equation (5.124) is good for a pendulum bob like this:
(https://theelectromagneticnatureofthings.com/img/i9/pendulum02.png)
Assuming we ignore the mass of the string and the bob axle and the axle has "friction-less" bearings.
Do I miss anything?
Post by: guest45734 on 21/06/2018 19:06:43
Sorry I have not fathomed this math editor yet to give a better explanation.
Post by: Jaaanosik on 21/06/2018 19:17:51
Your additional term (Bob) is a restatement of the first term. ie I = r^2 m, your differential term is your angular velocity w.Well, it is not a restatement.
Bob has its radius
and the pendulum string has the length 
.These are different.
The bob is rotating in the figure 5.8 but it is not rotating in the second image.
These two are different scenarios. They need two different equations.
Post by: alancalverd on 22/06/2018 09:15:15
https://en.wikipedia.org/wiki/Kater%27s_pendulum shows how a rigid compond pendulum can be designed to eliminate many of the unknowns.
If your freely-rotating bob is on a frictionless axle, there is no torque making it rotate with respect to the vertical, so
ω = 0
Post by: Jaaanosik on 22/06/2018 13:59:27
You are complaining that the equation for a simple pendulum (with a point mass) is not the same as that for a compound pendulum (with a distributed mass). Perfectly true. The art of real life is either to solve the complete equation (which is what engineers have to do when designing oscillating things like cars and bridges) or to design an experiment that eliminates the second-order terms (which is what physicists try to do when measuring things like g).
https://en.wikipedia.org/wiki/Kater%27s_pendulum shows how a rigid compond pendulum can be designed to eliminate many of the unknowns.
If your freely-rotating bob is on a frictionless axle, there is no torque making it rotate with respect to the vertical, so
ω = 0
The center of mass (center of oscillation) is
distance from the pivot in both instances. Let us assume the mass distribution is the same in both pendulum bobs.
The book example bob rotates the second image does not rotate.
Why the authors call their pendulum 'simple' if it is suppose to be compound then?
Where is the disconnect?
Post by: Jaaanosik on 25/06/2018 16:29:10
We can see that
... and let us call it 
because this 
belongs to equation 1.Let us consider
for the second image pendulum .Now we have two equations.
Eq.1:
Eq.2:
Comparing the two equations shows us that
from the strictly mathematical point of view.Nevertheless we have gravitational acceleration
on the right side of the equations that has abundance of potential energy.Would it be possible that gravity can make the pendulums fall at the same
and provide inertia to the bob on top of that at the same time?Post by: PmbPhy on 01/07/2018 10:54:40
In most situations one makes approximations, i.e. the use a model which helps the reader understand the physics at hand and to do that they simplify things. Notice in the photo how long the string is compared to the bob and that the authors mentioned only small oscillations? There's also no reason to assume that the bob rotates with the pendulum. It could have the same orientation in space as it moves. Rotational inertia would see to that. Otherwise it'd be called a compound pendulum.
Regarding models in physics see your text. Also use Google to find a good treatment about the use of models in spaceflight. For example; when NASA is plotting a launch of a rocket to Mars they need to know the orbit of Mars. They assume that Mars, the Earth and the Sun are all point objects.
Post by: Jaaanosik on 02/07/2018 17:32:13
The question as its written by the author implies that he's modeling the pendulum by taking into account only the translational energy of the bob and is taking the rotational energy of the bob to be insignificant when compared to it and thus ignorable I.e. he's using a model where the bob is being treated as a point particle.
In most situations one makes approximations, i.e. the use a model which helps the reader understand the physics at hand and to do that they simplify things. Notice in the photo how long the string is compared to the bob and that the authors mentioned only small oscillations? There's also no reason to assume that the bob rotates with the pendulum. It could have the same orientation in space as it moves. Rotational inertia would see to that. Otherwise it'd be called a compound pendulum.
