A disc rotating at a certain rpm (e.g. at 1000rpm) having a radial hole with a plunger or piston inside the hole. Say the mass of this piston is 1kg. Certainly this piston will move radially outward under the action of centrifugal force. What will be the speed or velocity of this piston in the radial direction? Or how much distance it can travel radially outward in 1 revolution of the disc?

The piston would be subjected to an acceleration of a = \omega ^2 r, where ω is the rotational velocity of the wheel in radians/second and r is the piston's position measured from the center of the wheel. This assumes that ω is constant and that there is no friction resisting the motion of the piston. From this equation you can see that as the piston slides outward it is subjected to an ever greater radial acceleration, so its speed will increase dramatically. The equation of motion for the piston will be of the form:


r = R_0 e^{\omega t}


Whether the piston slides out of the wheel one rotation or not depends on the initial position of the piston R_0 and the radius of the wheel. Note that the mass of the piston is immaterial here.

**EDIT** See improved solution in the following post

Update: upon refelection I realized the solution is a bit different from what I originally wrote. The differential equation for the motion of the piston is:


\frac {d^2r} {dr^2} = \omega^2 r


Asuming \omega is constant, the general solution for this is:


r(t) = A e ^{\omega t} + B e ^{-\omega t}


where A and B are constants, whose values can be determined from boundary conditions - namely the initial position and initial radial velocity of the piston at time t = 0. For example, if the piston is originally at position R_0, and its initial velocity is 0, then you get:


r(t=0) = A + B = R_0 \\
v(t=0) = \frac {dr(0)} {dt} = A\omega-B\omega = 0


From this the solution is:


r = \frac {R_0} 2 e^ {\omega t} + \frac {R_0} 2 e^{-\omega t }\ = \ R_0 \cosh(\omega t)

Thank you for your consideration. As an example, suppose that wheel is rotating at 1500 rpm, piston is initially at 100 mm from the wheel axis of rotation.
Q.1 How much distance the piston will move outward in 1 revolution of the wheel while rotating at 1500 rpm?
Q.2 If the piston has mass of 1 kg, it will exert centrifugal force of about 251 kg force at 100 mm radius. Is it correct? Please verify.

Q1.: Using the formula I gave you, you need to convert 1500 RPM into the equivalent values for ω and the time taken for one revolution, T:


\omega = 2 \pi \frac {Rad} {Rev} \ 1500 RPM \ \times \ \frac 1 {60} \frac {min} s \\
T = 60 \frac s {min} \times \frac 1 {1500 RPM}


So \omega T = 2 \pi or one revolution. Hence assuming there is no friction, the distance the piston slides in one revolution is:


r = R_0 \cosh(2 \pi) = R_0\ \times \ 267.7 = \ 0.1m\ \times \ 267.7 = 26.77 m.


Q2: when the piston is at r=0.1m, its centripedal acceleration is:

a = r \omega ^2 = 0.1m\ \times \ (2 \pi \frac {Rad} {Rev} \ \times \ 1500 RPM \ \times \ \frac {min} {60s})^2= 2467 \frac {m}{s^2}


And the force exerted is:


F = ma = 1 kg \ \times \ 2467 \frac m {s^2} = 2467 N.

If the movement of the piston is opposed say by some sort of hydraulic or spring force equal to half of the centrifugal force (i.e. 2467N), then what will be the speed of the piston at that particular point?

If the movement of the piston is opposed say by some sort of hydraulic or spring force equal to half of the centrifugal force (i.e. 2467N), then what will be the speed of the piston at that particular point?

If the spring is designed so that its opposing force is half of the force due to centripedal acceleration regardless of the pistons radial position, then the pistons radial acceleration is half of what I provided earlier. Hence its velocity at any point in time is half of what it would be without the spring.

But if the spring is designed to provide 2467N no matter what position the piston is in, then theoretically the piston is in equilibrium in its starting position of 0.100 m, so it doesn't move at all. However this is an unstable equilibrium, so the piston will actually either move inward or outward - it's impossible to tell. It's like balancing a pencil on its point and trying to figure out which way will it fall.

Mayya - perhaps you can explain why you're asking these questions - what exactly are you trying to design?

Thanks for your replies as they helped me a lot. Actually I am asking these questions on behalf of my father who is trying to design a mechanism for utilization of centrifugal force. With thanks and anticipation.

It can be confirmed by calculation that the piston having 1kg mass with its centre of gravity at 100 mm while rotating at 1500 rpm needs about 185 watts of input power/energy. Whereas it is generating a centrifugal force of 2467N just calculated by you. If this radial force can be diverted somehow and applied tangentially to drive another disk having 100 mm radius, it can produce tremendous power (about 19 kilowatt).

With reference to the same data, what will be the velocity of the piston in radial direction at particular distance from the center e.g. at 200 mm radius?

mayya - are you asking about the velocity of the unrestrained piston (no opposing force)? Go back to post #3, which gives the radial position of the piston as a function of time:


r =R_0 \cosh (\omega t)


If you set R_0 = 100 mm and use a rotational velocity of 1500 RPM = 157 rad/s, then the elapsed time for the piston to reach position r = 200 mm:


t = \frac 1 {\omega} arccosh(r/R_0) = \frac 1 {157 s^{-1}} arccosh(2) = 0.0083 s.


The radial velocity is the derivative of the position:


v = \frac {dr} {dt} = R_0 \omega \sinh (\omega t)


Substitute in the value for t that we found above:


v = 0.1m\ \times \ 157 s^{-1} sinh(157s^{-1} 0.0083s) = 27 m/s


Note that all this assumes that the wheel is rotating at a constant speed of 1500 RPM. Hope this helps.

