Say I have a pendulum 3m long with 2 x 1kg weights on it. The first weight is attached at 2m and the second weight is attached at 3m.
How do I calculate the velocity at 6 o'clock when started from 3 o'clock?
Pendulum calculation
Moderator: scott
re: Pendulum calculation
I will take a guess at it.
The center of mass of the two one kilogram masses is at 2.5 m.
A drop from 3 o’clock to 6 o’clock is 2.5 meters.
And d = ½ v²/a; so the center of mass will be traveling 7.00 m/sec.
The center of mass of the two one kilogram masses is at 2.5 m.
A drop from 3 o’clock to 6 o’clock is 2.5 meters.
And d = ½ v²/a; so the center of mass will be traveling 7.00 m/sec.
re: Pendulum calculation
Ok, the reason I ask is that i can't reconcile the math on the following:
You have a pendulum with a length of 3m. A 2kg weight attached to a linear bearing is attached at 2m. A second 2kg weight is attached at 3m also on a linear bearing.
The linear bearings allow both weights to traverse up and down the rod.
The 2 weights are then connected via pulleys so that as one moves the other moves in the opposite direction. See the attached image. The purpose of this is so that a difference in CF between the 2 weights will cause them to move during the swing but will not change the overall center of mass of the pedulum. If they move 50cm either way the COM will still remain at 2.5m
So, I thought, if we can get the weights to move during the swing without changing the COM we have gained energy as the pendulum will continue to swing as if nothing happened.
My gut instinct told me that CF would be stronger the further away from the pivot you go so I thought the bottom weight would have more CF than the top weight which would cause the bottom weight to move down and the top weight move up.
But then I ran some numbers:
First get the velocity of the whole unit. The average COM between the 2 weights is at 2.5m, so I used the r=2.5 and m=4. From that velocity calculates to 9.903. this can be verified at this website http://www.calctool.org/CALC/phys/newtonian/pendulum
Then we calculate the CF of each weight based on v=9.903
The top weight is at 2m, so r=2 m=2
CF = m x v^2/r
CF = 117 newtons = 10kg
The bottom weight is at 3m so r=3 m=2
CF = 59 newtons = 6.7kg
Therefore we have excess CF of 3.2kg - You beauty! But the maths is saying he opposite of what I originally thought. Its saying the top weight is exhibiting more CF than the bottom
But, then I build a test unit and the complete opposite happens. The bottom weight is exhibiting more CF than the top weight so the bottom weight moves down and the top weight moves up.
Where is my maths wrong?
You have a pendulum with a length of 3m. A 2kg weight attached to a linear bearing is attached at 2m. A second 2kg weight is attached at 3m also on a linear bearing.
The linear bearings allow both weights to traverse up and down the rod.
The 2 weights are then connected via pulleys so that as one moves the other moves in the opposite direction. See the attached image. The purpose of this is so that a difference in CF between the 2 weights will cause them to move during the swing but will not change the overall center of mass of the pedulum. If they move 50cm either way the COM will still remain at 2.5m
So, I thought, if we can get the weights to move during the swing without changing the COM we have gained energy as the pendulum will continue to swing as if nothing happened.
My gut instinct told me that CF would be stronger the further away from the pivot you go so I thought the bottom weight would have more CF than the top weight which would cause the bottom weight to move down and the top weight move up.
But then I ran some numbers:
First get the velocity of the whole unit. The average COM between the 2 weights is at 2.5m, so I used the r=2.5 and m=4. From that velocity calculates to 9.903. this can be verified at this website http://www.calctool.org/CALC/phys/newtonian/pendulum
Then we calculate the CF of each weight based on v=9.903
The top weight is at 2m, so r=2 m=2
CF = m x v^2/r
CF = 117 newtons = 10kg
The bottom weight is at 3m so r=3 m=2
CF = 59 newtons = 6.7kg
Therefore we have excess CF of 3.2kg - You beauty! But the maths is saying he opposite of what I originally thought. Its saying the top weight is exhibiting more CF than the bottom
But, then I build a test unit and the complete opposite happens. The bottom weight is exhibiting more CF than the top weight so the bottom weight moves down and the top weight moves up.
Where is my maths wrong?
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re: Pendulum calculation
What you haven't accounted for is an understanding of a body's Moment of Inertia - this changes even when the CoM remains at the same distance along the pendulum rod as per your description.
