Physics Quiz...

[Martin Brown]

On 29/04/2023 19:06, Fred Bloggs wrote:

On Friday, April 28, 2023 at 8:00:48 PM UTC-4, Jasen Betts wrote:
On 2023-04-24, Lasse Langwadt Christensen <lang...@fonz.dk> wrote:
mandag den 24. april 2023 kl. 20.12.14 UTC+2 skrev John Larkin:
On Mon, 24 Apr 2023 09:24:51 -0700 (PDT), Lasse Langwadt
Christensen <lang...@fonz.dk> wrote:

mandag den 24. april 2023 kl. 00.33.17 UTC+2 skrev John
Larkin:

The skater must use muscle power to pull her arms in. The
work done in pulling her arms in is converted to energy
stored in the spinning mass of her body, and is
recoverable.

angular momentum is the product of moment of inertia and
angular velocity

angular momentum can\'t just change (newtons 3rd) so when the
moment of inertia get smaller (pulling arms in) the angular
velocity has to increase

That statement is correct.

no energy added, just \"stored\" differently

It depends what you mean by stored differently. There is more kinetic
energy and less stored potential energy when she brings her arms in.

She is doing work against a force which spins her up faster converting
the potential energy stored in her arms into kinetic energy.

You can call it centrifugal or centripetal depending on your age and
chosen sign convention for workign in a roatating frame of reference.
Same thing that pins you to the sidewall in some fairground rides. The
force is real enough for someone in a rotating frame of reference.

She did work pulling her arms in, burned a bit of breakfast and
added energy to the rotating system.

sure she spend some energy but it doesn\'t add to the rotational
energy

Of course it does you idiot - that is the whole point.

The outstretched arms have stored potential energy so that she has to
work hard to bring them in - the resulting forces help to spin her up
and convert that potential energy into kinetic energy and a faster spin.

*Angular* momentum is the cleanly conserved quantity in this problem

Iw = iW


So that W = Iw/i

And her kinetic energy changes from

1/2Iw^2 to 1/2iW^2 = 1/2(I^2/i)w^2

potential energy decreases to maintain conservation of *TOTAL* energy.

Didn\'t add rotational momentum, but did add rotational kinetic
energy.

Angular kinetic energy is 1/2 I w^2, obvious notation, I moment of
inertia about axis of rotation. Bringing the arms in reduces I
forcing w to increase to maintain the same energy, which is ideally
conserved.

*IDIOT* You have got it so wrong that it has now become very funny.

Total Energy *is* conserved but to get the right answer you have to
include the stored potential energy in the spinning arms as well!

Think about how a mechanical engine governor works - as the shaft spins
up energy is stored as a mixture of kinetic energy in the moment of
inertia of the whole assembly *and* potential energy of the weights
moving outwards and rising upwards against the force of gravity.

https://en.wikipedia.org/wiki/Centrifugal_governor

--
Martin Brown

 

[John Larkin]

On Sat, 29 Apr 2023 11:06:49 -0700 (PDT), Fred Bloggs
<bloggs.fredbloggs.fred@gmail.com> wrote:

On Friday, April 28, 2023 at 8:00:48?PM UTC-4, Jasen Betts wrote:
On 2023-04-24, Lasse Langwadt Christensen <lang...@fonz.dk> wrote:
mandag den 24. april 2023 kl. 20.12.14 UTC+2 skrev John Larkin:
On Mon, 24 Apr 2023 09:24:51 -0700 (PDT), Lasse Langwadt Christensen
lang...@fonz.dk> wrote:

mandag den 24. april 2023 kl. 00.33.17 UTC+2 skrev John Larkin:
On Sun, 23 Apr 2023 15:19:40 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:
On Sunday, April 23, 2023 at 1:03:04?PM UTC-4, John Robertson wrote:
On 2023/04/23 4:01 a.m., Fred Bloggs wrote:
So a vehicle is driving down the road when an entire wheel comes off and continues its direction unchanged rolling alongside the vehicle. Question is why does the wheel end up accelerating, rolling much faster than its original speed, outpacing the vehicle significantly? Answer should be obvious, but you need practical insight.
The wheel can\'t accelerate without an external force being added to it -
it has no means of self propulsion. Laws of the conservation of momentum
apply as usual.

Think of a figure skater doing one of those stationary spins. Arms outstretched is one spin rate, arms brought into the side and they turn into a blur. No external forces come into play.

The skater must use muscle power to pull her arms in. The work done in
pulling her arms in is converted to energy stored in the spinning mass
of her body, and is recoverable.

angular momentum is the product of moment of inertia and angular velocity

angular momentum can\'t just change (newtons 3rd) so when the moment of inertia get smaller (pulling arms in)
the angular velocity has to increase

no energy added, just \"stored\" differently
She did work pulling her arms in, burned a bit of breakfast and added
energy to the rotating system.

sure she spend some energy but it doesn\'t add to the rotational energy

Didn\'t add rotational momentum, but did add rotational kinetic energy.

Angular kinetic energy is 1/2 I w^2, obvious notation, I moment of inertia about axis of rotation. Bringing the arms in reduces I forcing w to increase to maintain the same energy, which is ideally conserved.

The skater expends chemical energy, in her muscles, to pull her arms
in. Energy is conserved if you count her breakfast in the equation.

 

[Fred Bloggs]

On Monday, May 1, 2023 at 6:12:06 AM UTC-4, Martin Brown wrote:

On 29/04/2023 19:06, Fred Bloggs wrote:
On Friday, April 28, 2023 at 8:00:48 PM UTC-4, Jasen Betts wrote:
On 2023-04-24, Lasse Langwadt Christensen <lang...@fonz.dk> wrote:
mandag den 24. april 2023 kl. 20.12.14 UTC+2 skrev John Larkin:
On Mon, 24 Apr 2023 09:24:51 -0700 (PDT), Lasse Langwadt
Christensen <lang...@fonz.dk> wrote:

mandag den 24. april 2023 kl. 00.33.17 UTC+2 skrev John
Larkin:

The skater must use muscle power to pull her arms in. The
work done in pulling her arms in is converted to energy
stored in the spinning mass of her body, and is
recoverable.

angular momentum is the product of moment of inertia and
angular velocity

angular momentum can\'t just change (newtons 3rd) so when the
moment of inertia get smaller (pulling arms in) the angular
velocity has to increase
That statement is correct.

no energy added, just \"stored\" differently
It depends what you mean by stored differently. There is more kinetic
energy and less stored potential energy when she brings her arms in.

