Why is rigid body rotational energy not exactly applicable to fluids?


by Compressible
Tags: applicable, body, energy, fluids, rigid, rotational
Compressible
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#1
Mar14-13, 04:16 AM
P: 19
I was thinking about the rotational kinetic energy of fluids the other day and I realized that I have a huge gap in my knowledge of physics. Why doesn't rigid body rotational kinetic energy (KE = 1/2*I*ω^2) not apply to fluids or deformable bodies (it should at least be proportional to that equation)? Is it only because the moment of inertia is not constant or is there another underlying physics involved?
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mfb
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#2
Mar14-13, 04:50 AM
Mentor
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In fluids, different parts can have different ω, and you can have radial flow.
Compressible
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#3
Mar14-13, 12:08 PM
P: 19
Yeah but radial and tangential motion are perpendicular to each other, so they should be able to be assessed separately (similar to translational and rotational motion).

mikeph
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#4
Mar14-13, 12:53 PM
P: 1,199

Why is rigid body rotational energy not exactly applicable to fluids?


You are free to assign radial and tangential velocities, relative to some arbitrary point, to a fluid element, but the radial elements won't obey v = r*w because the fluid isn't a rigid rotor. I don't see any benefit in this. The total kinetic energy is the same, you're just calculating it in a more complicated and less generalised way.
Compressible
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#5
Mar14-13, 02:15 PM
P: 19
I'm not sure what you're trying to say. Why would the radial elements contribute to the rotational energy? They should be completely independent of tangential (v*r) motion.
olivermsun
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#6
Mar14-13, 02:28 PM
P: 498
The radial and tangential motions aren't independent. If some bits of fluid have a radial velocity, then that means their [itex]r[/itex] is changing and hence the total [itex]I[/itex] is changing. By the same token, a fluid parcel that is moving outward at constant (linear) velocity has decreasing [itex]\omega[/itex].
Compressible
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#7
Mar14-13, 03:16 PM
P: 19
Ah, I got you. So if we were to assume that a column of liquid that had no radial velocity and that all its parts were moving at the same tangential velocity (for example, a column of liquid jet exiting an infinitely long pipe where the flow has been fully established), then 0.5*I*ω^2 would give a good approximation to the bulk rotational kinetic energy. Am I correct in this assumption or am I missing something else?
mikeph
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#8
Mar15-13, 07:17 AM
P: 1,199
Quote Quote by Compressible View Post
Why would the radial elements contribute to the rotational energy?
I didn't say this.

Quote Quote by Compressible View Post
Ah, I got you. So if we were to assume that a column of liquid that had no radial velocity and that all its parts were moving at the same tangential velocity (for example, a column of liquid jet exiting an infinitely long pipe where the flow has been fully established), then 0.5*I*ω^2 would give a good approximation to the bulk rotational kinetic energy. Am I correct in this assumption or am I missing something else?
If tangential velocity is uniform then ω = v/r = ω(r), but your formula for rotational kinetic energy assumes ω is uniform.

The point of my first post was exactly this- you're applying a model to a scenario which does not satisfy the assumptions of the model. Rotational kinetic energy as 0.5*I*ω^2 is defined for a rigid body, because the formula implicitly assumes ω is a constant. Liquids are not rigid bodies because the relative distances between two elements in the liquid can change.

You can find a form of rotational kinetic energy from tangential velocity, but you're probably going to have to integrate over concentric ring elements to find it.
Compressible
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#9
Mar15-13, 08:35 PM
P: 19
The model isn't so far from actual physics though (in some scenarios). A forced vortex generally rotates at a constant angular velocity (assuming no turbulence).
olivermsun
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#10
Mar15-13, 10:23 PM
P: 498
If the entire body of water is acting exactly like a solid body, then you can use the solid body formula.
physics.cie
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#11
Mar16-13, 12:52 AM
P: 17
KE(total)=KE(translation)+KE(rotation) now here rotational KE is independent of translation KE, we can use KE(rot)=L_2/2w and as angular momentum(L) is constant then we can estimate KE(rot)
physics.cie
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#12
Mar16-13, 12:53 AM
P: 17
i agree


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