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Problems with calclulating power 
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#19
Jan1413, 06:10 PM

P: 446

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. 


#20
Jan1413, 06:58 PM

P: 18

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!!! 


#21
Jan1413, 07:08 PM

Sci Advisor
PF Gold
P: 2,645

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


#22
Jan1413, 10:09 PM

P: 446

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 [itex]x=0[/itex] and the arm of the lever is 2m long. Assume that the lever extends from zero to [itex]x=2[/itex]. Now, lets 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 [itex]\tau=1N\cdot1m=1Nm[/itex]. If you measure the force at the end of the lever, the TORQUE will be the same and therefore [itex]F=\frac{\tau}{r}=\frac12N[/itex]. 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, lets assume that applying the force of 1N allows this lever to move at a constant velocity [itex]v=1m/s[/itex] (maybe it is in some viscous liquid or whatever, the specifics aren't important). So we apply this force and we find that [itex]P=Fv=1N\cdot 1m/s=1W[/itex]. Since the torque and angular velocity are constant and defined as [itex]\tau=Fr,\ \omega=\frac{v}{r}[/itex], this is the same as saying [itex]P=\tau\omega=1Nm\cdot 1s^{1}[/itex]. If somebody else attempts to extract energy at the far end of this lever, by my calculations the power will be [itex]P=\frac12\cdot 2m/s=1W[/itex] 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 [itex]P=\tau r\omega[/itex] which gives the same power as input but at the end of the lever [itex]P=1Nm\cdot 2\cdot 1s^{1}=2W[/itex]. 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. 


#23
Apr413, 05:11 AM

P: 72

The best way to see this (IMO) is this:
If you accept (or look up a proof) that rotational K.E. is given by [itex]0.5Iω^{2}[/itex], and that [itex]T = I dw/dt[/itex]. If you think about the change in rotational K.E. : [itex]d(0.5Iω^{2})/dt = Iω dw/dt = Tω [/itex] 


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