She is doing work against a force which spins her up faster converting
the potential energy stored in her arms into kinetic energy.

You can call it centrifugal or centripetal depending on your age and
chosen sign convention for workign in a roatating frame of reference.
Same thing that pins you to the sidewall in some fairground rides. The
force is real enough for someone in a rotating frame of reference.
She did work pulling her arms in, burned a bit of breakfast and
added energy to the rotating system.

sure she spend some energy but it doesn\'t add to the rotational
energy
Of course it does you idiot - that is the whole point.

The outstretched arms have stored potential energy so that she has to
work hard to bring them in - the resulting forces help to spin her up
and convert that potential energy into kinetic energy and a faster spin.

*Angular* momentum is the cleanly conserved quantity in this problem

Iw = iW


So that W = Iw/i

And her kinetic energy changes from

1/2Iw^2 to 1/2iW^2 = 1/2(I^2/i)w^2

That\'s a pathetic expression one would expect from a confused pedant.

This childish expression 1/2(I^2/i)w^2 , MEANS 1/2(I/i)Iw^2 = (I/i) x original KE. Speed-up factor SQRT(I/i) due to reduced i when arms are pulled in.

Making E1/E2= I1/I2 x (w1/w2)^2 or w1/w2 = SQRT ( I2/I1) when energy is conserved (E1/E2=1), which it is in the simplified case of ignoring friction with ice and maybe more exact positioning of the arms, which contribute a very slight gravitational potential energy input if and when lowered.

potential energy decreases to maintain conservation of *TOTAL* energy.
Didn\'t add rotational momentum, but did add rotational kinetic
energy.

Angular kinetic energy is 1/2 I w^2, obvious notation, I moment of
inertia about axis of rotation. Bringing the arms in reduces I
forcing w to increase to maintain the same energy, which is ideally
conserved.
*IDIOT* You have got it so wrong that it has now become very funny.

Total Energy *is* conserved but to get the right answer you have to
include the stored potential energy in the spinning arms as well!

You have to be inebriated to dream up the potential energy stored in the kinetic energy bit.

Think about how a mechanical engine governor works - as the shaft spins
up energy is stored as a mixture of kinetic energy in the moment of
inertia of the whole assembly *and* potential energy of the weights
moving outwards and rising upwards against the force of gravity.

This is not at all like the spinning governor.

Interestingly Newton used that configuration to tease out a measurement of inertial to gravitational mass difference, and amazingly discovered they agreed to 6 significant figures. That would be the precision of his measurements, not the discovery.

https://en.wikipedia.org/wiki/Centrifugal_governor

--
Martin Brown

 

[Fred Bloggs]

On Monday, May 1, 2023 at 12:19:42 PM UTC-4, John Larkin wrote:

On Sat, 29 Apr 2023 11:06:49 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:
On Friday, April 28, 2023 at 8:00:48?PM UTC-4, Jasen Betts wrote:
On 2023-04-24, Lasse Langwadt Christensen <lang...@fonz.dk> wrote:
mandag den 24. april 2023 kl. 20.12.14 UTC+2 skrev John Larkin:
On Mon, 24 Apr 2023 09:24:51 -0700 (PDT), Lasse Langwadt Christensen
lang...@fonz.dk> wrote:

mandag den 24. april 2023 kl. 00.33.17 UTC+2 skrev John Larkin:
On Sun, 23 Apr 2023 15:19:40 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:
On Sunday, April 23, 2023 at 1:03:04?PM UTC-4, John Robertson wrote:
On 2023/04/23 4:01 a.m., Fred Bloggs wrote:
So a vehicle is driving down the road when an entire wheel comes off and continues its direction unchanged rolling alongside the vehicle. Question is why does the wheel end up accelerating, rolling much faster than its original speed, outpacing the vehicle significantly? Answer should be obvious, but you need practical insight.
The wheel can\'t accelerate without an external force being added to it -
it has no means of self propulsion. Laws of the conservation of momentum
apply as usual.

Think of a figure skater doing one of those stationary spins. Arms outstretched is one spin rate, arms brought into the side and they turn into a blur. No external forces come into play.

The skater must use muscle power to pull her arms in. The work done in
pulling her arms in is converted to energy stored in the spinning mass
of her body, and is recoverable.

angular momentum is the product of moment of inertia and angular velocity

angular momentum can\'t just change (newtons 3rd) so when the moment of inertia get smaller (pulling arms in)
the angular velocity has to increase

no energy added, just \"stored\" differently
She did work pulling her arms in, burned a bit of breakfast and added
energy to the rotating system.

sure she spend some energy but it doesn\'t add to the rotational energy

Didn\'t add rotational momentum, but did add rotational kinetic energy.

Angular kinetic energy is 1/2 I w^2, obvious notation, I moment of inertia about axis of rotation. Bringing the arms in reduces I forcing w to increase to maintain the same energy, which is ideally conserved.

The skater expends chemical energy, in her muscles, to pull her arms
in. Energy is conserved if you count her breakfast in the equation.

That little bit of muscular tension doesn\'t begin to account for increasing spinning up that body mass by at least a factor of 10.

Moment of inertia is a power law, goes by square of moment arm of the mass perpendicular to the axis. If the average of that mass was say 18 inches arms outstretched, and then gets pulled in to say 2 inches, that\'s (18/2)^2 factor of 81 reduction of inertia. Resulting speed increase will x sqrt =9..

