Ask Me Help Desk - Centrifugal force

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Special characters have been replaced by common ones and the matter has been reproduced under.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 'râ€ 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.

â€œMy 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 = Â½ x mass x square of linear velocity of center of gravity + Â½ mass moment of
inertia (I) x rotational velocity (w)
= Â½ mv^2 + Â½ I w^2
= Â½ m(2pi.rn)^2 + Â½(Â½ 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?â€

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.

Quote:

Originally Posted by mayya

Wip = Â½ x mass x square of linear velocity of center of gravity + Â½ mass moment of
inertia (I) x rotational velocity (w)
= Â½ mv^2 + Â½ I w^2
= Â½ m(2pi.rn)^2 + Â½(Â½ 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.

Quote:

Originally Posted by mayya

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).

Quote:

Originally Posted by mayya

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.

Quote:

Originally Posted by mayya

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.

Quote:

Originally Posted by mayya

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.

Quote:

Originally Posted by mayya

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.

Quote:

Originally Posted by mayya

... 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).

Quote:

Originally Posted by mayya

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?â€

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, â€œPower 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).â€

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, â€œ---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).â€

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.

Quote:

Originally Posted by mayya

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).

Quote:

Originally Posted by mayya

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

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?

Quote:

Originally Posted by mayya

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...

Quote:

Originally Posted by mayya

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.

Quote:

Originally Posted by mayya

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.

Quote:

Originally Posted by mayya

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.

Quote:

Originally Posted by mayya

... 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.

Quote:

Originally Posted by mayya

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?