# Problems with calclulating power

• losbellos
In summary, power is not just angular velocity multiplied by radius. You need to find the rotational energy of the system per its time period.
losbellos
Hej Guys,

I have seen on wikipedia, that power = Force x velocity, and this is fine but when they explain the same for circular moving force then they use power = Force (x arm) x Angular velocity (rads/sec). This cannot be, the velocity actually cannot be represented by only angular velocity, it needs the radius as well. For example distance = velocity x time = angular velocity * radius * time. So in this way the equation power = Force x velocity (for linear moving force) and for circular moving force the power will be = arm x force x angular velocity x radius .. in circular considering that the force is always perpendicular to the arm. If not then power would be > force x sin (Angle of force-arm) x arm x angular velocity x radius.

example considering the force and velocity as a point on a circle or a line...
circle circumference = 62.83m
powerL = force * vel = 100N * 14.13675 = 1413.675 watts
powerC = force * angular velocity * radius = 100N * 1,4137 rads * 10.0m = 1413.7 watts
so out of calculation error PowerC = PowerL

angular velocity != velocity so even in rotational systems it cannot be used alone without the radius. This way the power = torque x angular velocity is not true. To prove this like above one maybe divide the circle to line segments where the a value = force x arm x velocity x distance can be calculated for each segments, considering the torque, and these segments added will be much different from the torque x angular speed (rads/sec) x distance segments together. Naturally there will be some difference based on the number of segments. The more segments the closer it will be. Again torque x velocity != torque x angular velocity (rads/sec). The result is only same if the other equation is power = force x arm x angular velocity x radius = torque x angular velocity x radius.Am I right?
Thx.

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First, stop using Wikipedia especially if you are using it as your PRIMARY source.

Secondly, "Power" is defined as the amount of energy expanded or supplied per unit time. This means that you need to find the time rate of energy available or produced. That is the general principle.

Thus, find the rotational energy of the system per its time period.

Zz.

losbellos
POWER
Linear system P = F v , Power = force times velocity
rotational system P = T ω , Power = torque times angular velocity(rad/sec)

( if you want to analyze, simplistically ; knowing T = F r, then P = F r ω = F v )

Also, you stated
power = force x arm x angular velocity x radius = torque x angular velocity x radius.
Check your units to see if they come out correctly as something resembling power.

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Well they mean mechanical power. I usually deal with Newtons. Can you direct me to a good guider to torque and other other calculations regarding force moving in circular path, using arms? I mean real validated source.

But by the way 1 joule is also 1 Nm(lets say second as well) and 1 watt second, so they all unit of different types of power aren't they? The time in my example is always 1 second to make it simple... I edited my post to make it more understandable.

Zapper Z;4228976 said:
First, stop using Wikipedia especially if you are using it as your PRIMARY source.

Secondly, "Power" is defined as the amount of energy expanded or supplied per unit time. This means that you need to find the time rate of energy available or produced. That is the general principle.

Thus, find the rotational energy of the system per its time period.

Zz.

256bits said:
losbellos
POWER
Linear system P = F v , Power = force times velocity
rotational system P = T ω , Power = torque times angular velocity(rad/sec)

( if you want to analyze, simplistically ; knowing T = F r, then P = F r ω = F v )

Its not true, because you consider the r only once. You need it once for the torque and then one more time for the velocity because r x ω = velocity. Just see my example. I cannot state that the velocity I am doing linear is not the same the as the velocity travels on a circle. Both velocity travels the same distance by the time... circular or linear it doesn't matter. So defining velocity by angular speed is invalid, unless the r multiplies.. Its like you try to consider the circular move without actually the arm. We say torque because it have an arm and the force. That makes torque but this torque rotates with a speed. For the speed I need the r again, because velocity = ω x r . Without it, its like you consider only the force and you leave the arm out of the calculation. Thats not valid.

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Torque and energy might have to same units but you have to keep the concepts seperate, and the units are designated so to avoid confusion.

torque is the unit Newton-meter , N-m
energy is the unit joule

A torque of 1 N-m applied rotationally through a full revolution will require 2∏ joules of energy.

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losbellos said:
Its not true, because you consider the r only once. You need it once for the torque and then one more time for the velocity because r x ω = velocity. Just see my example. I cannot state that the velocity I am doing linear is not the same the as the velocity travels on a circle. Both velocity travels the same distance by the time... circular or linear it doesn't matter. So defining velocity by angular speed is invalid, unless the r multiplies.. Its like you try to consider the circular move without actually the arm. We say torque because it have an arm and the force. That makes torque but this torque rotates with a speed. For the speed I need the r again, because velocity = ω x r . Without it, its like you consider only the force and you leave the arm out of the calculation. Thats not valid.

You can argue about it all you want until the cows come home, but you are incorrect.
In any case in your calculation you have used r only once.
powerL = force * vel = 100N * 14.13675 = 1413.675 watts
powerC = force * angular velocity * radius = 100N * 1,4137 rads * 10.0m = 1413.7 watts
so you seem to be saying one thing but nonetheless, doing the calculations correctly.