The simplified first order analysis is to consider the external torques applied. There are none. No torques means no work done by or on the system, energy stays the same.

 

[whit3rd]

On Monday, May 1, 2023 at 9:57:07 AM UTC-7, Fred Bloggs wrote:

On Monday, May 1, 2023 at 12:19:42 PM UTC-4, John Larkin wrote:

The skater expends chemical energy, in her muscles, to pull her arms
in. Energy is conserved if you count her breakfast in the equation.
That little bit of muscular tension doesn\'t begin to account for increasing spinning up that body mass by at least a factor of 10.

Moment of inertia is a power law, goes by square of moment arm of the mass perpendicular to the axis. If the average of that mass was say 18 inches arms outstretched, and then gets pulled in to say 2 inches, that\'s (18/2)^2 factor of 81 reduction of inertia. Resulting speed increase will x sqrt =9.

The simplified first order analysis is to consider the external torques applied. There are none. No torques means no work done by or on the system, energy stays the same.

No torques means zero change in ANGULAR MOMENTUM. That\'s because angular momentum in a
rotor is proportional to the integral of the torque. Total energy and rotational energy are not identical,
and \'her breakfast\' becomes a bit of kinetic energy when the skater does the work. The skater
is not, however, changing the angular momentum, because she exerts no torque. Rotational
kinetic energy DOES change; the rotational kinetic energy goes as the square of the angular momentum
divided by the moment of inertia (which is altered by the arm configuration).

 

[Fred Bloggs]

On Monday, May 1, 2023 at 1:33:28 PM UTC-4, whit3rd wrote:

On Monday, May 1, 2023 at 9:57:07 AM UTC-7, Fred Bloggs wrote:
On Monday, May 1, 2023 at 12:19:42 PM UTC-4, John Larkin wrote:

The skater expends chemical energy, in her muscles, to pull her arms
in. Energy is conserved if you count her breakfast in the equation.
That little bit of muscular tension doesn\'t begin to account for increasing spinning up that body mass by at least a factor of 10.

Moment of inertia is a power law, goes by square of moment arm of the mass perpendicular to the axis. If the average of that mass was say 18 inches arms outstretched, and then gets pulled in to say 2 inches, that\'s (18/2)^2 factor of 81 reduction of inertia. Resulting speed increase will x sqrt =9.

The simplified first order analysis is to consider the external torques applied. There are none. No torques means no work done by or on the system, energy stays the same.
No torques means zero change in ANGULAR MOMENTUM. That\'s because angular momentum in a
rotor is proportional to the integral of the torque. Total energy and rotational energy are not identical,
and \'her breakfast\' becomes a bit of kinetic energy when the skater does the work. The skater
is not, however, changing the angular momentum, because she exerts no torque. Rotational
kinetic energy DOES change; the rotational kinetic energy goes as the square of the angular momentum
divided by the moment of inertia (which is altered by the arm configuration).

You\'re getting hung up on equations. Work done is torque x angular displacement. No torque= No work, and no work means no change in energy. Rotational kinetic energy does not change.

 

[John Larkin]

On Mon, 1 May 2023 09:57:03 -0700 (PDT), Fred Bloggs
<bloggs.fredbloggs.fred@gmail.com> wrote:

On Monday, May 1, 2023 at 12:19:42?PM UTC-4, John Larkin wrote:
On Sat, 29 Apr 2023 11:06:49 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:
On Friday, April 28, 2023 at 8:00:48?PM UTC-4, Jasen Betts wrote:
On 2023-04-24, Lasse Langwadt Christensen <lang...@fonz.dk> wrote:
mandag den 24. april 2023 kl. 20.12.14 UTC+2 skrev John Larkin:
On Mon, 24 Apr 2023 09:24:51 -0700 (PDT), Lasse Langwadt Christensen
lang...@fonz.dk> wrote:

mandag den 24. april 2023 kl. 00.33.17 UTC+2 skrev John Larkin:
On Sun, 23 Apr 2023 15:19:40 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:
On Sunday, April 23, 2023 at 1:03:04?PM UTC-4, John Robertson wrote:
On 2023/04/23 4:01 a.m., Fred Bloggs wrote:
So a vehicle is driving down the road when an entire wheel comes off and continues its direction unchanged rolling alongside the vehicle. Question is why does the wheel end up accelerating, rolling much faster than its original speed, outpacing the vehicle significantly? Answer should be obvious, but you need practical insight.
The wheel can\'t accelerate without an external force being added to it -
it has no means of self propulsion. Laws of the conservation of momentum
apply as usual.

Think of a figure skater doing one of those stationary spins. Arms outstretched is one spin rate, arms brought into the side and they turn into a blur. No external forces come into play.

The skater must use muscle power to pull her arms in. The work done in
pulling her arms in is converted to energy stored in the spinning mass
of her body, and is recoverable.

angular momentum is the product of moment of inertia and angular velocity

angular momentum can\'t just change (newtons 3rd) so when the moment of inertia get smaller (pulling arms in)
the angular velocity has to increase

no energy added, just \"stored\" differently
She did work pulling her arms in, burned a bit of breakfast and added
energy to the rotating system.

sure she spend some energy but it doesn\'t add to the rotational energy

Didn\'t add rotational momentum, but did add rotational kinetic energy.

Angular kinetic energy is 1/2 I w^2, obvious notation, I moment of inertia about axis of rotation. Bringing the arms in reduces I forcing w to increase to maintain the same energy, which is ideally conserved.

The skater expends chemical energy, in her muscles, to pull her arms
in. Energy is conserved if you count her breakfast in the equation.

That little bit of muscular tension doesn\'t begin to account for increasing spinning up that body mass by at least a factor of 10.