It can be confirmed by calculation that the piston having 1kg mass with its centre of gravity at 100 mm while rotating at 1500 rpm needs about 185 watts of input power/energy. Whereas it is generating a centrifugal force of 2467N just calculated by you. If this radial force can be diverted somehow and applied tangentially to drive another disk having 100 mm radius, it can produce tremendous power (about 19 kilowatt).

I'd like to see your calculation of this. It seems like you are thinking that you could build a perpetual motion machine, that with input power of 185 watts it could generate 19 kilowatts of output. Of course that's impossible, at least for continuous operation.

Many thanks for your helping efforts.

Regarding your question, reason for consideration of centrifugal force for free power generation is due to following calculations and logics.

Rotation of one kilogram mass with its center of gravity at 100 mm from axis of rotation at 1500 rpm (25rps) requires input power equal to its total kinetic energy. Using Steiner's parallel axis theorem this comes out to about 185 joules/sec (watt) and the tangential force involved for torque is just 11.78 N. On the other hand centrifugal force naturally created in this process is 2467 N in radial direction along with acceleration of 2467 m/s2. If this much force may be diverted to be applied tangentially at 50 mm radius to produce torque at 1500 rpm the power output comes out to be about 19379 joules/sec (watt), 104 times the input! With increase in radius output will increase proportionally. Keeping in view practical resistance/constraints whatsoever, if only 50% (52 times) or even 10% (10 times) of that is achieved then it may be a wonderful revolution.

A dynamic (moving) force always does the useful work. If the force is dead, that is not moving like hydraulic pressure force in a pressurized container, it does produce stresses but no useful work is done unless the pressurized fluid is moving. In spite of great confusion about reality of centrifugal force, it is a real, dynamic and feel-able force equal and opposite to centripetal force having same acceleration and velocity in magnitude but opposite in direction. That is why the piston under discussion travels radially outward with velocity yet to be analyzed. Suction of oil from the reservoir quite below the hydraulic pump, particularly of radial piston type, due to centrifugal force of outwardly extending pistons is another indication of reality and usefulness of centrifugal force.

Other than the torque that may be produced or not, radial travel of piston under discussion through certain distance with that much dynamic force at the expense of very little input force/energy, can do too much work accordingly. Certainly, continuous working and harness of centrifugal force with this sort of arrangement is impossible. But there may be other possible options to be discussed later. Oscillations produced by centrifugal force with surplus energy have already been worked upon a lot in manual way by Veljko Milković. The continuous motorized oscillation converted into rotation has been experienced and being tested myself in a different way. Seemingly more viable one wherein much of the peaks of radial centrifugal force may be directly converted into continuous torque/circular-motion requires authentication of radial velocity of radially moving mass under the influence of centrifugal force at certain radius. Working of perpetual motion machine or free energy generation just depends upon the fact whether radial velocity is how much more or less than tangential velocity at particular radius.

Now coming back to my questions and your calculation in this regard the following may kindly be reviewed.

Firstly, calculated distance traveled by the piston is about 27 (=26.77) meters radially outward in one revolution. This is covered in 1/25 seconds as there are 25 revolutions per second (1500 rpm). Therefore, average velocity/speed would be (27 x 25 =) 675 m/s.

Secondly, acceleration (a) at 100 mm radius (r) calculated above is 2467 m/s2 then acceleration at 200 mm radius would be 4934 as value of acceleration has been doubled.. If we start at the center, initial velocity/acceleration is zero then, radial velocity at 100mm & 200mm would be 2476 & 4934 m/s respectively and after one second time it would become quite higher further in proportion to radius occupied by that time.

Thirdly, calculated time taken by piston from 100 mm to reach 200 mm is 0.0083 second. Then average velocity at 200 mm radius would be acceleration x time = 4934 x 0.0083 = 41 meter.

Fourthly, quite big value of acceleration as well as the heavy centrifugal force as compared to tangential force involved for rotation of I kg mass at any radius, apart from complex calculus, logically indicates larger values of radial velocity increasing with increasing radius.

All above needs verification/better understanding of which value of average radial velocity/speed of piston, 27, 675 or 4934 m/s is correct.

It is quite clear that tangential velocity at any radius is equal to rotational velocity at that point converted into linear dimensions. If piston escapes rotation at any radius it will leave tangentially with the same velocity as centrifugal force disappears there instantly.

Kindly don't mind my poor understanding of complex mathematics/calculus as I need very simple and clear comparison of rotational/tangential and radial velocities at any radius. In spite of serious search I could not find relevant information either in available books or on the net.

Thanks and regards.

Many thanks for your helping efforts.

Regarding your question, reason for consideration of centrifugal force for free power generation is due to following calculations and logics.

Rotation of one kilogram mass with its center of gravity at 100 mm from axis of rotation at 1500 rpm (25rps) requires input power equal to its total kinetic energy. Using Steiner’s parallel axis theorem this comes out to about 185 joules/sec (watt) and the tangential force involved for torque is just 11.78 N.

You are confusing torque, energy, and power. This device has angular kinetic energy of 123 N-m . To maintain the rotational velocity of 1500 RPM actually requires no power at all - as long as there is not any friction nor any resisting force.


On the other hand centrifugal force naturally created in this process is 2467 N in radial direction along with acceleration of 2467 m/s2. If this much force may be diverted to be applied tangentially at 50 mm radius to produce torque at 1500 rpm the power output comes out to be about 19379 joules/sec (watt), 104 times the input!

Again - force is NOT power. Please show me your calculation of this so I can comment on the error you are making here.


