Rotational Power = torque (τ)*(ω) angular velocity? - Problem with units

Oct 2018
4
0
I set up an experiment with a pendulum (uniform beam) swinging on a hinge and the hinge attached to a small PMDC generator.

The generator had no load attached and its output voltage is proportional to its speed of rotation.

I swung the pendulum from 90 degrees and using a data logger I recorded the voltage of the generator (at 100 samples / second).

I approximated the output voltage (and speed) to a decaying sine wave.

To calculate mechanical rotating power using torque and angular velocity, can I simply multiply the instantaneous equations for both, or do I need to somehow do a dot product?

The equations are shown below including the units. I've clearly gone wrong some where since I end up with watts*radians^2.

(The units on the graph are not accurate because I've compressed/stretched various values so they nicely fit onto the same scale)




Interesting, I ran another test with a resistive load attached to the generator. Max power occurs at max voltage which occurs at max speed of rotation.

This maximum output electrical power actually occurs when there is apparently zero mechanical input power.

How can this be? Obviously the pendulum has the most angular momentum at the base of the swing. Is it that I'm not taking into account the rotor inertia (which in this case is so small it seems to barely effect the dynamics of the swing).
 

topsquark

Forum Staff
Apr 2008
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630
On the dance floor, baby!
I'm kind of on the run this afternoon but to answer one of your questions at least.

The unit "radian" can be defined as the angle through which a point on a circle travels divided by the radius. \(\displaystyle \theta = \frac{s}{r}\). This means that rad = m / m. So the unit radian is actually unitless. It's occasionally useful to let us know that there is some kind of radial motion so it's kept to illustrate it.

So, \(\displaystyle P = \omega \tau = W~rad^2 = W\)

-Dan

Addendum: The power equation is indeed a dot product. In this case the two vectors are either pointing in the same direction or pointed opposite of each other. So the dot product gives either a positive or negative result with the same scalar value.
 
Aug 2010
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I feel compelled to point out that \(\displaystyle 3\times 4.08= 12.24\), not 12.3! Even rounded to one decimal place it would be 12.2.
 
Oct 2017
576
295
Glasgow
Yes, it should be the dot product.
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
Remember that radians are dimensionless, so your power calculation which ends up with units of watts x radian^2 is the same as simply watts.
 
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Oct 2018
4
0
Thank you for the response benit.

How can I apply dot product of two functions?

To simplify, say w(t) = sin(3t) and τ(t) = cos(3t).

I’m only familiar with using dot product for sets such as a.b where a=[a1,a2,a3] and b=[b1,b2,b3] and a.b = a1b1+a2b2+a3b3.
 
Oct 2017
576
295
Glasgow
The vector that describes torque or angular momentum is one that points perpendicular to the plane that the torque is trying to instigate rotation or the plane of rotation (for the case of angular velocity). So, if you have a pendulum that doesn't swing all the way round (drawn on a diagram on a piece of paper), your angular rotation vector is going to switch periodically from pointing "out of the paper" and "into the paper" depending on which way it is swinging. The torque bestowed on the pendulum from gravity will also have this periodic behaviour.

So how will this look on paper? It will mean that your sine wave formula for torque and angular velocity will have a direction, say \(\displaystyle \hat{z}\). If you do a dot product in the same way you are used to, you will end up with for your power vector where the components of the dot product in the \(\displaystyle \hat{x}\) and \(\displaystyle \hat{y}\) directions won't contribute to the power. So, even though you've taken into account the direction, your dot product will end up giving you a straightforward multiplication.
 
Oct 2018
4
0
The vector that describes torque or angular momentum is one that points perpendicular to the plane that the torque is trying to instigate rotation or the plane of rotation (for the case of angular velocity). So, if you have a pendulum that doesn't swing all the way round (drawn on a diagram on a piece of paper), your angular rotation vector is going to switch periodically from pointing "out of the paper" and "into the paper" depending on which way it is swinging. The torque bestowed on the pendulum from gravity will also have this periodic behaviour.

So how will this look on paper? It will mean that your sine wave formula for torque and angular velocity will have a direction, say \(\displaystyle \hat{z}\). If you do a dot product in the same way you are used to, you will end up with for your power vector where the components of the dot product in the \(\displaystyle \hat{x}\) and \(\displaystyle \hat{y}\) directions won't contribute to the power. So, even though you've taken into account the direction, your dot product will end up giving you a straightforward multiplication.
I followed most of that and will try to visualize the directions changing via drawing it out. But essentially because of the relative directions, the dot product of torque and angular velocity becomes a straight forward multiplication meaning my initial equations in the original post are valid?

Do \(\displaystyle \hat{x}\), \(\displaystyle \hat{y}\) and \(\displaystyle \hat{z}\) represent the "unit vectors" in the respective axes? So multiplying a directionless function by any one of those will keep the magnitude the same but give it a direction? It's been a number of years since I've covered this kind of maths and I didn't think I would ever need it again.

Your help is invaluable and is really saving me from having a complete anxious breakdown.
 
Oct 2018
4
0
The vector that describes torque or angular momentum is one that points perpendicular to the plane that the torque is trying to instigate rotation or the plane of rotation (for the case of angular velocity). So, if you have a pendulum that doesn't swing all the way round (drawn on a diagram on a piece of paper), your angular rotation vector is going to switch periodically from pointing "out of the paper" and "into the paper" depending on which way it is swinging. The torque bestowed on the pendulum from gravity will also have this periodic behaviour.

So how will this look on paper? It will mean that your sine wave formula for torque and angular velocity will have a direction, say \(\displaystyle \hat{z}\). If you do a dot product in the same way you are used to, you will end up with for your power vector where the components of the dot product in the \(\displaystyle \hat{x}\) and \(\displaystyle \hat{y}\) directions won't contribute to the power. So, even though you've taken into account the direction, your dot product will end up giving you a straightforward multiplication.
Just to summarize, due to the directions of the vectors that are being dot producted (is that a term?), the initial equations I posted are valid?

It's been 5 years since I've covered this kind of math but, \(\displaystyle \hat{x}\), \(\displaystyle \hat{y}\) and \(\displaystyle \hat{z}\) are the unit vectors for the x, y and z axes? As in multiplying a directionless quantity but one of them will not change it's magnitude but will give it a direction?

Your help is invaluable and is saving me from a lot of anxiety. Thank you