They skate a curve pre-spin to start, pushing against the ice with
muscle power, and then the real mass that\'s pulled in is her legs and
even torso and head. But it still takes chemical energy pull mass
towards the axis of rotation to spin up.

https://www.youtube.com/watch?v=0RVyhd3E9hY

There are a lot of youtubes about this.

 

[John Larkin]

On Mon, 1 May 2023 10:49:07 -0700 (PDT), Fred Bloggs
<bloggs.fredbloggs.fred@gmail.com> wrote:

On Monday, May 1, 2023 at 1:33:28?PM UTC-4, whit3rd wrote:
On Monday, May 1, 2023 at 9:57:07?AM UTC-7, Fred Bloggs wrote:
On Monday, May 1, 2023 at 12:19:42?PM UTC-4, John Larkin wrote:

The skater expends chemical energy, in her muscles, to pull her arms
in. Energy is conserved if you count her breakfast in the equation.
That little bit of muscular tension doesn\'t begin to account for increasing spinning up that body mass by at least a factor of 10.

Moment of inertia is a power law, goes by square of moment arm of the mass perpendicular to the axis. If the average of that mass was say 18 inches arms outstretched, and then gets pulled in to say 2 inches, that\'s (18/2)^2 factor of 81 reduction of inertia. Resulting speed increase will x sqrt =9.

The simplified first order analysis is to consider the external torques applied. There are none. No torques means no work done by or on the system, energy stays the same.
No torques means zero change in ANGULAR MOMENTUM. That\'s because angular momentum in a
rotor is proportional to the integral of the torque. Total energy and rotational energy are not identical,
and \'her breakfast\' becomes a bit of kinetic energy when the skater does the work. The skater
is not, however, changing the angular momentum, because she exerts no torque. Rotational
kinetic energy DOES change; the rotational kinetic energy goes as the square of the angular momentum
divided by the moment of inertia (which is altered by the arm configuration).

You\'re getting hung up on equations. Work done is torque x angular displacement. No torque= No work, and no work means no change in energy. Rotational kinetic energy does not change.

The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

If she had her arms fully extended, 90 degrees from the axis, and then
just relaxed her arm muscles, her arms would naturally droop to some
angle, and she would speed up and her spin energy would increase. Then
she could add muscle power to pull her arms the rest of the way in and
spin even faster.

So there are actually two phases of energy transfer, gravitational and
chemical.

 

[Fred Bloggs]

On Monday, May 1, 2023 at 2:35:48 PM UTC-4, John Larkin wrote:

On Mon, 1 May 2023 10:49:07 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:

On Monday, May 1, 2023 at 1:33:28?PM UTC-4, whit3rd wrote:
On Monday, May 1, 2023 at 9:57:07?AM UTC-7, Fred Bloggs wrote:
On Monday, May 1, 2023 at 12:19:42?PM UTC-4, John Larkin wrote:

The skater expends chemical energy, in her muscles, to pull her arms
in. Energy is conserved if you count her breakfast in the equation..
That little bit of muscular tension doesn\'t begin to account for increasing spinning up that body mass by at least a factor of 10.

Moment of inertia is a power law, goes by square of moment arm of the mass perpendicular to the axis. If the average of that mass was say 18 inches arms outstretched, and then gets pulled in to say 2 inches, that\'s (18/2)^2 factor of 81 reduction of inertia. Resulting speed increase will x sqrt =9.

The simplified first order analysis is to consider the external torques applied. There are none. No torques means no work done by or on the system, energy stays the same.
No torques means zero change in ANGULAR MOMENTUM. That\'s because angular momentum in a
rotor is proportional to the integral of the torque. Total energy and rotational energy are not identical,
and \'her breakfast\' becomes a bit of kinetic energy when the skater does the work. The skater
is not, however, changing the angular momentum, because she exerts no torque. Rotational
kinetic energy DOES change; the rotational kinetic energy goes as the square of the angular momentum
divided by the moment of inertia (which is altered by the arm configuration).

You\'re getting hung up on equations. Work done is torque x angular displacement. No torque= No work, and no work means no change in energy. Rotational kinetic energy does not change.
The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

If she had her arms fully extended, 90 degrees from the axis, and then
just relaxed her arm muscles, her arms would naturally droop to some
angle, and she would speed up and her spin energy would increase. Then
she could add muscle power to pull her arms the rest of the way in and
spin even faster.

So there are actually two phases of energy transfer, gravitational and
chemical.

I vaguely recall the skaters fold their arms over their chests to turn into a blur. As the arms move closer to the axis, the centripetal force decreases considerably, requiring less work on the skater. But it makes no difference whatsoever because just bringing the arms inward in no way shape or form applies a force tangential to the radius connecting the axis, which is what you need to torque up the rotational velocity. So why does the skater mass speed up? The answer is Coriolis. I\'ve been trying to tell people this all along.

https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/12%3A_Non-inertial_Reference_Frames/12.08%3A_Coriolis_Force#:~:text=An%20interesting%20application%20of%20the,external%20forces%20which%20is%20correct.

 

[Fred Bloggs]

On Monday, May 1, 2023 at 1:33:28 PM UTC-4, whit3rd wrote:

On Monday, May 1, 2023 at 9:57:07 AM UTC-7, Fred Bloggs wrote:
On Monday, May 1, 2023 at 12:19:42 PM UTC-4, John Larkin wrote:

The skater expends chemical energy, in her muscles, to pull her arms
in. Energy is conserved if you count her breakfast in the equation.
That little bit of muscular tension doesn\'t begin to account for increasing spinning up that body mass by at least a factor of 10.

Moment of inertia is a power law, goes by square of moment arm of the mass perpendicular to the axis. If the average of that mass was say 18 inches arms outstretched, and then gets pulled in to say 2 inches, that\'s (18/2)^2 factor of 81 reduction of inertia. Resulting speed increase will x sqrt =9.