256bits said:
Torque and energy might have to same units but you have to keep the concepts seperate, and the units are designated so to avoid confusion.

torque is the unit Newton-meter , N-m
energy is the unit joule

A torque of 1 N-m applied rotaionally through a full revolution will require 2∏ joules of energy.

I think not, its 1 joule if the rotation is done by one second. Even though Nm doesn't have time value assigning it that the torque did "work" for one sec (for one revolution) then it will be one joule. Otherwise it is not known. It could be 10 sec or 5 sec or anything therefore I cannot say 2Pi joule. Also why would i use radians instead of velocity?

For simplicity let's consider the arm and the force as a point. This force point rotates around the center of the circle on the cirle. It won't matter that it rotates or moving on a linear path,velocity is what defines the power. So why I should use angular speed instead of velocity.
Its like I want to degrade the output power because its moving circular. Seeing the prove above, that divide a circle into points or lines it would result very different values...

256bits said:
so you seem to be saying one thing but nonetheless, doing the calculations correctly.
I explain there the force moving on a circle it doesn't matter that it is considered as a torque or else. This is what I am tryin to say. I will explain the problem in a graphic then maybe it will be clearer why I would say that Power including torque would be T x ω x r

So here is a diagram explains the problem.
There is an arm in between A and B points.
This arm is attached to the smaller circle which rotates with 3.21 m/s. This rotation is due the force being applied at the end of the arm at B point.

Now let's start with the forces; The torque at point A would be arm length x force x sin (angle in between arm and force) . THe torque is 6.84m x 100N x sin(147.995
) = 362.5 NM

Basically at that point 362NM "rotates" with 3.21 m/s or with 1.04 rad. So the power would be P=T x ω by the physics but hej, this force - because torque is basically a force with a support arm - why would I use uu instead of velocity. And why ω would be different in this sense from the actual velocity? What is the reason for this? So why not use instead P = ω x r x T Can you explain this to me ?

Also this torque force could be just a liner move of force with velocity, must be same... (lets not consider the angle of the force or consider it always parallel to the arm)

So actually here it would be valid to say that power in other way would be: Power = Force x Distance / Time . This way the 362 N moves 1 meter in one second. So the power is 362Watts.
Well since I don't point exactly out where the force actually acting so I consider the arm as a One ARM lifter (dont know it in english) and consider the weight I have to lift and the rotation point on the other side at A point and the Force at B point. So just simply multiply with the length of the arm.

This arm is attached to the smaller circle which rotates with 3.21 m/s
A point on the circumference of the smaller circle rotates at a tangential velocity of 3.21 m/s

torque at point A would be arm length x force x sin (angle in between arm and force)
are you using the correct arm length from the axis of rotation to point B. There are no dimensions so impossible to tell.

Basically at that point 362NM "rotates" with 3.21 m/s or with 1.04 rad
i think you mean 1.04 rad/sec, the angular velocity.

So the power would be P=T x ω by the physics
correct

why would I use uu instead of velocity. And why ω would be different in this sense from the actual velocity.
One reason is that it is easier to do calculations using ω. All points on a body rotating about an axis have the same angular velocity ω. The tangential velocity of points vary from a value 0 at the axis to a value v = r ω at the circumference.
For your problem, Point A and B have the same angular velocity ω ( rad/s ), but the tangentiial velocity v ( m/s ) of A is less than that of B.

So why not use instead P = ω x r x T
Check the units and you will see that ωrT does not come out as watts, or power.

losbellos said:
because torque is basically a force with a support arm - why would I use ω instead of velocity. And why ω would be different in this sense from the actual velocity? What is the reason for this? So why not use instead P = ω x r x T Can you explain this to me?

Perhaps you should think about it coming from the linear definition.
We define $P=Fv$, but in a rotational setting, $v=r\omega$ so we find that $P=Fr\omega$. But now we notice that $Fr=\tau$ so we must have $P=\tau\omega$ and not $P=\tau\omega r$ as you said.

The reason for using torque and angular velocity is, as 256 says, it is often easier to use if you know that a system is rotating. It is significantly easier when you have a constant angular velocity. In this case, using the linear velocity is complicated since it is continuously changing.

hej, well, őőőőő I don't what to say.

again as you state, tangential speed is equal to ω x r
this the velocity. - ω x r
what is the power. Power is Force x velocity. true BUT HEJ THERE IS AN ARM THERE! SO THE FORCE WILL BE...
So power is P = Force x arm length < (BECAUSE YOU CANT TAKE THIS AWAY FROM IT) x ω x r .

If it would be force only then you would disregard the arm... But we can't do that... Because it is an arm and it makes the force much stronger on the smaller circle. Nobody doubts it I hope! So this is why i don't understand why we say p = T x ω ... You understand my problem with this? Force acting there call it torque or not it forces the cylinder to rotate... Force basically travels there and keeps the cylinder rotating with certain velocity.

