Pendulum into RPM???????????


by mikej4823
Tags: pendulum
mikej4823
mikej4823 is offline
#1
Nov23-12, 07:02 AM
P: 4
Hi

I'm trying to throw together a concept design for a swing set which can convert kinetic energy into mechanical energy. In short I would like a swing set, using one way roller bearings to drive an output shaft but am unsure where to start.

If I have a mass (m), suspended at a length (L) with a angle of swing (θ), does anyone know how I would get to a point to calculate the RPM of a the shaft.

I am thinking I need to calculate the Kinetic Energy, Time taken for one complete swing (θmax), Torque and RPM???

Any help would be much appeciated. (I havn't included numerical units as they don't yet exisit, I'm trying to get the step by step process in place first)

Thanks

Mike
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mfb
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#2
Nov23-12, 09:10 AM
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Can you add a sketch of your setup?
Does it help to determine the (linear) speed of the pendulum at its lowest point? You can get this via energy conservation.
mikej4823
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#3
Nov23-12, 09:36 AM
P: 4
swing.pdf

Here is a rough sketch. Hope it helps a little.

As I say I am trying to work towards an output shaft RPM.

Please note all dimensions etc are ball park.

Thanks

Mike

mfb
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#4
Nov23-12, 10:05 AM
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P: 10,864

Pendulum into RPM???????????


Ah.
If damping from your shaft is not too strong (=> the amplitude difference within one swing is small), you can use the common equations for a pendulum (wikipedia, ...). If damping is significant, the equations get a bit more complicated, and they can depend on the type of damping.
mikej4823
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#5
Nov23-12, 10:17 AM
P: 4
I'm assuming minimal damping. I will just use conservative numbers later on.

Are you referring to T=2∏√L/g ???
mfb
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#6
Nov23-12, 11:16 AM
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P: 10,864
That is one of those formulas.

Angular velocity follows a sine curve:
$$\omega(t)=\omega_{max}\sin(\sqrt{\frac{g}{l}}t) $$
Energy conservation: ##mgl(1-\cos(30)=\frac{1}{2}m \omega_{max}^2 l^2##
$$\omega_{max}=\sqrt{\frac{g}{l}(2-\sqrt{3})}$$

If you don't care about the instantaneous angular velocity: Each timestep of T, you get 60 rotation of you shaft, which corresponds to one revolution per 6T.
mikej4823
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#7
Nov23-12, 11:19 AM
P: 4
Thats brilliant thank you. I will spend some time throwing some numbers into it.

Much appreciated from a Mechanical engineer ha


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