The simplified first order analysis is to consider the external torques applied. There are none. No torques means no work done by or on the system, energy stays the same.
No torques means zero change in ANGULAR MOMENTUM. That\'s because angular momentum in a
rotor is proportional to the integral of the torque. Total energy and rotational energy are not identical,
and \'her breakfast\' becomes a bit of kinetic energy when the skater does the work. The skater
is not, however, changing the angular momentum, because she exerts no torque. Rotational
kinetic energy DOES change; the rotational kinetic energy goes as the square of the angular momentum
divided by the moment of inertia (which is altered by the arm configuration).

My apologies. I neglected consideration of the Coriolis effect.

 

[John Larkin]

On Mon, 1 May 2023 14:12:58 -0700 (PDT), Fred Bloggs
<bloggs.fredbloggs.fred@gmail.com> wrote:

On Monday, May 1, 2023 at 2:35:48?PM UTC-4, John Larkin wrote:
On Mon, 1 May 2023 10:49:07 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:

On Monday, May 1, 2023 at 1:33:28?PM UTC-4, whit3rd wrote:
On Monday, May 1, 2023 at 9:57:07?AM UTC-7, Fred Bloggs wrote:
On Monday, May 1, 2023 at 12:19:42?PM UTC-4, John Larkin wrote:

The skater expends chemical energy, in her muscles, to pull her arms
in. Energy is conserved if you count her breakfast in the equation.
That little bit of muscular tension doesn\'t begin to account for increasing spinning up that body mass by at least a factor of 10.

Moment of inertia is a power law, goes by square of moment arm of the mass perpendicular to the axis. If the average of that mass was say 18 inches arms outstretched, and then gets pulled in to say 2 inches, that\'s (18/2)^2 factor of 81 reduction of inertia. Resulting speed increase will x sqrt =9.

The simplified first order analysis is to consider the external torques applied. There are none. No torques means no work done by or on the system, energy stays the same.
No torques means zero change in ANGULAR MOMENTUM. That\'s because angular momentum in a
rotor is proportional to the integral of the torque. Total energy and rotational energy are not identical,
and \'her breakfast\' becomes a bit of kinetic energy when the skater does the work. The skater
is not, however, changing the angular momentum, because she exerts no torque. Rotational
kinetic energy DOES change; the rotational kinetic energy goes as the square of the angular momentum
divided by the moment of inertia (which is altered by the arm configuration).

You\'re getting hung up on equations. Work done is torque x angular displacement. No torque= No work, and no work means no change in energy. Rotational kinetic energy does not change.
The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

If she had her arms fully extended, 90 degrees from the axis, and then
just relaxed her arm muscles, her arms would naturally droop to some
angle, and she would speed up and her spin energy would increase. Then
she could add muscle power to pull her arms the rest of the way in and
spin even faster.

So there are actually two phases of energy transfer, gravitational and
chemical.

I vaguely recall the skaters fold their arms over their chests to turn into a blur. As the arms move closer to the axis, the centripetal force decreases considerably, requiring less work on the skater. But it makes no difference whatsoever because just bringing the arms inward in no way shape or form applies a force tangential to the radius connecting the axis, which is what you need to torque up the rotational velocity.

If no tangential force is applied to her body when she pulls her arms
in, how does her body know to spin faster?

A faster spinning body has more rotational energy than a slow spinning
one. Where did that energy come from?

If her arms suddenly detatched at the shoulder and flew away, would
her body spin faster?

 

[whit3rd]

On Monday, May 1, 2023 at 10:49:11 AM UTC-7, Fred Bloggs wrote:

On Monday, May 1, 2023 at 1:33:28 PM UTC-4, whit3rd wrote:
On Monday, May 1, 2023 at 9:57:07 AM UTC-7, Fred Bloggs wrote:
On Monday, May 1, 2023 at 12:19:42 PM UTC-4, John Larkin wrote:

The skater expends chemical energy, in her muscles, to pull her arms
in. Energy is conserved if you count her breakfast in the equation.
That little bit of muscular tension doesn\'t begin to account for increasing spinning up that body mass by at least a factor of 10.

Moment of inertia is a power law, goes by square of moment arm of the mass perpendicular to the axis. If the average of that mass was say 18 inches arms outstretched, and then gets pulled in to say 2 inches, that\'s (18/2)^2 factor of 81 reduction of inertia. Resulting speed increase will x sqrt =9.

The simplified first order analysis is to consider the external torques applied. There are none. No torques means no work done by or on the system, energy stays the same.
No torques means zero change in ANGULAR MOMENTUM. That\'s because angular momentum in a
rotor is proportional to the integral of the torque. Total energy and rotational energy are not identical,
and \'her breakfast\' becomes a bit of kinetic energy when the skater does the work. The skater
is not, however, changing the angular momentum, because she exerts no torque. Rotational
kinetic energy DOES change; the rotational kinetic energy goes as the square of the angular momentum
divided by the moment of inertia (which is altered by the arm configuration).

You\'re getting hung up on equations. Work done is torque x angular displacement. No torque= No work, and no work means no change in energy. Rotational kinetic energy does not change.

I\'m wise to the equations, not hung up. Your \'torque x angular displacemet\' is work on a rotating solid, when
the torque is applied on-axis. There\'s NOT a rotating solid here, the rotating item changes its angular moment
of inertia, without any torque, because it is changing its mass distribution. Angular momentum
being conserved, and torque being absent, means that the spin frequency goes up.

Rotational kinetic energy certainly does change when frequency rises, in this case, in the proportion that
my equations show.

 

[RichD]

On May 1, Fred Bloggs wrote:

The skater expends chemical energy, in her muscles, to pull her arms

Work done is torque x angular displacement. No torque= No work, and no work means
no change in energy. Rotational kinetic energy does not change.