... A dynamic (moving) force always does the useful work. If the force is dead, that is not moving like hydraulic pressure force in a pressurized container, it does produce stresses but no useful work is done unless the pressurized fluid is moving. In spite of great confusion about reality of centrifugal force, it is a real, dynamic and feel-able force equal and opposite to centripetal force having same acceleration and velocity in magnitude but opposite in direction. That is why the piston under discussion travels radially outward with velocity yet to be analyzed. Suction of oil from the reservoir quite below the hydraulic pump, particularly of radial piston type, due to centrifugal force of outwardly extending pistons is another indication of reality and usefulness of centrifugal force.

This is the first time you have mentioned a resistant force - namely sucking oil from some sort of reservoir. The problem is that the resistant force of the oil causes your piston to move MUCH slower radially than previously calculated. The work done by the priston is the force being applied by it (the 2467N figure) times the radial velocity of the piston, which is yet to be determined.


... Now coming back to my questions and your calculation in this regard the following may kindly be reviewed.

Firstly, calculated distance traveled by the piston is about 27 (=26.77) meters radially outward in one revolution. This is covered in 1/25 seconds as there are 25 revolutions per second (1500 rpm). Therefore, average velocity/speed would be (27 x 25 =) 675 m/s.

Wrong. As I've said before this is true ONLY if there is no resistive force (i.e the piston does no work). Once you introduce the resistive force of the oil, so that the piston is actually doing some work, this velocity is greatly reduced.


Secondly, acceleration (a) at 100 mm radius (r) calculated above is 2467 m/s2 then acceleration at 200 mm radius would be 4934 as value of acceleration has been doubled.. If we start at the center, initial velocity/acceleration is zero then, radial velocity at 100mm & 200mm would be 2476 & 4934 m/s respectively and after one second time it would become quite higher further in proportion to radius occupied by that time.

Yes, but only if there is no resistive force (how many times do I have to say this? )


Thirdly, calculated time taken by piston from 100 mm to reach 200 mm is 0.0083 second. Then average velocity at 200 mm radius would be acceleration x time = 4934 x 0.0083 = 41 meter.

Fourthly, quite big value of acceleration as well as the heavy centrifugal force as compared to tangential force involved for rotation of I kg mass at any radius, apart from complex calculus, logically indicates larger values of radial velocity increasing with increasing radius.

All above needs verification/better understanding of which value of average radial velocity/speed of piston, 27, 675 or 4934 m/s is correct.


None of these values are correct for the machine you've proposed.


It is quite clear that tangential velocity at any radius is equal to rotational velocity at that point converted into linear dimensions. If piston escapes rotation at any radius it will leave tangentially with the same velocity as centrifugal force disappears there instantly.

Kindly don’t mind my poor understanding of complex mathematics/calculus as I need very simple and clear comparison of rotational/tangential and radial velocities at any radius. In spite of serious search I could not find relevant information either in available books or on the net.

Thanks and regards.

The key here is that if the piston does work against the oil reservoir (and ultimately in turning a generator) its kinetic energy is decreased by an equivalent amount. This means its radial velocity decreases as well. In the extreme, if the radial velocity is 0 then the piston can do no work at all, EVEN THOUGH IT'S TANGENTIAL VELOCITY IS STILL QUITE HIGH. If you try to convert that tangental velocity into work, the machine will slow down, or else require additional power input to keep turning.

There is nothing here that suggests any sort of "break through" with respect to designing a perpetual motion machine. Thinking through the physics of a machine such as this can is a good intellectual exercise that can help you build a better undertanding of how work and energy are traded off against each other. But one thing is for sure: if your conclusion is that you can design a machine that produces more energy than it consumes, you have made an error in your thinking.

A lot of mix-up and confusion has taken place. I shall show you detailed calculation a little bit later. Actually, I expected calculation based on this data yourself and confirmation of my calculations thereby.

I have, though may not be perfect, quite sufficient concept of force, torque, work, energy and power etc. All discussion pertains to same data from the very beginning under ideal conditions. I did never mention motion of piston in radial direction under the influence of centrifugal force against any resistance. Example of intake of oil by the piston was quoted just to emphasize the existence, dynamicity and ability of centrifugal force to work. As you mentioned earlier, distance and velocity of radial travel of piston against any resistance will certainly decrease in proportion to the resistance and become zero for 2467 N. Also, I am not talking about extracting centrifugal force energy of the piston mass for free power generation. It is quite a different story and mechanism to be discussed in detail later. Of course, over unity machine is impossible but excitation and incorporation of any natural energy cannot be denied.

Question at the moment is just to explore exact and confirm value of radial velocity of piston or any radially slid-able mass under the influence of centrifugal force at certain radius. This will help analytical comparison of the same with tangential velocity (= rotational velocity in linear dimensions) at certain radius whereas practical confirmation needs a lot of R & D and investment. What I have observed physically through my practical experiments, radial velocity obviously seems to be much higher than tangential velocity at certain radius. Relevant values calculated or estimated logically so far are confusing as they do not tally each other as mentioned in my previous query and reproduced hereunder.

“Firstly, calculated distance traveled by the piston is about 27 (=26.77) meters radially outward in one revolution. This is covered in 1/25 seconds as there are 25 revolutions per second (1500 rpm). Therefore, average velocity/speed would be (27 x 25 =) 675 m/s.

Secondly, acceleration (a) at 100 mm radius (r) calculated above is 2467 m/s2 then acceleration at 200 mm radius would be 4934 as value of radius has been doubled.. If we start at the center, initial velocity/acceleration is zero then, radial velocity at 100mm & 200mm would be 2476 & 4934 m/s respectively and after one second time it would become quite higher further in proportion to radius occupied by that time.

Thirdly, calculated time taken by piston from 100 mm to reach 200 mm is 0.0083 second. Then average velocity at 200 mm radius would be acceleration x time = 4934 x 0.0083 = 41 meter.