When the weights don't move [given a frictionless scenario] the pendulum will fall & also rise to the same Potential Energy [of Position @ 9 o'cl].
But when you let them move according to Cf's then something else enters the equation - first the outer weight will override the inner & the weights will move further apart & not closer together as you summized.
Next, as the weights move apart by Cf's the pendulums Moment of Inertia changes i.e. more inertia is experienced - this results in a slower swing & also the pendulum [in a frictionless environment] won't reach 9 o'cl i.e. restore its Pe.
This loss in Pe [Energy] is proportional to the Cf's produced.
N.B. Force x Distance = Work Done [Joules] which is the same units as Pe lost.
In essence Cf's generate a force but when you try to use that force to do real work it results in an equivalent loss in Energy of Position i.e. zero sum game ala conservative, IMO.
When the weights don't move [given a frictionless scenario] the pendulum will fall & also rise to the same Potential Energy [of Position @ 9 o'cl].
But when you let them move according to Cf's then something else enters the equation - first the outer weight will override the inner & the weights will move further apart & not closer together as you summized.
Next, as the weights move apart by Cf's the pendulums Moment of Inertia changes i.e. more inertia is experienced - this results in a slower swing & also the pendulum [in a frictionless environment] won't reach 9 o'cl i.e. restore its Pe.
This loss in Pe [Energy] is proportional to the Cf's produced.
N.B. Force x Distance = Work Done [Joules] which is the same units as Pe lost.
In essence Cf's generate a force but when you try to use that force to do real work it results in an equivalent loss in Energy of Position i.e. zero sum game ala conservative, IMO.
re: Pendulum calculation
Thanks Fletcher, what you've described is exactly what occurred. The weights moved apart and the pendulum slowed down.
re: Pendulum calculation
I put in 2.5 m at your site and they verified the 7.00237 m/sec. Where did you get 9.903. I guess I missed something.
re: Pendulum calculation
You plugged in a 5 meter drop (9.905 m/sec). Why did you do that?
re: Pendulum calculation
Pequaide is of course right - if you use 2.5 m CoM radius.
Where I think you went wrong perhaps is in using the erroneous 9.903 m/s velocity instead of the 7.0 m/s.
As pequaide said previously you can calculate the "Ideal" velocity by rearranging the standard formula's.
Force = Mass x Acceleration [g] ... & ... Work Done = Force x Distance
Therefore d = 1/2V^2/a .. where a = 'g'
This gives 7.0 m/s velocity at 6 o'cl when dropped from 3 o'cl at 2.5 CoM radius [better IMO to learn & understand the longhand way than use a calculator].
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The problem arises in your calc's of Cf's because you have used the CoM velocity a 2.5 m radius - you actually have to use the velocity for each weight at its own radius which can be calculated longhand as above & then the Cf results will conform to your intuition, IINM.
N.B. to clarify - increasing the Moment of Inertia by moving weights apart [due to Cf's] results in increased resistance to change of motion - therefore the pendulum velocity is slowed resulting in a loss of Pe [in an ideal situation] - this loss of Pe Joules equals the Work capability of the Cf's generated i.e. zero sum, though some would argue differently.
Conversely starting with weights apart & moving them together along the pendulum shaft [whilst maintaining CoM position] reduces Moment of Inertia resulting in an increase in pendulum velocity & in an ideal world would see an increase in height to above 9 o'cl i.e. a gain in Pe Joules - however Cf's work outwards & not inwards when two weights are on the same radial as per your example so some other form of energy is required to make this shift together happen - the result is that the energy required [from whatever source] equals the Pe gain in Joules.
Where I think you went wrong perhaps is in using the erroneous 9.903 m/s velocity instead of the 7.0 m/s.
As pequaide said previously you can calculate the "Ideal" velocity by rearranging the standard formula's.
Force = Mass x Acceleration [g] ... & ... Work Done = Force x Distance
Therefore d = 1/2V^2/a .. where a = 'g'
This gives 7.0 m/s velocity at 6 o'cl when dropped from 3 o'cl at 2.5 CoM radius [better IMO to learn & understand the longhand way than use a calculator].
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The problem arises in your calc's of Cf's because you have used the CoM velocity a 2.5 m radius - you actually have to use the velocity for each weight at its own radius which can be calculated longhand as above & then the Cf results will conform to your intuition, IINM.