The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

As the arms move closer to the axis, the centripetal force decreases considerably, requiring
less work on the skater. But it makes no difference whatsoever because just bringing the
 arms inward in no way shape or form applies a force tangential to the radius connecting
the axis, which is what you need to torque up the rotational velocity. So why does the skater
mass speed up? The answer is Coriolis.

https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/12%3A_Non-inertial_Reference_Frames/12.08%3A_Coriolis_Force#:~:text=An%20interesting%20application%20of%20the,external%20forces%20which%20is%20correct.

That article is instructive, but his final claim is incorrect.  
Retracting the arms injects kinetic energy, from the muscles.
The author\'s claim is that this eventually adds to the total
rotational energy, as the arms merge with the torso.

The human body is too complex (distrust any simplistic physics
analysis of human mobility).  Try a simpler analog:

A rotating platform, with a ball attached to the rim. A golfer standing
at that point strokes a putt straight at the center. The ball rolls radially.
And, the floor has tangential static friction, such that the ball continues
to rotate with the platform, at the same angular speed as the platform.

Thus, seen from the center, it rolls straight in; an external observer sees
it follow a spiral path. The ball eventually drops into the hole... then
what? Does it revolve around the center? Does its linear energy convert
into rotational energy? That\'s silly. It bounces off the walls, then settles.
The energy dissipates as heat.

Analogously, the arms end up crashing into her hips, they HALT
(radially) at that point. Their kinetic energy goes into heat.
They continue to rotate with her body, but that\'s merely the
remnant of their initial energy, which they contained while outstretched.

--
Rich

 

[Jasen Betts]

On 2023-05-08, RichD <r_delaney2001@yahoo.com> wrote:

On May 1, Fred Bloggs wrote:
The skater expends chemical energy, in her muscles, to pull her arms

Work done is torque x angular displacement. No torque= No work, and no work means
no change in energy. Rotational kinetic energy does not change.

The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

As the arms move closer to the axis, the centripetal force decreases considerably, requiring
less work on the skater. But it makes no difference whatsoever because just bringing the
 arms inward in no way shape or form applies a force tangential to the radius connecting
the axis, which is what you need to torque up the rotational velocity. So why does the skater
mass speed up? The answer is Coriolis.

https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/12%3A_Non-inertial_Reference_Frames/12.08%3A_Coriolis_Force#:~:text=An%20interesting%20application%20of%20the,external%20forces%20which%20is%20correct.

That article is instructive, but his final claim is incorrect.  
Retracting the arms injects kinetic energy, from the muscles.
The author\'s claim is that this eventually adds to the total
rotational energy, as the arms merge with the torso.

The human body is too complex (distrust any simplistic physics
analysis of human mobility).  Try a simpler analog:

A rotating platform, with a ball attached to the rim. A golfer standing
at that point strokes a putt straight at the center. The ball rolls radially.
And, the floor has tangential static friction, such that the ball continues
to rotate with the platform, at the same angular speed as the platform.

so as-if the ball was on a track ?

Thus, seen from the center, it rolls straight in; an external observer sees
it follow a spiral path. The ball eventually drops into the hole... then
what? Does it revolve around the center? Does its linear energy convert
into rotational energy?

While approaching the hole it does, that\'s why the rotating observer
sees it slow as it travels inwards. This is caused by an effect
called centrifugal force.

That\'s silly. It bounces off the walls, then settles.
The energy dissipates as heat.

That would happen for any left over energy.

--
Rich

--
Jasen.
🇺🇦 Слава Україні

 

[John Larkin]

On Mon, 8 May 2023 13:55:00 -0700 (PDT), RichD
<r_delaney2001@yahoo.com> wrote:

On May 1, Fred Bloggs wrote:
The skater expends chemical energy, in her muscles, to pull her arms

Work done is torque x angular displacement. No torque= No work, and no work means
no change in energy. Rotational kinetic energy does not change.

The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

As the arms move closer to the axis, the centripetal force decreases considerably, requiring
less work on the skater. But it makes no difference whatsoever because just bringing the
 arms inward in no way shape or form applies a force tangential to the radius connecting
the axis, which is what you need to torque up the rotational velocity. So why does the skater
mass speed up? The answer is Coriolis.

https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/12%3A_Non-inertial_Reference_Frames/12.08%3A_Coriolis_Force#:~:text=An%20interesting%20application%20of%20the,external%20forces%20which%20is%20correct.

That article is instructive, but his final claim is incorrect.  
Retracting the arms injects kinetic energy, from the muscles.
The author\'s claim is that this eventually adds to the total
rotational energy, as the arms merge with the torso.

He did the math and is correct. When the skater pulls her arms in, the
work she does adds energy to the spinning system. That energy is
recovered when she extends her arms back out, slowing the spin.

She can\'t chemically store the recovered energy, but another mechanism
could. A lossless robotic skater could spin up and down an infinite
number of times without needing an external energy source.

The human body is too complex (distrust any simplistic physics
analysis of human mobility).  Try a simpler analog:

A rotating platform, with a ball attached to the rim. A golfer standing
at that point strokes a putt straight at the center. The ball rolls radially.
And, the floor has tangential static friction, such that the ball continues
to rotate with the platform, at the same angular speed as the platform.

Thus, seen from the center, it rolls straight in; an external observer sees
it follow a spiral path. The ball eventually drops into the hole... then
what? Does it revolve around the center? Does its linear energy convert
into rotational energy? That\'s silly. It bounces off the walls, then settles.
The energy dissipates as heat.

Analogously, the arms end up crashing into her hips, they HALT
(radially) at that point. Their kinetic energy goes into heat.

She shouldn\'t crash her arms into her hips. She could get bruised and
the judges would think it inelegant.

They continue to rotate with her body, but that\'s merely the
remnant of their initial energy, which they contained while outstretched.

 

[Martin Brown]

On 08/05/2023 21:55, RichD wrote:

On May 1, Fred Bloggs wrote:
The skater expends chemical energy, in her muscles, to pull her arms

Work done is torque x angular displacement. No torque= No work, and no work means
no change in energy. Rotational kinetic energy does not change.

