- #1
1000_Words
- 10
- 0
Good day.
This is a reformulation of another post which I made with the physics guys at the Forum ;)
OK. Here's the scoop:
Through some helpful input from another source, I have found that the following handy equation
Fc (N) = 4 m pi2 n2 r / 602
can be used to determine the restraining force (N) required in the following example:We have a square tube 1" in diameter X 2' long spinning at 3,000 RPM (n); filled with H20.
Metric: The spinning radius is .3 meters (r); the watermass weight (for the .3 meter radial segment) is .1944 kg (m).
On the basis of the preceding equation used to figure centripetal force vs. RPM, we get a figure of about 5,750 N as the result.
However, this basic arithmetic assumes that all of the mass in question is concentrated at the distal end of r; giving nothing to the fact that the mass is uniformally distributed along r from the center of rotation outward.
To solve for this, I have been informed that one merely needs to adjust for the distributed mass within the tube by dividing the result (5,750 N) by 2 to obtain the actual distal restraining force (2,875 N) required at the outer end of the spun water column at n RPM.Assuming that all of this is correct (yes?), I now can get to my question.
How would we use our original equation,
Fc (N) = 4 m pi2 n2 r / 602
to solve for a homogeneous mass of water centrifugally slung against the outer wall of a rotating cylindrical housing (think spinning barrel; not tube) at n RPM? Could this setting for the math conceivably involve evaluating for a triangular section?
As a quick working example along these lines, let's take the case of a flat, closed cylinder 1" high by 2' in diameter, filled with water; and spinning at 3,000 RPM.
In this case, could someone take a one square inch (outer wall footprint) "pie slice" triangular segment of the homogeneous fluidmass in question and solve for that one element using the cited equation? Would this accurately represent the local restraining force required of the outer wall? If so, how would we correct for the distributed mass in this situation?Sorry if this has been a bit prolix -- Thanks for any help.
This is a reformulation of another post which I made with the physics guys at the Forum ;)
OK. Here's the scoop:
Through some helpful input from another source, I have found that the following handy equation
Fc (N) = 4 m pi2 n2 r / 602
can be used to determine the restraining force (N) required in the following example:We have a square tube 1" in diameter X 2' long spinning at 3,000 RPM (n); filled with H20.
Metric: The spinning radius is .3 meters (r); the watermass weight (for the .3 meter radial segment) is .1944 kg (m).
On the basis of the preceding equation used to figure centripetal force vs. RPM, we get a figure of about 5,750 N as the result.
However, this basic arithmetic assumes that all of the mass in question is concentrated at the distal end of r; giving nothing to the fact that the mass is uniformally distributed along r from the center of rotation outward.
To solve for this, I have been informed that one merely needs to adjust for the distributed mass within the tube by dividing the result (5,750 N) by 2 to obtain the actual distal restraining force (2,875 N) required at the outer end of the spun water column at n RPM.Assuming that all of this is correct (yes?), I now can get to my question.
How would we use our original equation,
Fc (N) = 4 m pi2 n2 r / 602
to solve for a homogeneous mass of water centrifugally slung against the outer wall of a rotating cylindrical housing (think spinning barrel; not tube) at n RPM? Could this setting for the math conceivably involve evaluating for a triangular section?
As a quick working example along these lines, let's take the case of a flat, closed cylinder 1" high by 2' in diameter, filled with water; and spinning at 3,000 RPM.
In this case, could someone take a one square inch (outer wall footprint) "pie slice" triangular segment of the homogeneous fluidmass in question and solve for that one element using the cited equation? Would this accurately represent the local restraining force required of the outer wall? If so, how would we correct for the distributed mass in this situation?Sorry if this has been a bit prolix -- Thanks for any help.
Last edited: