# Variable inertia flywheel

by miljacev
Tags: flywheel, inertia, variable
 Share this thread:
 P: 4 Hello, My question is: how much energy is needed for changing inertia of flywheel? example: on rotary shaft we have linear actuator which is moving some object (mass). by that movement, radius of flywheel is changing. ...if there is energy input to flywheel, mass is moving away form shaft in order to maintain constant flywheel speed. If flywheel is sending energy to generator we need to move that mass towards the shaft in order to maintain speed. m=10kg r=100mm speed= 3000rpm at that moment, there is 10kN of centrifugal force on that mass. ...do I need to apply more than 10kN of force on that mass in order to change radius? Thanks
Sci Advisor
HW Helper
P: 6,679
 Quote by miljacev Hello, My question is: how much energy is needed for changing inertia of flywheel? example: on rotary shaft we have linear actuator which is moving some object (mass). by that movement, radius of flywheel is changing. ...if there is energy input to flywheel, mass is moving away form shaft in order to maintain constant flywheel speed. If flywheel is sending energy to generator we need to move that mass towards the shaft in order to maintain speed. m=10kg r=100mm speed= 3000rpm at that moment, there is 10kN of centrifugal force on that mass. ...do I need to apply more than 10kN of force on that mass in order to change radius? Thanks
Welcome to PF!

Assuming the flywheel is of uniform thickness and density, the moment of inertia is:

$I = \frac{1}{2}mr^2$

The kinetic energy of the spinning flywheel is:

$KE = \frac{1}{2}I\omega^2$ where ω is the angular speed in radians / sec.

If you are applying torque to the flywheel changing it rotational speed, the work done is the change in kinetic energy:

$W = \frac{1}{2}I({\omega_f^2 - \omega_i^2})$

If you are changing the moment of inertia of the flywheel to maintain constant speed, then you will have to apply a radial force to the mass on the wheel. You would have to tell us the mass distribution on the wheel and how you change that. But the change in energy is just:

$\Delta KE = \frac{1}{2} \Delta(I\omega^2)$

AM
P: 4
 Quote by Andrew Mason Welcome to PF! Assuming the flywheel is of uniform thickness and density, the moment of inertia is: $I = \frac{1}{2}mr^2$ The kinetic energy of the spinning flywheel is: $KE = \frac{1}{2}I\omega^2$ where ω is the angular speed in radians / sec. If you are applying torque to the flywheel changing it rotational speed, the work done is the change in kinetic energy: $W = \frac{1}{2}I({\omega_f^2 - \omega_i^2})$ If you are changing the moment of inertia of the flywheel to maintain constant speed, then you will have to apply a radial force to the mass on the wheel. You would have to tell us the mass distribution on the wheel and how you change that. But the change in energy is just: $\Delta KE = \frac{1}{2} \Delta(I\omega^2)$ AM
...thanks for welcome and replay.
To make things more clear, I've made a simplified sketch in paint, and I think that it will be a good idea to explain little bit more that concept in my head.

That flywheel should have function of a battery. energy will be pumped into flywheel in cycles, but generator is hooked up to flywheel and it requires constant rotation speed.

My idea is to maintain constant speed of flywheel with variable radius. More power is pumped into flywheel than it is taken by generator, radius is increasing. Les power is pumped into flywheel than it is taken by generator, radius is decreasing.

My question is: how much energy I need for changing radius?

tnx

http://cdn.imghack.se/images/5af9e9c...b9a234c9ce.jpg

P: 134
Variable inertia flywheel

 Quote by miljacev ...thanks for welcome and replay. To make things more clear, I've made a simplified sketch in paint, and I think that it will be a good idea to explain little bit more that concept in my head. That flywheel should have function of a battery. energy will be pumped into flywheel in cycles, but generator is hooked up to flywheel and it requires constant rotation speed. My idea is to maintain constant speed of flywheel with variable radius. More power is pumped into flywheel than it is taken by generator, radius is increasing. Les power is pumped into flywheel than it is taken by generator, radius is decreasing. My question is: how much energy I need for changing radius? tnx http://cdn.imghack.se/images/5af9e9c...b9a234c9ce.jpg
You can calculate work from

