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

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Oct 15th 2018, 08:30 AM #1 Junior Member   Join Date: Oct 2018 Posts: 2 Rotational Power = torque (τ)*(ω) angular velocity? - Problem with units 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).
 Oct 16th 2018, 08:57 AM #2 Senior Member   Join Date: Aug 2010 Posts: 383 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 16th 2018, 10:18 AM #3 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 268 Yes, it should be the dot product.
 Oct 16th 2018, 10:32 AM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,324 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. aberrant likes this.
 Oct 16th 2018, 10:39 AM #5 Junior Member   Join Date: Oct 2018 Posts: 2 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 17th 2018, 03:07 AM #6 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 268 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.

 Tags τω, angular, inertia, pendulum, power, problem, rotational, torque, units, velocity

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