N.B. to clarify - increasing the Moment of Inertia by moving weights apart [due to Cf's] results in increased resistance to change of motion - therefore the pendulum velocity is slowed resulting in a loss of Pe [in an ideal situation] - this loss of Pe Joules equals the Work capability of the Cf's generated i.e. zero sum, though some would argue differently.
Conversely starting with weights apart & moving them together along the pendulum shaft [whilst maintaining CoM position] reduces Moment of Inertia resulting in an increase in pendulum velocity & in an ideal world would see an increase in height to above 9 o'cl i.e. a gain in Pe Joules - however Cf's work outwards & not inwards when two weights are on the same radial as per your example so some other form of energy is required to make this shift together happen - the result is that the energy required [from whatever source] equals the Pe gain in Joules.
re: Pendulum calculation
You could have put in 180° to get your 9.903 m/sec, but that would be a drop from 12 o'clock.
re: Pendulum calculation
Why does the MOI change if the weights move equally apart? I would have thought the weight moving inwards would reduce the inertial properties just as much as the weight moving outwards increases it.
re: Pendulum calculation
Pequade, you are correct, I had run calcs for a 12 o'clock drop and a 3 o'clock drop and transposed the wrong figure from my spreadsheet.
The figures end up being v= 7.002
Top weight CF = 5kg
Bottom weight CF = 3.33kg
Which of course is still wrong as my test device shows the opposite. The bottom weight displays more CF than the bottom weight so they move apart rather than towards.
The figures end up being v= 7.002
Top weight CF = 5kg
Bottom weight CF = 3.33kg
Which of course is still wrong as my test device shows the opposite. The bottom weight displays more CF than the bottom weight so they move apart rather than towards.
Re: re: Pendulum calculation
Using this method I get the following:Fletcher wrote:
The problem arises in your calc's of Cf's because you have used the CoM velocity a 2.5 m radius - you actually have to use the velocity for each weight at its own radius which can be calculated longhand as above & then the Cf results will conform to your intuition, IINM.
Top weight v=6.26 CF = 2 * 6.26^2 / 2 = 4kg
Bottom weight v=7.67 CF = 2 * 7.67^2 / 3 = 4kg
No difference in CF? My experiment clearly shows the weights moving apart. Could it be caused by the weight of the rod?
re: Pendulum calculation
If we drop from 90° or 3 o’clock then the distance dropped will always be equal to the length of the pendulum. So a 10 meter pendulum will drop 10 meters from 3 o’clock (or 90°). And a 2.5 meter pendulum would drop 2.5 meters from 90°. So if we plug these pendulum lengths and drops into the program we find that a 10 meter pendulum from 3 o’clock has a velocity of 14.0047 m/sec velocity at the down swing point and a 2.5 meter pendulum from 3 o’clock has 7.00237 m/sec. Isn’t this interesting; something moving only twice as fast will rise 4 times as height. This is twice the momentum but four times the energy.
A 1 kilogram mass would have a velocity of 6.26 m/sec if dropped 2 meters. And another 1 kilogram mass would have a velocity of 7.67 m/sec if dropped 3 meters. This is a total momentum of about 14.00 m/sec at 13.93; or you could just drop them both 2.5 meters for a total momentum (mv) of 14.0047.
What if we transferred the momentum of both of the one kilogram masses to only one of the 1 kilogram masses? That one kilogram mass would then have all the motion and would be traveling 14.0047 m/sec. At 14.0047 m/sec it would rise 10 meters. One kilogram at 10 meters would have 98.0667 joules of energy. One kilogram at 3 meters is 29.43 joules of energy, and one kilogram at 2 meters is 19.62 joules of energy. Or two kilogram at 2.5 meters is (24.5166 + 24.5166) 49.03 joules.
Find a way of consolidating the motion of two kilograms into one kilogram and you have an energy source. 98.0667 joules is greater than (19.62 + 29.43) 49.05 joules.
So instead of using your pulley pendulum; use a ‘cylinder and spheres’ or a wheel trebuchet.
A 1 kilogram mass would have a velocity of 6.26 m/sec if dropped 2 meters. And another 1 kilogram mass would have a velocity of 7.67 m/sec if dropped 3 meters. This is a total momentum of about 14.00 m/sec at 13.93; or you could just drop them both 2.5 meters for a total momentum (mv) of 14.0047.