Fourthly, quite big value of acceleration as well as the heavy centrifugal force as compared to tangential force involved for rotation of I kg mass at any radius, apart from complex calculus, logically indicates larger values of radial velocity increasing with increasing radius.

All above needs verification/better understanding of which value of average radial velocity/speed of piston, 27, 675 or 4934 m/s or even else is correct.�

Firstly, calculated distance traveled by the piston is about 27 (=26.77) meters radially outward in one revolution. This is covered in 1/25 seconds as there are 25 revolutions per second (1500 rpm). Therefore, average velocity/speed would be (27 x 25 =) 675 m/s.

Yes, that would be the average veocity. But the actual velocity at t = 0.04 seconds turns out to be 4205 m/s.



Secondly, acceleration (a) at 100 mm radius (r) calculated above is 2467 m/s2 then acceleration at 200 mm radius would be 4934 as value of acceleration has been doubled.. If we start at the center, initial velocity/acceleration is zero then, radial velocity at 100mm & 200mm would be 2476 & 4934 m/s respectively and after one second time it would become quite higher further in proportion to radius occupied by that time.

Yes - clearly as time elapses and the piston slides further outward its acceleration increases exponentially. In fact, given an ideal piston with no friction forces, IF the disc is large enouh and IF you could actually maintain 1500 RPM as the piston slides outward then believe it or not within 1 second the piston would be moving FASTER THAN THE SPEED OF LIGHT! Clearly there is an issue here - see below.


Thirdly, calculated time taken by piston from 100 mm to reach 200 mm is 0.0083 second. Then average velocity at 200 mm radius would be acceleration x time = 4934 x 0.0083 = 41 meter.

Acceleration is not a constant, so your average calculation doesn't mean very much. As already noted, the actual velocity for r = 200mm is 2467 m/s, and the actual accelaration for r = 200 mm is 4934 m/s^2


Fourthly, quite big value of acceleration as well as the heavy centrifugal force as compared to tangential force involved for rotation of I kg mass at any radius, apart from complex calculus, logically indicates larger values of radial velocity increasing with increasing radius.

Yes, but... As noted below, you have set up an impossible task.


All above needs verification/better understanding of which value of average radial velocity/speed of piston, 27, 675 or 4934 m/s is correct.

The term "average velocity" really has no meaning.


It is quite clear that tangential velocity at any radius is equal to rotational velocity at that point converted into linear dimensions. If piston escapes rotation at any radius it will leave tangentially with the same velocity as centrifugal force disappears there instantly.

Agree. It's a great way to make a catapult.


Kindly don’t mind my poor understanding of complex mathematics/calculus as I need very simple and clear comparison of rotational/tangential and radial velocities at any radius. In spite of serious search I could not find relevant information either in available books or on the net.

As I have noted in previous posts, the formulas that govern the motion of the piston under ideal conditions where (a) the rotational veocity is constant 1500RPM, (b) the piston starts at r = 100mm, and (c) there is no friction, is as follows:


r = R_0 \cosh(\omega t) \\
v_r = R_0 \omega \sinh (\omega t) \\
a_r = R_0 \omega^2 \cosh(\omega t)


Note that these are all exponential functions - and as I will show below the numbers become outrageously large very quickly. But what really makes this whole project impossible is that it assumes that the disc can be spun at a constant rate of 1500 RPM, even as the piston slides outward. But this isi bver y difficult, and quickly becomes absolutely impossible as the piston slides further out. The moment of inertia of the piston increases as it slides outward, and so to maintain a constant spin rate more power is needed to turn the disc. This is because in addition to the radial velocity and acceleration that we calculated earlier, there is also a tangential acceleration component, which in turn introduces correolis forces into the mechanism. The way it works is this - the tangential acceleration of the piston is:


a_t = 2 \omega v_t


The force that the piston applies to the side walls of its cylinder is therefore:


F_t = ma_t = 2 m \omega v_t


and the torque and power required to keep the disc spinning at a constant RPM is:


T = Fr = 2mR_0^2 \omega^2 \cosh(\omega t)\sinh(\omega t) \\
P = T \omega = 2mR_0^2 \omega^3 \cosh(\omega t)\sinh(\omega t)


Now, to crunch some numbers: at t = 0.0083 seconds you get:

r =0.2m
v = 27.2 m/s
a_r = 4934 m/s^2
a_t = 8547 m/s^2
T = 1709 N-m
P = 268522 watts

You see the problem? You need an impossibly large motor to keep this mechanism spinning at 1500 RPM. To take it ridiculous extremes - using these classical physics calculations, if you wanted a disc that was of sufficient size for the piston to take 1 second to reach the edge, the disc would need to be 8.3 x 10^66 meters across! And the piston's velocity at one second would be 1.3 x 10^69 m/s! Obviously this calculation is in error as it does not take into account the effects of relativity. In brief as the piston approaches the speed of light its mass would increases to near infinity and the power required to keep the disc turning becomes infinite as well.

Bottom line - it is impractical for the mechanism to maintain a constant RPM as the piston slides outward.

Mayya - these graphs show the piston's position, velocity and acceleration for the first 0.0083 seconds. Note that the tangential acceleration becomes huge - and hence the power needed to keep the disc turning becomes huge as well.

My assumption and calculation regarding amplification of input energy by utilizing centrifugal force are as under for your review and comments.

Consider a body having one kilogram mass (m) with its center of gravity at 0.10 meter radius (r) from axis rotating at 1500 rpm (25rps). It will require work input (Wip) or energy equal to its total kinetic energy comprising kinetic energy of translation plus kinetic energy of rotation.

