# How to calculate how much power required to turn a flywheel?

How to calculate how much power required to turn a flywheel? Suppose I have a flywheel of 24 inch diameter, width 5 inches,50kg. How do I calculate the required power (HP) to turn this or any other with different values?

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The information given is not enough to find the power.
First what do you mean by "to turn"? To accelerate from rest to a given speed? Or just to keep a given speed constant?

The information given is not enough to find the power.
First what do you mean by "to turn"? To accelerate from rest to a given speed? Or just to keep a given speed constant?
To accelerate from rest to 2000rpm

A.T.
To accelerate from rest to 2000rpm
In what time?

In what time?
In 10-12 seconds. :-)

Then find the kinetic energy of the rotating flywheel at the required rotation rate and divide by the time.
You can approximate the flywheel with a cylinder when looking up the moment of inertia.

Then find the kinetic energy of the rotating flywheel at the required rotation rate and divide by the time.
You can approximate the flywheel with a cylinder when looking up the moment of inertia.
Ring's KE(joules): 101932.01
Time is 10 second
101932/10=10193.2
Is it the power required to rotate that flywheel? Please elaborate, I really don't know how to calculate? I have attached flywheel's energy calculations.

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Ring's KE(joules): 101932.01
Time is 10 second
101932/10=10193.2
Is it the power required to rotate that flywheel? Please elaborate, I really don't know how to calculate? I have attached flywheel's energy calculations.

i got some values, please check
values @ 2000rpm, 50 kg, 0.3048m radius
I = 0.5 m r^2 = 0.5*50*0.3048*0.3048 = 2.3225;
E = 0.5 I ω ^2 = 0.5*2.3225*209.33*209.33 = 50884.8705 Joules; <==

torque = mass moment of inertia * rotational acceleration rate = I*a= 2.3225*(20.933)=48.6168 Nm;
instantaneous power consumption = torque * max rotation rate = 48.6168*209.33= 10176.9741 watt; <==

is this the correct method to calculate power required to rotate flywheel?

You don't need the torque. Once you have the kinetic energy, the average power will be that energy divided by the time.

You don't need the torque. Once you have the kinetic energy, the average power will be that energy divided by the time.
So in this case, power required to rotate 50 kg, 0.3048m radius flywheel @ 2000 rpm for 10 sec is
101932/10=10193.2 watts. Is this correct?

Power = KE/time;
What does the time means? Is it the total time for which we r rotating the flywheel? What will be the value if I want to rotate that flywheel for hours? Please

Last edited:
Maybe you should ignore power at first and use torque (T):
(the torque is deemed to be constant)
calculate the acceleration rate (α) in rad/sec/sec
(ive attached a sheet to use)
then calculate the torque T ( in N-m ) from:
T = I * α
( I = flywheel moment of inertia (kg - m^2) , α = rotational acceleration rate (rad/sec/sec) )
With the torque value (T) you can then calculate the power (Watts) at any rad/ sec ( ω ) from:
Power = T * ω

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jbriggs444
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So in this case, power required to rotate 50 kg, 0.3048m radius flywheel @ 2000 rpm for 10 sec is
101932/10=10193.2 watts. Is this correct?
No. This is not correct.

The figure of 10 kilowatts is the startup power requirement. While the wheel is being spun up from rest to its desired rotation rate, you must provide "power". The faster you want to reach the steady-state RPM, the more power you must supply. Remember that in the physics vernacular, "power" is the rate at which energy is delivered. The amount of energy you must supply to spin the wheel up from rest is fixed. The rate at which you must supply that energy depends on how much time you are willing to spend getting the wheel up to speed. One Watt of power means you are delivering energy at a rate of one Joule per second.

The steady state power requirement to maintain the flywheel at 2000 rpm is zero. In principle, it take no power at all to keep a flywheel spinning forever. In practice, the power requirement will depend on how good your bearings are and how much air resistance the wheel is subject to.

In addition it is overly optimistic to quote the computed result to 6 significant figures when the inputs (mass and time interval) were known to only one significant figure.

A.T.