How can I determine the moment of inertia of an irregular object?

AI Thread Summary
To determine the moment of inertia of an irregular object, the discussion outlines a method involving measuring the torque required to move the armature of a motor. The proposed approach involves spinning the armature at a constant frequency and then cutting the power to observe the deceleration, using the work-kinetic energy theorem to relate work done to kinetic energy. The participant seeks clarification on integrating the kinetic energy formula, specifically how to handle the changing angular velocity during deceleration. They conclude that by measuring the initial and final angular velocities, they can isolate the moment of inertia as the unknown variable. This method provides a practical way to assess the moment of inertia for their motor project.
Xyius
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Hello everyone, (This might be a little long winded but I would very much appreciate some help on this!)

Today while in psychology class, I was thinking about a much more important subject, namely a project I want to start working on. Basically what I want to do is build a motor from the ground up. Like a real motor that would be in a machine. I want to start from complete theory and them move to actually building it once. (I am a senior undergraduate in Physics and I want to get into theoretical physics in my future.)

So my first thought would be, how would I determine how much current I need to run through each of the windings to get the motor to move? I then realized that I would need to know the torque required to move the armature, and this would be the LEAST amount of torque I would need. (I would obviously want more torque than this so the motor can actually drive things. And I would actually need a little more than this to get it moving due to friction effects.) Here is something I came up with to measure this torque.

Have an external force spin the armature to a constant frequency ω. When it reaches this point, cut the power to the armature (The force would be coming from some external motor) and see how long it takes it to stop spinning. The torque will be approximately equal to..

T≈\frac{ΔL}{Δt}=\frac{I(ω_2-ω_1)}{Δt}

So I would need to determine the rotational inertia of the armature first. It is spinning about one of its principle axis so it will not be a tensor. To determine the rotational inertia, the idea I came up with would be very similar to determining the torque. I would again drive the armature to spin at a constant frequency ω and then cut the power. But right when I cut the power, have a "negligible" string tied to one end of the armature. This string is attached to a block of known weight and will consequently do work on the block by moving it linearly. From the work-kinetic energy theorem we can find the kinetic energy.
W=ΔK

So here is the part I need help on. So for a rotating body the kinetic energy is..

K=\frac{1}{2}Iω^2
But ω would not be constant since the object is negatively accelerating (slowing down). So I would have..
dK=\frac{1}{2}I(dω)^2
Or..
∫dK=K=∫\frac{1}{2}I(dω)^2=\frac{1}{2}I∫(dω)^2

I have taken so many advanced physics and mathematics courses, I find it funny that I do not know how to integrate this. I have never worked with a squared differential before. How would I go about doing this?

Also, if anyone has any other ways about measuring the moment of inertia of something, or any comments about my way, feel free to let me know!

Thanks
 
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Xyius said:
dK=\frac{1}{2}I(dω)^2
That does not work.
dK=\frac{1}{2}Id(\omega^2) = I\omega d\omega
This can be integrated, but where is the point in taking the derivative if you integrate afterwards?
 
mfb said:
That does not work.
dK=\frac{1}{2}Id(\omega^2) = I\omega d\omega
This can be integrated, but where is the point in taking the derivative if you integrate afterwards?

Thank you for your reply.

So that would mean that I can simply take the final and initial ω values. I will already have K from the work done by the block, and the values of ω, so the only unknown in the equation would be the moment of inertia.

Thanks! Hope this is a correct analysis. :]
 
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