Wip = ½ x mass x square of linear velocity of center of gravity + ½ mass moment of
inertia (I) x rotational velocity (ω)
= ½ mv^2 + ½ I ω^2
= ½ m(2 π.r.n)^2 + ½(½ m.r^2)(2 π.n)^2
= 2m(Ï€ .r.n)^2 + m(Ï€.r.n)^2
= 3m(Ï€.r.n)^2 Nm, joules

When n is in number of revolutions per second then power input (Pip) becomes

Pip = 3m(Ï€.r.n)^2 Nm/s, joules/s, watt
= 3 x 1.0(Ï€ x 0.10 x 25)^2
= 185 watt

As power equals to torque (T) x rotational velocity and torque equals force (F) x radius,

Pip = Tip.ω and Tip = Fip.r

So Fip = Pip/r.ω
= 3m(π.r.n)^2/r.ω
= 3m(π.r.n)^2/2 π.r.n
= 3/2m. π .r.n
= 3/2 x 1.0 x π x 0.10 x 25
= 11.78 N

The centrifugal force naturally created in this process, I would like to call it the output force (Fop), at the expense of 11.78 N will be

Fop = m.r.ω^2
= m.r(2Ï€.n)^2
= 4 m.r(Ï€.n)^2
= 4 x 1.0 x 0.10(Ï€ x 25)^2
= 2467.40 N


This output centrifugal force (Fop) has also acceleration of 2467.40 m/s2 directed radially outward.

If this much force having big acceleration as well may somehow be applied tangentially at 50 mm radius to produce torque at 1500 rpm the power output (Pop) comes out to be

Pop = T.ω
= Fop.r.ω
= 4 m.r(π.n)^2. r. 2 π.n
= 8 m.r^2. π^3.n^3
= 8 x 1.0 x 0.05^2 x π^3 x 25^3
= 19378.92 watt

If Fop is applied at 100 mm radius

Pop = 8 x 1.0 x 0.10^2 x π^3 x 25^3
= 38757.84 watt.

Thus the ratio output force to input force becomes

Fop/Fip = 2467.40/11.78
= 104.75

And ratio of output power to input power becomes

Pop/ Pip = 19378.92/185
= 104.75 for 50 mm radius.

and = 38757.84/185
= 209.5 for 100 mm radius

and similarly varies in direct proportion to value of radius.


It may be reminded here that I am not talking about extraction of centrifugal force energy of radially sliding piston mass for free power generation. It is quite a different mechanism to be discussed in detail later. The case of piston has been quoted just to analyze the radial velocity of piston or any radially slid-able mass under the influence of centrifugal force at certain radius.

As you agreed to most of my point raised in last posting, high values of centrifugal force associated with huge acceleration at even smaller radii is also suggestive of high radial velocity accordingly as compared to tangential linear velocity at relevant radius. So piston can still move against reasonable resistance to give output work/energy at velocity comparable with tangential velocity.

Keeping all the complex calculus aside, let us consider just one simple logic for the time being. Suppose the piston initially occupies a position with its center of gravity at zero radius or at any minimum (a few millimeter) radius but prevented to slide outward while the disc is rotating at its speed of 1500 rpm. So, initial acceleration and velocity of piston will be zero. Then piton is suddenly released to slide. In just fraction of a second it will reach 50 or 100 mm radius with centrifugal force and acceleration as calculated above. Then velocity just after a fraction of second or after one complete second will be equal to the value of acceleration by that time which comes out to be much higher than relevant tangential velocities there. Is it correct or not?

It is noted that query prepared in word document when posted does not show some of the characters or operators like omega, pi, etc. in actual format in relevant formulae. Kindly advise what to do to solve this problem.

It is noted that query prepared in word document when posted does not show some of the characters or operators like omega, pi, etc. in actual format in relevant formulae. Kindly advise what to do to solve this problem.

Mayya - please note that theis site can not display the special characters that you are using in your post, so it is impossible to read much of what you wrote. However, it seems that your first equation is in error. It seems that you are saying that the Kinetic Energy of a rotating mass is:


KE = \frac 1 2 m v^2 + \frac 1 2 I \omega ^2


where v is the tangential velocity of the mass. But this is incorrect - if the mass is restricted from sliding outward then its KE is simply


KE = \frac 1 2 I \omega ^2


However, if the mass is allowed to slide outward radially, and it is sliding at velocity v relative to the center of the roating wheel, then its KE is indeed \frac 1 2 m v^2 + \frac 1 2 I \omega ^2 but here v is its radial velocity, not tangential velocity. Hence v \ne r \omega.

The rest of your equations I can't decipher. I suggest that either use Latex like I have, or copy in an image file of your equations.

Special characters have been replaced by common ones and the matter has been reproduced under. It may also be noted that these calculations have no concern with radially sliding piston. Rather these are for a mass with its center of gravity fixed at 100 mm radius from axis of rotation. On the other hand, value of 'r� for sliding piston will be changing continuously and so every thing (Wip, Fip, Pip, Fop, Pop, etc.) will also be varying accordingly.

Also, please be sure that formula used for total kinetic energy is authentic for the case. However if it is incorrect and the other one that you have mentioned would be used then values for Wip, Fip and Pip will decrease and ratios Fop/Fip and Pop/ Pip will increase further.

“My assumption and calculation regarding amplification of input energy by utilizing centrifugal force are as under for your review and comments.

Consider a body having one kilogram mass (m) with its center of gravity at 0.10 meter radius (r) from axis rotating at 1500 rpm (25rps). It will require work input (Wip) or energy equal to its total kinetic energy comprising kinetic energy of translation plus kinetic energy of rotation.