The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

As the arms move closer to the axis, the centripetal force decreases considerably, requiring
less work on the skater. But it makes no difference whatsoever because just bringing the
 arms inward in no way shape or form applies a force tangential to the radius connecting
the axis, which is what you need to torque up the rotational velocity. So why does the skater
mass speed up? The answer is Coriolis.

https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/12%3A_Non-inertial_Reference_Frames/12.08%3A_Coriolis_Force#:~:text=An%20interesting%20application%20of%20the,external%20forces%20which%20is%20correct.

That article is instructive, but his final claim is incorrect.
Retracting the arms injects kinetic energy, from the muscles.
The author\'s claim is that this eventually adds to the total
rotational energy, as the arms merge with the torso.

No he is right as is John. That is *exactly* what happens.

The skater does real work by moving her arms inwards in a rotating frame
of reference and it creates a torque that spins her up faster. She has
more kinetic energy and less potential energy with her arms in.

It is much easier to understand the physics by considering conservation
of angular momentum on an isolated system.

Kinetic energy on its own is *NOT* conserved. *TOTAL* energy *IS*.

The human body is too complex (distrust any simplistic physics
analysis of human mobility).  Try a simpler analog:

A rotating platform, with a ball attached to the rim. A golfer standing
at that point strokes a putt straight at the center. The ball rolls radially.
And, the floor has tangential static friction, such that the ball continues
to rotate with the platform, at the same angular speed as the platform.

Thus, seen from the center, it rolls straight in; an external observer sees
it follow a spiral path. The ball eventually drops into the hole... then
what? Does it revolve around the center? Does its linear energy convert
into rotational energy? That\'s silly. It bounces off the walls, then settles.
The energy dissipates as heat.

OK lets actually try an analogue that is relevant to the problem in
hand. Take a nice long pole attached to a bearing in the middle and put
two equal weights either side of the centre with a catch mechanism to
hold them in place. A playground roundabout will do.

Spin the whole assembly up and then release the catch. The weights will
spontaneously accelerate away from their initial position near hte
centre of the rod and the spin rate will slow down as a result.

Beware that when the weights drop off the ends of the pole they will be
travelling quickly and in a straight line tangential to the pole.

Analogously, the arms end up crashing into her hips, they HALT
(radially) at that point. Their kinetic energy goes into heat.
They continue to rotate with her body, but that\'s merely the
remnant of their initial energy, which they contained while outstretched.

When her arms were outstretched they had significant stored potential
energy that the dancer converts into kinetic energy by her actions.

If you really think that kinetic energy on its own is a conserved
quantity I suggest you throw a brick up in the air and stand under it!

--
Martin Brown

 

[Jeff Layman]

On 09/05/2023 14:50, John Larkin wrote:

On Mon, 8 May 2023 13:55:00 -0700 (PDT), RichD
r_delaney2001@yahoo.com> wrote:

On May 1, Fred Bloggs wrote:
The skater expends chemical energy, in her muscles, to pull her arms

Work done is torque x angular displacement. No torque= No work, and no work means
no change in energy. Rotational kinetic energy does not change.

The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

As the arms move closer to the axis, the centripetal force decreases considerably, requiring
less work on the skater. But it makes no difference whatsoever because just bringing the
 arms inward in no way shape or form applies a force tangential to the radius connecting
the axis, which is what you need to torque up the rotational velocity. So why does the skater
mass speed up? The answer is Coriolis.

https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/12%3A_Non-inertial_Reference_Frames/12.08%3A_Coriolis_Force#:~:text=An%20interesting%20application%20of%20the,external%20forces%20which%20is%20correct.

That article is instructive, but his final claim is incorrect.
Retracting the arms injects kinetic energy, from the muscles.
The author\'s claim is that this eventually adds to the total
rotational energy, as the arms merge with the torso.

He did the math and is correct. When the skater pulls her arms in, the
work she does adds energy to the spinning system. That energy is
recovered when she extends her arms back out, slowing the spin.

She can\'t chemically store the recovered energy, but another mechanism
could. A lossless robotic skater could spin up and down an infinite
number of times without needing an external energy source.

Surely all it has to do with is conservation of angular momentum
(<https://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum>).
You don\'t have to add the complication of \"added energy\" from muscles.
It\'s the same reason that neutron stars spin so fast when the star they
are formed from collapses.

--

Jeff

 

[John Larkin]

On Tue, 9 May 2023 17:18:20 +0100, Jeff Layman <Jeff@invalid.invalid>
wrote:

On 09/05/2023 14:50, John Larkin wrote:
On Mon, 8 May 2023 13:55:00 -0700 (PDT), RichD
r_delaney2001@yahoo.com> wrote:

On May 1, Fred Bloggs wrote:
The skater expends chemical energy, in her muscles, to pull her arms

Work done is torque x angular displacement. No torque= No work, and no work means
no change in energy. Rotational kinetic energy does not change.

The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

As the arms move closer to the axis, the centripetal force decreases considerably, requiring
less work on the skater. But it makes no difference whatsoever because just bringing the
 arms inward in no way shape or form applies a force tangential to the radius connecting
the axis, which is what you need to torque up the rotational velocity. So why does the skater
mass speed up? The answer is Coriolis.

https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/12%3A_Non-inertial_Reference_Frames/12.08%3A_Coriolis_Force#:~:text=An%20interesting%20application%20of%20the,external%20forces%20which%20is%20correct.

That article is instructive, but his final claim is incorrect.
Retracting the arms injects kinetic energy, from the muscles.
The author\'s claim is that this eventually adds to the total
rotational energy, as the arms merge with the torso.

He did the math and is correct. When the skater pulls her arms in, the
work she does adds energy to the spinning system. That energy is
recovered when she extends her arms back out, slowing the spin.

She can\'t chemically store the recovered energy, but another mechanism
could. A lossless robotic skater could spin up and down an infinite
number of times without needing an external energy source.

