We have a massless disk carrying one arm.
The mass of the arm is 1kg.
The arm length is 1m.
The center to the top of the arm is 11m.
We let it start at rest as the left figure. Let it turn 1 complete turn clockwise and end up with figure on the right.
We can calculate the final speed of the ball by the GPE difference (2 meters of GPE).
1kg x 10 x 2m = 20 joules
The final speed with 20 joules - .5mv^2 = 20J ; v= 6.3 m/s
But this answer is wrong. The ball gained more energy.
MT13 single arm energy analysis
Moderator: scott
MT13 single arm energy analysis
Last edited by Leafy on Sun Feb 13, 2022 9:44 pm, edited 1 time in total.
I would trade everything to see her again, even a perpetual motion machine…
Re: MT13 single arm energy analysis
You need to do the analasys with a second balancing weight, either a stationary weight or duplicate mechanism.
Re: MT13 single arm energy analysis
Hi Leafy .. something doesn't seem right with your thought experiment.
My 2 cents .. the arm is 1 meter in length, and it has a sphere or ball on the end. You say that the 1 meter arm has a mass of 1 kg. Presumably the sphere has zero mass ? That means the Center of Mass (COM) of the arm is half way along its length (assuming even distribution of mass). IOW's it's COM is at 1/2 meter from pivot point.
After rotation where it appears the arm first rests against a stop until after 6 o'cl when it then is free to translate (hang vertically) into the position you show in the second drawing.
That would mean the arm COM only loses 1 meter of GPE (mgh), and not 2 as per your experiment.
You can simplify this thought experiment considerably. Do away with the massless 10 meter radius backboard and just make a study of a vertical pendulum arm of 1 meter length. Let it swing down etc.
At any vertical height its speed can be calculated by rearranging the formulas mgh = m1/2v^2 to give v = sqrt(2gh).
## this assumes there are no frictions in the pivot or windage (which are non conservative forces and waste energy - in this case velocity and KE achieved).
** you can also do the experiment in at least two other ways. Just attach a ball to a rim at radius 1 meter etc etc. Or let a ball or sphere run down a slope (aka Galileo) and be able to calculate its speed at any height against v= sqrt(2gh) etc.
%% .. first thing to do is know where your COM is located !!! Very important to know in advance IMO !
My 2 cents .. the arm is 1 meter in length, and it has a sphere or ball on the end. You say that the 1 meter arm has a mass of 1 kg. Presumably the sphere has zero mass ? That means the Center of Mass (COM) of the arm is half way along its length (assuming even distribution of mass). IOW's it's COM is at 1/2 meter from pivot point.
After rotation where it appears the arm first rests against a stop until after 6 o'cl when it then is free to translate (hang vertically) into the position you show in the second drawing.
That would mean the arm COM only loses 1 meter of GPE (mgh), and not 2 as per your experiment.
You can simplify this thought experiment considerably. Do away with the massless 10 meter radius backboard and just make a study of a vertical pendulum arm of 1 meter length. Let it swing down etc.
At any vertical height its speed can be calculated by rearranging the formulas mgh = m1/2v^2 to give v = sqrt(2gh).
## this assumes there are no frictions in the pivot or windage (which are non conservative forces and waste energy - in this case velocity and KE achieved).
** you can also do the experiment in at least two other ways. Just attach a ball to a rim at radius 1 meter etc etc. Or let a ball or sphere run down a slope (aka Galileo) and be able to calculate its speed at any height against v= sqrt(2gh) etc.
%% .. first thing to do is know where your COM is located !!! Very important to know in advance IMO !
Last edited by Fletcher on Mon Feb 14, 2022 12:14 am, edited 1 time in total.
Re: MT13 single arm energy analysis
Hi Fletcher,
I meant the sphere is 1 kg. So the COM is the sphere. The 1m arm is massless.
The reason I let it travel one turn is because from 6 o’clock back to 12 o’clock the ball also move from outer rim to inner him. This is the ice skater effect and something( gravity) has to add energy to it.
I meant the sphere is 1 kg. So the COM is the sphere. The 1m arm is massless.
