Escape velocity of a spoked wheel.

[yuiop]

Consider a spoked wheel of radius r.
The instantaneous tangential velocity of a point on the rim is v.
A test particle in the form of a small bead that can freely slide along a spoke when released, is placed as close to the axis as possible so that for the sake of argument its radial displacent is aproximately zero. If the bead is released what would the terminal velocity of the bead be when it it hit the rim of the wheel, in terms of r and v?

Please ignore any slowdown of the wheel due to consevation of angular momentum, as the test particle is considered to have negligable mass relative to the mass of the large wheel.

 

[cesiumfrog]

Have you yet attempted to integrate the centrifugal force, and change frame?

 

[yuiop]

Have you yet attempted to integrate the centrifugal force, and change frame?

This is my attempt to integrate the centrifugal acceleration (g) expressed as:

g = r \omega ^2

I have used angular velocity (omega) rather than tangential velocity because omega is independent of radius.

v_{e} = \int_0^r \, {r \omega^2} \, dr = \frac{r^2 \omega^2}{2}

Substituting the tangential velocity expression v = r \omega back in gives:

v_{e}= \frac{v^2}{2}

However this answer is not entirely satisfactory to me because of something for reasons to do with the EP principle in relativity and secondly I have to question whether the sliding bead really leaves the end of the spoke at a velocity that is the square of the rim velocity? That is before I add in the component of velocity tangential to the spoke that the bead will acquire. Any thoughts?

 

[cesiumfrog]

v_{e} = [..] (r \omega^2) \, dr

Since when does acceleration multiplied by distance give velocity? Dimension check fail.

Hint: force x distance gives?

 

[yuiop]

Since when does acceleration multiplied by distance give velocity? Dimension check fail.

Hint: force x distance gives?

OK, fair comment. I'll try again.

KE = \int_0^r \, {mr \omega^2} \, dr = \frac{mr^2 \omega^2}{2}

Substituting the tangential velocity expression v = r \omega back in gives:

KE= \frac{mv^2}{2}

OK, it would appear that the kinetic energy acquired by the sliding test particle as it reaches the end of the spoke is the same as the kinetic energy of a similar test particle at rest on the rim of the wheel. Does this mean that the potential energy of a particle on the rim has the same value and this represents the work required to "raise" the particle back up to the axis of the wheel again?

Also, does this mean the final outward velocity of the sliding test particle in the frame of the rotating wheel is the same as the tangential rim velocity of a similar test particle in the non rotating frame outside the wheel?

Basically, I am trying to figure out the analogues for potential energy and "escape velocity" in a gravitational field are, in the context of a rotating disc. This is relevant to a discussion about the Equivalence Principle in another forum.

 

[cesiumfrog]

The work done on the particle in the rotating frame tells you its final velocity (which clearly is in the radial direction) in the rotating frame. This leaves one of the simplest exercises in trigonometry to calculate the final velocity in the inertial frame, and hence the multiple by which this bead (released from the axis) would be flung faster (presuming its path is not blocked at the rim) than it would if it were instead just released directly from the rim.

OK, it would appear that the kinetic energy acquired by the sliding test particle as it reaches the end of the spoke is the same as the kinetic energy of a similar test particle at rest on the rim of the wheel. Does this mean that the potential energy of a particle on the rim has the same value and this represents the work required to "raise" the particle back up to the axis of the wheel again?

Huh? What form of potential? You do realize energy is frame dependent?

Basically, I am trying to figure out the analogues for potential energy and "escape velocity" in a gravitational field are, in the context of a rotating disc. This is relevant to a discussion about the Equivalence Principle in another forum.

Link please? (Also: is that forum moderated? science-specific?)

 

[yuiop]

The work done on the particle in the rotating frame tells you its final velocity (which clearly is in the radial direction) in the rotating frame. This leaves one of the simplest exercises in trigonometry to calculate the final velocity in the inertial frame, and hence the multiple by which this bead (released from the axis) would be flung faster (presuming its path is not blocked at the rim) than it would if it were instead just released directly from the rim.

The combined radial and tangential velocity is \sqrt{(2v^2)} so "the multiple by which this bead (released from the axis) would be flung faster (presuming its path is not blocked at the rim) than it would if it were instead just released directly from the rim" is simply \sqrt{2} but that is not really what I was trying to find out. I am more interested in the purely radial velocity as that is all that would be measured by an observer at rest with the rotating wheel frame, while the bead remains on the spoke.

Huh? What form of potential? You do realize energy is frame dependent?

I mean the equivalent of gravitational potential bu as applied to a rotating disc in flat space. The equivalence principle of General Relativity implies that there is an equivalence between a rotating frame and a gravitational frame and it is that relationship that I am trying to explore.

Most of the time I spend in physicsforums is in the Special & General relativity section, so yes, I am vaguely aware that some quantities are frame dependent.

Link please? (Also: is that forum moderated? science-specific?)

https://www.physicsforums.com/showthread.php?t=374346&page=2 in particular post #23.

"That forum" is the Special & General relativity section of physicsforums, which I am fairly sure is moderated and science-specific, although my posts might not always qualify, but I plead ignorance.