Surely all it has to do with is conservation of angular momentum
(<https://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum>).
You don\'t have to add the complication of \"added energy\" from muscles.

There are many cases where a lot of nasty math, calculus and
differenial equations and hard stuff, can be eliminated by simply
invoking Conservation Of Energy. That happens a lot in electronic
design.

This is such a case.

Imagine a very simple version of that robotic skater. It doesn\'t even
need a battery.

 

[Phil Hobbs]

On 2023-05-09 13:10, John Larkin wrote:> On Tue, 9 May 2023 17:18:20
+0100, Jeff Layman <Jeff@invalid.invalid>

wrote:

On 09/05/2023 14:50, John Larkin wrote:
On Mon, 8 May 2023 13:55:00 -0700 (PDT), RichD
r_delaney2001@yahoo.com> wrote:

On May 1, Fred Bloggs wrote:
The skater expends chemical energy, in her muscles, to pull
her arms

Work done is torque x angular displacement. No torque= No work,
and no work means
no change in energy. Rotational kinetic energy does not change.

The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

As the arms move closer to the axis, the centripetal force
decreases considerably, requiring
less work on the skater. But it makes no difference whatsoever
because just bringing the
arms inward in no way shape or form applies a force tangential
to the radius connecting
the axis, which is what you need to torque up the rotational
velocity. So why does the skater
mass speed up? The answer is Coriolis.


https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/12%3A_Non-inertial_Reference_Frames/12.08%3A_Coriolis_Force#:~:text=An%20interesting%20application%20of%20the,external%20forces%20which%20is%20correct.

That article is instructive, but his final claim is incorrect.
Retracting the arms injects kinetic energy, from the muscles.
The author\'s claim is that this eventually adds to the total
rotational energy, as the arms merge with the torso.

He did the math and is correct. When the skater pulls her arms in, the
work she does adds energy to the spinning system. That energy is
recovered when she extends her arms back out, slowing the spin.

She can\'t chemically store the recovered energy, but another mechanism
could. A lossless robotic skater could spin up and down an infinite
number of times without needing an external energy source.

Surely all it has to do with is conservation of angular momentum

(<https://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum>).
You don\'t have to add the complication of \"added energy\" from muscles.

There are many cases where a lot of nasty math, calculus and
differenial equations and hard stuff, can be eliminated by simply
invoking Conservation Of Energy. That happens a lot in electronic
design.

This is such a case.

Imagine a very simple version of that robotic skater. It doesn\'t even
need a battery.

It\'s easier than that, even. Since there are (by hypothesis) no
external torques applied to the skater, her angular momentum I Omega is
conserved. (I is the moment of inertia and Omega is the angular velocity.)

Thus after her arms are pulled in, I_1 and Omega_1 obey

I_1 Omega_1 = I_0 Omega_0

so the angular velocities obey

Omega_1 / Omega_0 = I_0 / I_1.

The work done by her arm muscles is what\'s required to supply the
increase in the rotational kinetic energy:

W = I_1 Omega_1**2 / 2 - I_0 Omega_0**2 / 2

Cheers

Phil Hobbs

 

[John Larkin]

On Tue, 9 May 2023 14:59:34 -0400, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

On 2023-05-09 13:10, John Larkin wrote:> On Tue, 9 May 2023 17:18:20
+0100, Jeff Layman <Jeff@invalid.invalid
wrote:

On 09/05/2023 14:50, John Larkin wrote:
On Mon, 8 May 2023 13:55:00 -0700 (PDT), RichD
r_delaney2001@yahoo.com> wrote:

On May 1, Fred Bloggs wrote:
The skater expends chemical energy, in her muscles, to pull
her arms

Work done is torque x angular displacement. No torque= No work,
and no work means
no change in energy. Rotational kinetic energy does not change.

The skater did work to pull her arms and legs in towards the spin
axis. Where did that energy come from and where did it go?

As the arms move closer to the axis, the centripetal force
decreases considerably, requiring
less work on the skater. But it makes no difference whatsoever
because just bringing the
arms inward in no way shape or form applies a force tangential
to the radius connecting
the axis, which is what you need to torque up the rotational
velocity. So why does the skater
mass speed up? The answer is Coriolis.


https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/12%3A_Non-inertial_Reference_Frames/12.08%3A_Coriolis_Force#:~:text=An%20interesting%20application%20of%20the,external%20forces%20which%20is%20correct.

That article is instructive, but his final claim is incorrect.
Retracting the arms injects kinetic energy, from the muscles.
The author\'s claim is that this eventually adds to the total
rotational energy, as the arms merge with the torso.

He did the math and is correct. When the skater pulls her arms in, the
work she does adds energy to the spinning system. That energy is
recovered when she extends her arms back out, slowing the spin.

She can\'t chemically store the recovered energy, but another mechanism
could. A lossless robotic skater could spin up and down an infinite
number of times without needing an external energy source.

Surely all it has to do with is conservation of angular momentum

(<https://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum>).
You don\'t have to add the complication of \"added energy\" from muscles.

There are many cases where a lot of nasty math, calculus and
differenial equations and hard stuff, can be eliminated by simply
invoking Conservation Of Energy. That happens a lot in electronic
design.

This is such a case.

Imagine a very simple version of that robotic skater. It doesn\'t even
need a battery.


It\'s easier than that, even. Since there are (by hypothesis) no
external torques applied to the skater, her angular momentum I Omega is
conserved. (I is the moment of inertia and Omega is the angular velocity.)

Thus after her arms are pulled in, I_1 and Omega_1 obey

I_1 Omega_1 = I_0 Omega_0

so the angular velocities obey

Omega_1 / Omega_0 = I_0 / I_1.

The work done by her arm muscles is what\'s required to supply the
increase in the rotational kinetic energy:

W = I_1 Omega_1**2 / 2 - I_0 Omega_0**2 / 2

Cheers

Phil Hobbs

Let\'s design the robot!