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24 April 2024 04:45
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Question |
| Asked by: |
Luis Gonzalez |
| Subject: |
A Droll question |
| Question: |
At the risk of demonstrating how dull I can be here is a basic and somewhat naïve question.
Start with a very wide bodied boat in space (no gravity).
On each side of the boat we have oars with weights of considerable mass attached to their outer ends.
We start with the weighted end of the oars pointing straight to the front of the wide boat.
Scenario “A” (pure centrifugal-force):
1- In Scenario “A” both oars are driven mechanically to orbit at a fast-steady speed, around their respective pivots in a synchronized manner so that both point simultaneously back, and point simultaneously forward.
(The boat is wide enough to accommodate this motion without collision of the two masses.)
2- Both oars continue to turn at the same steady rate on and on…
Scenario “B” - Starts fom the same psition as Scenario “A” (combines angular-acceleration and centrifugal-force):
1- Ostart, both oars are operated mechanically to accelerate in a backward arc (similar to rowing).
2- The acceleration continues to increase all the way until the masses are pointing straight back.
3- When the oars are pointing back, we remove the force and allow the momentum to carry both masses along their respective circular path.
4- When the two masses arrive where they are pointing toward each other (across the boat), we intervene and slowdown the masses during the last quarter of their circular path.
5- This way both masses are forced to come to a complete stop when they are pointing straight forward again.
6- Immediately upon coming to a full stop, a new exact cycle starts, over and over…
Now for the naïve questions:
How will the results of these 2 experiments differ?
Will scenario “A” perform the exact same motions as scenario “B” (simply moving forward and backwards), albeit with different lengths of displacement?
Or, does scenario “B” behave differently?
In a nutshell, is there a difference in the effects of centrifuge vs. the effects of angular acceleration?
Regards,
LuisG |
| Date: |
17 January 2013
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Answers (Ordered by Date)
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| Answer: |
Blaze - 17/01/2013 23:24:32
| | | Hi Luis. Just a quick thought about your question. Scenarios A and B will act the same as far as the boat travel distance goes (the boat will just move forward and backward but not get anywhere). The speed at which that happens will be faster for scenario B. The acceleration in scenario B only makes the movement quicker and the deceleration makes it slower.
In your scenarios, the angular acceleration in scenario B increases the centifugal force which is why the movements happens more quickly than in scenario A (I am assuming that the speed at the end of acceleration is higher than for scenario A). The first part of the backwards movement (that is the 3rd 1/4 turn which is at high speed) will also be quick because of the high centrifugal force. The motion during the last 1/4 turn will have a slower average movement but should take the same as the the first 1/4 turn, assuming the acceleration and deceleration rates are the same.
just my thoughts,
Blaze
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| Answer: |
Luis Gonzalez - 19/01/2013 18:34:40
| | | Thanks Blaze,
We appear to see things nearly the same in this initial mental-experiment, which gets us closer to the real question.
What if we adjusted the angular velocities of “Scenario A” (which uses purely centrifugal-force) so that its cycle coincides with the cycle of “Scenario B” (which uses angular accelerations and decelerations)?
In other words, we synchronize both experiments so they get round to the start point at the exact same time for all subsequent cycles.
How will the results of these 2 experiment scenarios differ, with is added caveat?
Will scenario “A” perform the exact same motions as scenario “B”?
Or, does scenario “B” behave differently?
The bottom line is to determine whether the effects of accelerations in of angular velocity are in any way different from steady rotation’s centrifugal force?
Regards,
LuisG
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| Answer: |
Harry K. - 20/01/2013 13:22:41
| | | Hi Luis,
I agree with Blaze that there will be a forward and backward movement in both scenarios.
Let's define the positions of the oars with the numbers on a clock:
The front start position of the oars is at 6 o'clock and the rear position at 12 o'clock.
The horizontal positons are at 3 and 9 o'clock.
Defination of direction of rotation:
Both oars will rotate counter clockwise, for instance the left oar rotates clockwise and the right oar rotates counter clockwise.
1. Resulting movement in scenario "A":
There will act a forward movement in the area between 3 and 9 o'clock. The minimum (zero) will be at 3 and 9 o'clock and the maximum at 6 o'clock.
The forward movement is caused by centrifugal force. The necessary counter force in form of centripedal force is caused by the counter direction of rotation at both pivots of oars, which are fixed with the boat.
In the area between 9 and 3 o'clock there will act a rearward movement, also with the minimum (zero) at 3 and 9 o'clock but the maximum at 12 o'clock. The cause of this movement has the same explanation as stated before.
The resulting movement will be an oscillating forward - backward movement in a straight line. The strength of the oscillation varies in a sinusoidal manner between zero and maximum.
1. Resulting movement in scenario "B":
In this case I see it a bit different to Blaze.
During acceleration of the oars between 6 and 12 o'clock there will be no net movement at all, because the acceleration torque will cause a counter torque and thus at each position of the oars there will be a counter force (centripedal) with exactly equal size.
If acceleration torque will be removed at 12 o'clock, both oars behave in the same manner as stated in scenario "A", i.e. there will act a rearward movement in the areas between 12 and 3 o'clock (left oar) and respectively 12 and 9 o'clock (right oar).
During the deceleration phase in the areas between 3 and 6 o'clock (left oar) and respectively 6 and 9 o'clock (right oar) the stored energy in form of angular momentum must be eliminated by a counter torque which moves the boat straight back to the front to the initial start position.
The oscillating movement also acts in a straight line. The strength of the forces depending on accelerating and decelerating torques.
Regards,
Harry
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| Answer: |
Luis Gonzalez - 20/01/2013 18:49:26
| | | Thanks Harry,
Your perspective on the initial 180 degrees of acceleration in scenario “B” is very interesting.
Do you mean that zero motion occurs during the acceleration stage, or that any gain near the front will be lost near the rear (within the initial 180 degrees)?
Regards,
LuisG
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| Answer: |
Luis Gonzalez - 20/01/2013 18:55:32
| | | Harry,
To make it more clear:
Do you mean that zero motion occurs during the acceleration stage (within the initial 180 degrees)?
Or do you mean that any gain near the front will be lost near the rear, within the initial 180 degrees?
Regards,
LuisG
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| Answer: |
Harry K. - 20/01/2013 20:20:27
| | | Luis, I mean the zero motion occurs during acceleration stage because the counter torques to accelerate the oars are acting in counter direction. Think on a drilling machine where you have to hold the drilling machine against the counter torque if you drill a hole in the wall. Both counter torques of the both oars will canceling each other out during acceleration stage, but the counter centripedal forces are still acting with the result of zero movement of the system (boat).
Regards,
Harry
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| Answer: |
Luis Gonzalez - 21/01/2013 17:50:54
| | | Harry,
We have frequently parsed and spared at length regarding exact meaning of statements.
Not wanting to let the statement “Zero Motion”, lead us into an unwelcome expedition, I will only make a simple statement.
When the oar masses are Half-Way between front and rear (during rearward acceleration of scenario “B”), the oars are aligned straight and all centripetal-centrifugal forces are in Balance.
At that very same spot rearward acceleration of the masses is still occurring.
The NET effect of over-all front to rear acceleration may be zero; however, the boat experiences some motions over the length of this stage.
Hi Blaze,
Do you have any thoughts regarding the experiment with the caveat I presented in response to yours?
Regards,
LuisG
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