I don't know where else to talk about it so please don't stop the discussion about it.

losbellos said:
hej, well, őőőőő I don't what to say.

again as you state, tangential speed is equal to ω x r
this the velocity. - ω x r
what is the power. Power is Force x velocity. true BUT HEJ THERE IS AN ARM THERE! SO THE FORCE WILL BE...
So power is P = Force x arm length < (BECAUSE YOU CANT TAKE THIS AWAY FROM IT) x ω x r .

If it would be force only then you would disregard the arm... But we can't do that... Because it is an arm and it makes the force much stronger on the smaller circle. Nobody doubts it I hope! So this is why i don't understand why we say p = T x ω ... You understand my problem with this? Force acting there call it torque or not it forces the cylinder to rotate... Force basically travels there and keeps the cylinder rotating with certain velocity.

You are wrong. There isn't much discussion about it. The force is the same whether or not there is an arm. The torque will be different depending on the length of the arm. If I apply a force of 1N to a lever, it is 1N no matter where I push. Depending on how far from the pivot I push, it will have a different torque, but the force is 1N. Torque is not force. We are not calling a force a torque for fun. We are calling torque a torque. Spend a little time thinking/reading about torque and then come back with questions.

As a side note, I feel like your tone is little too confident since nobody agrees with you and you don't have any sources. I think most people are more likely to discuss this if you ask questions without assuming that you are correct.

got no clue what to say, when I think that you guys acknowledge it that If I have an arm and use my muscles on the end and the force will be less on the other end of the arm because its is less than 1 meter.. then I am sorry, must be some sort of planetary influence taking over the physics forum..

I am sorry I find an other community.

take care!

losbellos said:
got no clue what to say, when I think that you guys acknowledge it that If I have an arm and use my muscles on the end and the force will be less on the other end of the arm because its is less than 1 meter.. then I am sorry, must be some sort of planetary influence taking over the physics forum..

I am sorry I find an other community.

take care!

its unfortunate that you won't accept the facts :(

you are not going to find any different answer on any other reputable physics site
as we all live in the same universe

Dave

Okay, let me try to make this clearer (please excuse typos, I've had a few beers)

Consider a lever. There is a pivot at $x=0$ and the arm of the lever is 2m long. Assume that the lever extends from zero to $x=2$. Now, let's say we have a mechanism of some sort that can exert a force of 1N. If we apply that at the mid point (always at right angles to the arm in order to keep this as simple as possible), the torque will be $\tau=1N\cdot1m=1Nm$. If you measure the force at the end of the lever, the TORQUE will be the same and therefore $F=\frac{\tau}{r}=\frac12N$. So the measured force is less, but the torque, which is NOT the same as the force, is the same.

Now we want to consider the power. To keep things simple, let's assume that applying the force of 1N allows this lever to move at a constant velocity $v=1m/s$ (maybe it is in some viscous liquid or whatever, the specifics aren't important). So we apply this force and we find that $P=Fv=1N\cdot 1m/s=1W$. Since the torque and angular velocity are constant and defined as $\tau=Fr,\ \omega=\frac{v}{r}$, this is the same as saying $P=\tau\omega=1Nm\cdot 1s^{-1}$. If somebody else attempts to extract energy at the far end of this lever, by my calculations the power will be $P=\frac12\cdot 2m/s=1W$ since the force is one half that of the midpoint and the velocity (linear) is twice that of the midpoint. This clearly shows that the work extracted from the end point in any given time interval is equal to the work put in.

By your calculations, the power is $P=\tau r\omega$ which gives the same power as input but at the end of the lever $P=1Nm\cdot 2\cdot 1s^{-1}=2W$. So, by your calculation, every second this simple lever allows one to freely extract twice as much energy as one puts in. This would solve a lot of problems if it were true, but it is not.

The best way to see this (IMO) is this:

If you accept (or look up a proof) that rotational K.E. is given by $0.5Iω^{2}$, and that $T = I dw/dt$.

If you think about the change in rotational K.E. :

$d(0.5Iω^{2})/dt = Iω dw/dt = Tω$

## What are some common problems with calculating power?

There are several common problems that can arise when calculating power, including inaccurate data, incorrect assumptions about effect size, violation of statistical assumptions, and small sample sizes.

## Why is it important to accurately calculate power?

Accurately calculating power is important because it allows researchers to determine the appropriate sample size for their study, ensuring that they have enough participants to detect a meaningful effect if it exists.

## What is the formula for calculating power?

The formula for calculating power varies depending on the statistical test being used, but generally involves the sample size, effect size, and significance level.

## What are some strategies for improving power in a study?

Some strategies for improving power in a study include increasing the sample size, using more sensitive measures, reducing measurement error, and selecting appropriate statistical tests.

## What should be done if power is found to be low?

If power is found to be low, researchers may need to revise their study design, such as increasing the sample size or using a more sensitive measure. It is also important to interpret the results with caution and consider the potential for type II errors.

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