$\mathtt{\ W\ =\ \int_{r_2}^{r_1}\ F_{(r)}dr}\ =\mathtt{ \int_{r_2}^{r_1} m\ \omega^{2}rdr}$
Sci Advisor
HW Helper
P: 6,679
 Quote by miljacev ...thanks for welcome and replay. To make things more clear, I've made a simplified sketch in paint, and I think that it will be a good idea to explain little bit more that concept in my head. That flywheel should have function of a battery. energy will be pumped into flywheel in cycles, but generator is hooked up to flywheel and it requires constant rotation speed. My idea is to maintain constant speed of flywheel with variable radius. More power is pumped into flywheel than it is taken by generator, radius is increasing. Les power is pumped into flywheel than it is taken by generator, radius is decreasing. My question is: how much energy I need for changing radius?
Just use: $ΔKE = \frac{1}{2}\Delta (I\omega^2) = \frac{1}{2}\omega^2 ΔI$

As the flywheel gives up energy it will slow down so to maintain rotational speed, mass must be pulled in toward the centre. As you add energy to the wheel, you have to less mass out to maintain speed.

If the mass distribution was always uniform throughout the disc:

$ΔKE = \frac{1}{2}\omega^2 ΔI = \frac{m\omega^2}{4} (R_f^2-R_i^2)$

If the disk is not uniform, which this sounds like, you have to provide the details of the mass distribution in the wheel in order to calculate I and ΔI.

AM
 P: 4 let me just comment: if I have X joules stored in flywheel with max radius, I need to spend all my energy of flywheel in order to move that mass to minimum radius!? My idea is to maintain constant speed when flywheel is receiving and sending energy by changing inertia (radius). ... if I understood correctly from above formulas, that is not possible without taking a lot more energy from flywheel? If flywheel loses ΔKE of its energy, his ω will also drop down. If I want to increase ω to the previous value (when energy was bigger, I would like to do it only by changing radius to smaller one. tnx
 P: 58 This could be complex. For a start, as stated by AM, if the distribution of mass of the flywheel is not equal then the calculations change. How much energy is expended pulling the mass in or letting it out (relative to the KE of the system) is a dependent on other (non-disclosed) engineering decisions as to how you do this.
P: 4
 Quote by mic* This could be complex. For a start, as stated by AM, if the distribution of mass of the flywheel is not equal then the calculations change. How much energy is expended pulling the mass in or letting it out (relative to the KE of the system) is a dependent on other (non-disclosed) engineering decisions as to how you do this.
...I don't know why I did not simply made this drawing immediately, but now it is here: http://cdn.imghack.se/images/fb100c7...39b8b775df.jpg

By moving that pink linear bearing (up and down) I would like to regulate flywheel inertia.
Is it possible to calculate how much energy I will need in order to move that bearing along the shaft, and how much force I will need in order to keep it in place?
Sci Advisor
HW Helper
P: 6,679
 Quote by miljacev ...I don't know why I did not simply made this drawing immediately, but now it is here: http://cdn.imghack.se/images/fb100c7...39b8b775df.jpg By moving that pink linear bearing (up and down) I would like to regulate flywheel inertia. Is it possible to calculate how much energy I will need in order to move that bearing along the shaft, and how much force I will need in order to keep it in place?
I assume that the wheel will be balanced so that there is an equal mass diametrically opposite that undergoes the same change in radius.

You add energy to the wheel by applying a torque through an angle to the wheel. This causes the mass in the wheel to increase its tangential speed. If you do not change the distance of that mass from the centre of rotation, the rotational speed has to increase. However, if you then increase its moment of inertia by increasing that radius of rotation, you decrease the rotational speed (back to its original rotational speed) but maintain that increased tangential speed.

When you take energy from the wheel, the reverse applies. The wheel applies a torque through an angle which causes the mass to lose tangential speed. If you do not change the distance of that mass from the centre of rotation, the rotational speed has to decrease. If you then decrease the radius, you increase the rotational speed back to the original and maintain that decreased tangential speed.

If you make these changes in radius at the same as the wheel is slowing or speeding up, you don't use any energy in changing the radius. The force you will need for each m is the centripetal force: mω2r where ω=2πf where f = number of rotations per unit of time.

AM
 Sci Advisor PF Gold P: 2,253 The description from the op remind me of a centrifugal governor, used to regulate the speed of steam engines. But perhaps I am missing something?

 Related Discussions Classical Physics 1 Advanced Physics Homework 6 Engineering, Comp Sci, & Technology Homework 3 Introductory Physics Homework 2 Mechanical Engineering 3