Regarding models in physics see your text. Also use Google to find a good treatment about the use of models in spaceflight. For example; when NASA is plotting a launch of a rocket to Mars they need to know the orbit of Mars. They assume that Mars, the Earth and the Sun are all point objects.
Quote
...that the authors mentioned only small oscillations?...
This is not the case here. The opposite is true.
From the book: "For a maximum amplitude
large enough so that 
"This is what is discussed here... then we turn to conservation of energy.
An example: the pendulum bob is a solid sphere, radius
, mass 
, length 
, 
, 
We will go with the book equation (5.124) at the moment and when
at the bottom we get
We can see the rotational kinetic energy is 0.1% of the translational kinetic energy.
The ignorance of the rotational kinetic energy is an approximation that can be OK in some instances.
There might be cases where this approach is not acceptable.
Is there any physics book that treats the simple pendulum in the proper/correct way?
Post by: Jaaanosik on 10/07/2018 15:40:45
Because it is one approach how to determine velocity of the cycloid motion.
We can obtain a cycloid motion when a wheel rotates and translates at the same time.
Here is an example of a horizontally oriented wheel in a train car:
(https://theelectromagneticnatureofthings.com/img/i9/wheel02.png)
The wheel axle is pointing down to the center of the Earth.
There are two reference systems: train and the ground. The simple rotation in the train reference system is a cycloid in the ground reference system.
Let us assume the train moves at
and the wheel is stationary at time 
(
).The initial pendulum bob positions are
- slightly to the left from the top of the cycloid, 
- slightly to the right from the bottom of the cycloid and the left pendulum 
is at 
where 
is at the bottom of the cycloid 
and 
is at the top of the cycloid 
.The pendulum bobs are mounted on an elastic rod and
. The wheel acceleration starts at
and in time 
the wheel turns exactly 
so the pendulum bobs are in , 
, positions at 
.The pendulum bobs are undergoing the same acceleration in the moving reference frame but this is not the case according to conservation of energy analysis of the cycloid.
Here are cycloid plots to explain conservation of energy accelerations and velocities for the
of the moving/train reference frame after the first 
.(https://theelectromagneticnatureofthings.com/img/i9/w0.01r1v0.1_r.png)
Plot 1: Radius of the cycloid motion
(https://theelectromagneticnatureofthings.com/img/i9/w0.01r1v0.1_v.png)
Plot 2: Velocity of the cycloid motion
(https://theelectromagneticnatureofthings.com/img/i9/w0.01r1v0.1_w.png)
Plot 3: Omega of the cycloid motion
Here we can see that at points
and some energy goes to the rotational kinetic energy and at point all energy goes to the translational kinetic energy. The result will be different velocities at points , , in the moving/train reference frame and the ground reference frame as well when compared to Plot 2.The reason is that the velocity as shown in the Plot 2 is done by a simple Galilean transformation and it ignores the conservation of energy acceleration analysis of the higher reference frame.
This is IT!!!
We have an experiment to detect constant linear motion without any signal from the outside.
I left out lots of details about the acceleration, ... how pendulum bobs will bend the elastic rods, ... this post is just to layout a starting point for the further discussion.
Post by: Jaaanosik on 11/07/2018 14:48:20
(https://theelectromagneticnatureofthings.com/img/i9/book01.png)
(https://theelectromagneticnatureofthings.com/img/i9/book02.png)
The equation (17.106) ?!?
If the pendulum bob is cut from the string then it flies away with kinetic energy:
and not
Wow!!!
This is very bad!!!
Post by: mrsmith2211 on 12/07/2018 01:12:13
Post by: Jaaanosik on 12/07/2018 03:47:22
Because inertia and mass are not relative in the case of the pendulum?What is this sentence related to?
Post by: Jaaanosik on 17/07/2018 15:15:51
The relativity is not correct/precise.
We have an experiment to detect straight constant linear motion without any signal from the outside.
Nobody is screaming that I am wrong.
What is going on?