Wip = ½ x mass x square of linear velocity of center of gravity + ½ mass moment of
inertia (I) x rotational velocity (w)
= ½ mv^2 + ½ I w^2
= ½ m(2pi.rn)^2 + ½(½ m.r^2)(2 pi.n)^2
= 2m(pi.rn)^2 + m(pi.rn)^2
= 3m(pi.rn)^2 Nm, joules

When 'n' is number of revolutions per second then power input (Pip) becomes

Pip = 3m(pi.rn)^2 Nm/s, joules/s, watt = 3 x 1.0(pi x 0.10 x 25)^2
= 185 watt

As power equals to torque(T) x rotational velocity(w) and torque equals force(F) x radius(r),

Pip = Tipw and Tip = Fipr

So Fip = Pip/rw = 3m(Ï€rn)^2/rw = 3m(pi.rn)^2/2pi.rn = 3/2m.pi.rn
= 3/2 x 1.0 x pi x 0.10 x 25
= 11.78 N

The centrifugal force naturally created in this process, I would like to call it the output force
(Fop), at the expense of only 11.78 N will be

Fop = mrw^2 = mr(2pi.n)^2 = 4 mr(pi.n)^2
= 4 x 1.0 x 0.10(pi x 25)^2
= 2467.40 N
This huge output centrifugal force (Fop) has also acceleration of 2467.40 m/s^2 directed radially outward. If this much force may somehow be applied tangentially at 50 mm radius to produce torque at 1500 rpm then power output (Pop) comes out to be


Pop = Tw = Fop.rw = 4 mr(pi.n)^2r.2pi.n = 8 mr2pi^3n^3
= 8 x 1.0 x 0.05^2 x pi^3 x 25^3
= 19378.92 watt




If Fop is applied at 100 mm radius

Pop = 8 x 1.0 x 0.10^2 x pi^3 x 25^3
= 38757.84 watt.

Thus the ratio output force to input force becomes

Fop/Fip = 2467.40/11.78 = 104.75

And ratio of output power to input power becomes

Pop/ Pip = 19378.92/185 = 104.75 for 50 mm radius.

and = 38757.84/185 = 209.5 for 100 mm radius

Power ratio will similarly vary in direct proportion to value of radius.

It may be reminded here that I am not talking about extraction of centrifugal force energy of radially sliding piston mass for free power generation. It is quite a different mechanism to be discussed in detail later. The case of piston has been quoted just to analyze the radial velocity of piston or any radially slid-able mass under the influence of centrifugal force at certain radius.

As you agreed to most of my point raised in last posting, high values of centrifugal force, associated with huge acceleration even at smaller radii, is suggestive of high radial velocity accordingly as compared to tangential linear velocity at relevant radius. So piston can move against reasonable resistance to give output work/energy at velocity comparable with tangential velocity.

Keeping all the complex calculus aside, let us consider just one simple logic for the time being. Suppose the piston initially occupies a position with its center of gravity at zero radius
or at any minimum (just a few millimeters) radius but prevented to slide outward while the disc is rotating at speed of 1500 rpm. So, initial acceleration and velocity of piston will be zero. Then piton is suddenly released to slide radially outward. In just fraction of a second it will reach 50 or 100 mm radius with centrifugal force and acceleration as calculated above. Then velocity just after a fraction of second or after one complete second will be equal to the value of acceleration by that time which comes out to be much higher than relevant tangential velocities there. Is it correct or not?�

Mayya: Sorry, but you continue to make the same errors - first in having incorrect equations for the objects's kinetic energy (as I pointed out in my last post), second for confusing energy and power (as I pointed out in an earlier response), third for confusing the direction of the centripedal acceleration and force on the piston, and finally for not taking account the fact that the piston can't slide outward at the velocity you predict without a lot more input power to keep the disk turning - the very action of the piston sliding outward acts like a brake on the rotating disk (like when an ice skater in a spin raises her arms to slow down her spin rate). Below I will attempt to pont out the specific errors.


Wip = ½ x mass x square of linear velocity of center of gravity + ½ mass moment of
inertia (I) x rotational velocity (w)
= ½ mv^2 + ½ I w^2
= ½ m(2pi.rn)^2 + ½(½ m.r^2)(2 pi.n)^2
= 2m(pi.rn)^2 + m(pi.rn)^2
= 3m(pi.rn)^2 Nm, joules


You have again written that:


W_{ip} = \frac 1 2 m v^2 + \frac 1 2 I \omega^2


As I pointed out earlier this is incorrect. The correct value for the kinetic energy (and hence the amount of work required to get the disc to spin at this speed) is


W_{ip} = KE = \frac 1 2 I \omega^2 = \frac 1 2 (mr^2) (2 \pi n)^2 = 2m(\pi r n)^2


This is in units of Kg-m^2/s^2, or joules. It is equal to the amount of work necessary to spin the disc up from a speed of 0 rev/second to its final speed of 25 rev/s. At this point you have said nothing about how long it takes to attain this amount of energy, so we know nothing yet about input power required to attain this spin rate.


When 'n' is number of revolutions per second then power input (Pip) becomes

Pip = 3m(pi.rn)^2 Nm/s, joules/s, watt = 3 x 1.0(pi x 0.10 x 25)^2
= 185 watt


Wrong. You have simply copied the equation for kinetic energy and now called it power. But the units are wrong. Power is a measure of the change of kinetic energy per second. It takes power to get the disc up to speed, but once the disc is spinning at its final rate of 25 rev/s no additional power required to keep it spinning at that rate (as long as the piston is constrained from sliding outward).


As power equals to torque(T) x rotational velocity(w) and torque equals force(F) x radius(r),

Pip = Tipw and Tip = Fipr

So Fip = Pip/rw = 3m(Ï€rn)^2/rw = 3m(pi.rn)^2/2pi.rn = 3/2m.pi.rn
= 3/2 x 1.0 x pi x 0.10 x 25
= 11.78 N


Again, your units are wrong. You have divided kinetic energy (not power) by r \omega and ended up with something that has units of Kg m/s. This is not a unit of force.