What if we transferred the momentum of both of the one kilogram masses to only one of the 1 kilogram masses? That one kilogram mass would then have all the motion and would be traveling 14.0047 m/sec. At 14.0047 m/sec it would rise 10 meters. One kilogram at 10 meters would have 98.0667 joules of energy. One kilogram at 3 meters is 29.43 joules of energy, and one kilogram at 2 meters is 19.62 joules of energy. Or two kilogram at 2.5 meters is (24.5166 + 24.5166) 49.03 joules.
Find a way of consolidating the motion of two kilograms into one kilogram and you have an energy source. 98.0667 joules is greater than (19.62 + 29.43) 49.05 joules.
So instead of using your pulley pendulum; use a ‘cylinder and spheres’ or a wheel trebuchet.
Re: re: Pendulum calculation
The reason is [& I'm sure your capable of researching it] is that Inertia is a function of mass - when, as in this example, a mass is forced to move in an arc that mass is changing direction [this is an acceleration] - to have the lowest MoI you would use a point mass like "." - in reality mass takes up volume so we will represent it by "O" for explanation purposes - if we now take some atoms from that disk we can simplify to say an "X" shape - now imagine an equal distribution of atoms along the arms of the "X" instead of an "O".Tarsier79 wrote:Why does the MOI change if the weights move equally apart? I would have thought the weight moving inwards would reduce the inertial properties just as much as the weight moving outwards increases it.
Ok .. now that "X" represents the pendulum bob & it is fixed in orientation to the pendulum rod i.e. not free pivoted.
When the pendulum swings down in an arc the "X" is forced to change spacial orientation i.e. turn CW as you view it [although it doesn't change orientation to the shaft].
That means the atoms on the arms of the "X" have to move in space as well as loose height.
Now compare what happens to our hypothetical case where the bob is represented by a small "x" as compared to the same mass represented by a large "X" - in the same falling distance the atoms on the tips of the large "X" have to move further than those on the small "x" - since atoms have inertia & they have to move comparatively further then they have more Inertia to overcome.
Upshot:
1. When mass is spread wide around a pivot point it will have more Inertia than a smaller spread mass or point mass.
2. The way you pivot the mass also makes a difference to the Inertia i.e. fixed point has greatest Inertia whilst free to pivot fixing has less Inertia - this is because the free to pivot fixing means the bob can move somewhat independent of its orientation to the shaft & like water will take the path of least Inertial Resistance - that is it will not rotate if it doesn't have to.
3. When making a dual mass pendulum the distance that the masses are apart is proportional to the pendulums Inertia - this applies to any orientation e.g. along the radial line or perpendicular to it like a 'T' for example.
Kaine .. I suggest you make some examples in WM2D & measure velocity at 6 o'cl - the ones with least velocity have greatest MoI & they are the ones with fixed connections of the bob to shaft & the ones with a greatest mass spread in the bob or arrangement.
Re: re: Pendulum calculation
Clay .. see this previous post in another thread.clay973 wrote:Using this method I get the following:Fletcher wrote:
The problem arises in your calc's of Cf's because you have used the CoM velocity a 2.5 m radius - you actually have to use the velocity for each weight at its own radius which can be calculated longhand as above & then the Cf results will conform to your intuition, IINM.
Top weight v=6.26 CF = 2 * 6.26^2 / 2 = 4kg
Bottom weight v=7.67 CF = 2 * 7.67^2 / 3 = 4kg
No difference in CF? My experiment clearly shows the weights moving apart. Could it be caused by the weight of the rod?
http://www.besslerwheel.com/forum/viewt ... 2612#82612
In short Force = Mass x Acceleration
The Acceleration in pendulum bobs is given by v^2/r
therefore Cf = Mass x Acceleration
=> Cf = mv^2/r
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Logically, the velocity of the two weights are quite different because they are on the same radial & cover different distances [arcs] in the same amount of time - however the radius for the inner is 2 m & the outer 3 m.
When you do the calc's you will find that the Cf of the outer weight is substantially higher than the inner weight.
Hint:
The velocity is proportional to the radius, therefore a weight at say 1 m will have 1/2 the velocity of a weight at 2 m radius & 1/3rd the velocity of a weight at 3 m.
Since velocity is linear at different radii & the centrifugal Acceleration is by the square of velocity then an outer weight has far more Cf than an inner.