The reason I let it travel one turn is because from 6 o’clock back to 12 o’clock the ball also move from outer rim to inner him. This is the ice skater effect and something( gravity) has to add energy to it.
I would trade everything to see her again, even a perpetual motion machine…
Re: MT13 single arm energy analysis
I understand now. 6.3m/s is rounded off....19.845 joules.
The energy isn't more, I assume the back disk is massless. 6.3m/s is the speed of the mass, It is not the speed of the rim of the wheel or the pivot. If however the mass magically hangs perfectly downward through rotation, 6.3 is the speed of the pivot.
The energy isn't more, I assume the back disk is massless. 6.3m/s is the speed of the mass, It is not the speed of the rim of the wheel or the pivot. If however the mass magically hangs perfectly downward through rotation, 6.3 is the speed of the pivot.
Re: MT13 single arm energy analysis
I think T79 is correct. Once a pendulum is free to hang down and translate, its weight is felt at the pivot from which it hangs (after 6 o'cl). Assuming little or no pivot frictions.
The Ice Skater Effect comes into play IINM when an ice skater lets their arms out (increases MOI and slows rpm) .. and when they use muscle energy to pull their arms in closer to the Center of Rotation, thus decreasing MOI and increasing rpm. The important consideration is that energy is added in the second instance acting against linear (tangential) inertia of the arms etc.
In your thought experiment the pendulum translates and altho it does come inwards to a closer radius no energy was invested to achieve this - and IINM the velocity of the bob is exactly the same as the variations of the experiment I suggested in my earlier post.
Why don't you down load Algodoo (free) and try these calibration experiments. It's a good way to learn the program and why scientists insist that gravity force is conservative i.e. mgh = m1/2v^2 and the Work Energy Equivalence Principle. It's a good place to start imo.
The Ice Skater Effect comes into play IINM when an ice skater lets their arms out (increases MOI and slows rpm) .. and when they use muscle energy to pull their arms in closer to the Center of Rotation, thus decreasing MOI and increasing rpm. The important consideration is that energy is added in the second instance acting against linear (tangential) inertia of the arms etc.
In your thought experiment the pendulum translates and altho it does come inwards to a closer radius no energy was invested to achieve this - and IINM the velocity of the bob is exactly the same as the variations of the experiment I suggested in my earlier post.
Why don't you down load Algodoo (free) and try these calibration experiments. It's a good way to learn the program and why scientists insist that gravity force is conservative i.e. mgh = m1/2v^2 and the Work Energy Equivalence Principle. It's a good place to start imo.
Re: MT13 single arm energy analysis
I think gravity do work to keep the mass hang downward, otherwise, it would spin outward due to the CF.
I don’t know if Algodoo works on IPad, thanks for the suggestion.
I don’t know if Algodoo works on IPad, thanks for the suggestion.
I would trade everything to see her again, even a perpetual motion machine…
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Re: MT13 single arm energy analysis
The kinetic energy increases by the amount the rotational inertia decreases. The potential energy is converted to kinetic energy (+loss) so the total energy is conserved. Kinetic gain due to potential lost.The angular momentum is conserved. It ain't right to think energy is being added during revolutions of the disk or the skater because of increased velocity. Energy is being converted. It comes from somewhere, but not gravity in either case - in the muscles, in both cases, if you built mt13.
The world record for skating rpms is 342.
The world record for skating rpms is 342.
Re: MT13 single arm energy analysis
It is obvious that the MT13 is a good candidate for a permanent unbalance wheel. If you can get the weight up at noon for free, I still think it's possible, but I don't know how. :)
A++
A++
Last edited by thx4 on Tue Feb 15, 2022 7:36 am, edited 1 time in total.
The world is perfect.
Re: MT13 single arm energy analysis
I think Bessler thought the same thing as us initially, lifting weight for free or less. It makes sense.
I realized that we could look at it in a different way. Change the MOI to speed up the wheel and reset the MOI without slowing the wheel down. I think this is the right way looking at it and why lifting weight for free is wrong.
How to reset the MOI could be Bessler's secret.
I would trade everything to see her again, even a perpetual motion machine…