The centrifugal force naturally created in this process, I would like to call it the output force
(Fop), at the expense of only 11.78 N will be

Fop = mrw^2 = mr(2pi.n)^2 = 4 mr(pi.n)^2
= 4 x 1.0 x 0.10(pi x 25)^2
= 2467.40 N


OK - so F_{op} is a measure of the centripedal force appled to the piston held at distance r and spinning at \omega radians/second. Remember that this force is applied inward on the piston to keep it from sliding.


This huge output centrifugal force (Fop) has also acceleration of 2467.40 m/s^2 directed radially outward.

NO! The centripedal force on the mass is directed INWARD. This is what keeps the piston in its circular path around the center of the disc.


If this much force may somehow be applied tangentially at 50 mm radius to produce torque at 1500 rpm then power output (Pop) comes out to be


Pop = Tw = Fop.rw = 4 mr(pi.n)^2r.2pi.n = 8 mr2pi^3n^3
= 8 x 1.0 x 0.05^2 x pi^3 x 25^3
= 19378.92 watt


The trick here is how do you convert a radial force into a tangential force? If it was possible to do that, then you could do things like take the vertical component of force due to gravity, convert it to a horizontal force and use it to accelarate a car, and voilà! Free energy to make your car go. Unfortunately this is impossible.


... As you agreed to most of my point raised in last posting, high values of centrifugal force, associated with huge acceleration even at smaller radii, is suggestive of high radial velocity accordingly as compared to tangential linear velocity at relevant radius. So piston can move against reasonable resistance to give output work/energy at velocity comparable with tangential velocity.

Now you're talking about the piston sliding outward. None of your previous calculations mention that. Keep in mind that as the piston does work the overall kinetic energy of the system must decrease a comparable amount - and hence the disk slows its rotation (again, like the skater raising her arms as she spins).


Keeping all the complex calculus aside, let us consider just one simple logic for the time being. Suppose the piston initially occupies a position with its center of gravity at zero radius
or at any minimum (just a few millimeters) radius but prevented to slide outward while the disc is rotating at speed of 1500 rpm. So, initial acceleration and velocity of piston will be zero. Then piton is suddenly released to slide radially outward. In just fraction of a second it will reach 50 or 100 mm radius with centrifugal force and acceleration as calculated above. Then velocity just after a fraction of second or after one complete second will be equal to the value of acceleration by that time which comes out to be much higher than relevant tangential velocities there. Is it correct or not?�

It is certainly true that the radial velocity of the piton will be quite large - as we have discussed in previous posts. But one of two things must occur: eiether (a) the disc slows as the piston slides outward, or (b) more power is applied to the disc to maintain its rotational velocity as the piston slides.

Ok. Using correct formula for total kinetic energy the amount of work required to get the disc to spin at speed of 25 rev/s

Wip = KE = 1/2 IW^2 = 2m(pi.rn)^2 kgm^2/s^2, Nm, joules
= 2x1.0(pi. X 0.10 x 25)^2 = 123.37 joules

While I have previously calculated input work 185 watt. So, as I have already mentioned, said ratios and output force/power magnification will further increase for this current value of 123.37 joules.

Quoting Ebaines, “Power is a measure of the change of kinetic energy per second. It takes power to get the disc up to speed, but once the disc is spinning at its final rate of 25 rev/s no additional power required to keep it spinning at that rate (as long as the piston is constrained from sliding outward).�

This is not understandable. How the disk can continue rotating without continuous input energy?

Also as power is the rate of work done per unit time, I think it right to say that 123.37 joules per second will be required to keep the disc rotating at 25 rev/s continuously and hence power input Pip becomes 123.37 watt (= joules per second). That is why, in last post, I have taken value of input work Wip directly equal to rate of work done or power input Pip.

If it is not true then please help me calculate the power required only for continuous rotation of said 1kg fixed mass, and not for the sliding piston.

Quoting Ebaines, “---confusing the direction of the centripedal acceleration and force on the piston, and finally for not taking account the fact that the piston can't slide outward at the velocity you predict without a lot more input power to keep the disk turning - the very action of the piston sliding outward acts like a brake on the rotating disk (like when an ice skater in a spin raises her arms to slow down her spin rate).�

It is a well known fact that reactive centrifugal force/acceleration is just equal and opposite to centripetal force/acceleration and hence directed radially outward. It is evident that piston slides outward purely under influence of centrifugal force as long as it is rotating with the disc. Also, I have experimentally pressurized hydraulic oil outside the sliding piston with pressure displayed in accordance with centrifugal force.

Of course, input work/energy/power will vary according to radial position of mass due to its inertia. Input power for rotation/torque is affected only if there is tangential resistance (parallel to axis of rotation) opposite to direction of rotation at certain radius from center of rotation. Whereas, any force acting radially through the axis of rotation, like centrifugal force does not affect the input power. Also as mass is rotating at the speed of input drive, centrifugal force is also rotating along with the mass at the same rotational speed without any resistance to rotation requiring no additional input power. Hence, braking action of sliding piston have either never been experienced practically nor understandable logically.

This is not understandable. How the disk can continue rotating without continuous input energy? ....
... If it is not true then please help me calculate the power required only for continuous rotation of said 1kg fixed mass, and not for the sliding piston.


A spinning object will keep spinning unless a resisting torque is applied, such as friction. For a well designed mechanism the amount of friction should be pretty low, so the power required to keep it spinning is low - perhaps one or two watts. This assumes that air friction is not much of an issue (the disc has a nice smooth aerodynamic shape).


Input power for rotation/torque is affected only if there is tangential resistance (parallel to axis of rotation) opposite to direction of rotation at certain radius from center of rotation. ... Hence, braking action of sliding piston have either never been experienced practically nor understandable logically.

There are two issues here, both of which cause the disc to slow down its rotation as the piston slides outward:

1. Conservation of rotational inertia. If there is no external torque applied, the inertia of the disc remains constant:


I \omega \ = \ Constant


As the piston slides out the moment of inertia I increases; consequently the rotational velocity \omega decreases by a comparable amount. In the example I gave earlier about the skater this is what causes her spin to slow as she extends her arms outward.

2. As pointed out back in post #16, the outward radial velocity of the piston causes a coriolis force that opposes the rotation of the disc. The magnitide of that force is


F = 2m \omega v_r


where v_r is the piston's outward radial velocity. This is a resistive force applied by the piston to the piston cylinder wall that opposes the disc's rotation, and hence is a resistive torque on the mechanism. You can see an example of the coriolis effect here: YouTube - The Coriolis Force (http://www.youtube.com/watch?v=_36MiCUS1ro)

I think it's interesting that you think a significant amount of power is needed to keep the disc spinning at 25 revolutions/second when the piston is stationary, but no additional power is needed to keep the disc spinning at that rate if the piston is allowed to slide outward. This is precisely backwards.

Sorry, I am poor at understanding the things properly, so, asking questions again and again to conclude the matter.

Firstly, it may be reminded here that mechanism being designed for extraction of centrifugal force energy for free power generation is quite different from radially sliding piston mass. Even if the piston motion is considered for the purpose, its radial movement may be limited to only 200 to 300 mm from center of rotation.

WELL. AS NO CONSIDERABLE ENERGY IS REQUIRED FOR CONTINUOUS SPINNING OF THE DISC AT 25 REV/S THEN IT MEANS THAT HUGE DYNAMIC CENTRIFUGAL FORCE WILL BE AVAILABLE ALMOST FOR FREE AND, THEREFORE, OUTPUT/INPUT RATIOS I AM TAKING ABOUT WILL TREMENDOUSLY INCREASE FURTHER. IS IT CORRECT OR NOT?

But the Coriolis force has brought in a new problem regarding clear picture of the case.

Perhaps there is also a phenomenon that a body can rotate about a single center at a time. If it is forced to rotate about more than one center, it adopts a new single center again. Here the piston is bound to rotate with the disc at the same rotational velocity in straight radial slot about one and the same center, which is center of the disc. So, total/net/pure centrifugal force will be directed radially outward.

On the other hand, what I could hardly understand so far, Coriolis force also sometimes called complementary centrifugal force, occurs in the case of a body while moving away from the center in a rotating frame due to frictional contact, not at the rotational speed of the frame, follows a curved path instead of a straight radial one. In this case, centripetal force = Coriolis force + centrifugal force. Due to inclination of Coriolis force slightly opposing the rotation, total/compound/resultant centrifugal force (= Coriolis force + centrifugal force) will also be directed slightly inclined to radial direction opposing/resisting the rotation to certain extent.

THEN, IS IT NOT SO THAT CORIOLIS FORCE MAY NOT BE THERE IN CASE OF PISTON WHICH IS ENTIRELY DIFFERENT FROM THE CASE OF BODY MOVING AWAY FROM THE CENTER ON A ROTATING FRAME OR THE EARTH AND PLANETS?

AS NO CONSIDERABLE ENERGY IS REQUIRED FOR CONTINUOUS SPINNING OF THE DISC AT 25 REV/S THEN IT MEANS THAT HUGE DYNAMIC CENTRIFUGAL FORCE WILL BE AVAILABLE ALMOST FOR FREE

yes...


AND, THEREFORE, OUTPUT/INPUT RATIOS I AM TAKING ABOUT WILL TREMENDOUSLY INCREASE FURTHER. IS IT CORRECT OR NOT?

I have no idea. The ratios you talked about before were ratios of POWER input and output. I have no idea how you plan to covert the centrifugal force to power, so don't know what your power output would be. I maintain that power output = 0 if the piston is not actually moving outward.


But the Coriolis force has brought in a new problem regarding clear picture of the case.... Perhaps there is also a phenomenon that a body can rotate about a single center at a time. If it is forced to rotate about more than one center, it adopts a new single center again. Here the piston is bound to rotate with the disc at the same rotational velocity in straight radial slot about one and the same center, which is center of the disc. So, total/net/pure centrifugal force will be directed radially outward.

I don't understand what you are getting at, so I can't comment.


On the other hand, what I could hardly understand so far, Coriolis force also sometimes called complementary centrifugal force, occurs in the case of a body while moving away from the center in a rotating frame due to frictional contact

No - not due to friction, but rather due to conservation of total angular momentum. Again - this is a force that will be applied by the piston as a normal force against the piston cylinder wall, in the direction opposite rotation.


... centripetal force = Coriolis force + centrifugal force. ... Due to inclination of Coriolis force slightly opposing the rotation, total/compound/resultant centrifugal force (= Coriolis force + centrifugal force) will also be directed slightly inclined to radial direction opposing/resisting the rotation to certain extent.

Sorry, I'm not following you on this - I don't understand why you are adding the coriolis force (which the piston exerts on the wall of the piston cylinder, in a direction that opposes rotation) to the centripedal force needed to keep the piston from moving outward. Remember - these two forces are orthogonal. You can only add them as vectors.


THEN, IS IT NOT SO THAT CORIOLIS FORCE MAY NOT BE THERE IN CASE OF PISTON WHICH IS ENTIRELY DIFFERENT FROM THE CASE OF BODY MOVING AWAY FROM THE CENTER ON A ROTATING FRAME OR THE EARTH AND PLANETS?

If the piston is not moving relative to the center of rotation, there is no coriolis force. Is that what you mean?