Calculating the amount of energy for rotation

Here we have got in high orbit the satellite-electro motor that rotates the rod:

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I would like to know how it is possible to calculate the amount of energy (received from the Sun by means of solar panels) and time needed for accelerating the motionless rod to some certain speed (let’s say rod’s end’s circular speed should be 1 000m/sec). We know the mass of the rod-1 000 kg (equally distributed to both sides, 500 kg at left side (from the electro motor’s shaft) and 500 kg-at right side), its length (if needed)-1 000 meters, the amount of energy received from the Sun-1300 W/m^2 and the coefficient of efficiency-20 %. So, is there formula for this purpose?

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http://en.wikipedia.org/wiki/Rotational_energy

One question: what stops the main body from rotating?

 

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I think-drag will

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Well you didn't really give how large the solar panel is so I'm assuming 1 meter square. Basically you want to know how long it will take a power source of 260W to rotate a rod of 1000kg and 1000m long to a tangential velocity of 1000m/s. The short answer is a long time!

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Maybe I'll work it up...

 

If it's in orbit (and therefore not anchored) the torque will work on BOTH bodies (the rod and main body).
As a hint: why do you think helicopters have a tail rotor?

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I really hope you don't work for China's version of NASA, Eagle.
Actually, in the spirit of national competition, maybe I do!

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RJBeery

I would like to know how to calculate this exactly

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And-how much energy is needed for this…

Dywyddyr

Yes, I know that for balancing the satellite the other, opposite rotation will be needed, but for now I would like to know how to calculate this

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RJBeery

I am from Europe, not China

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Rotational kinetic energy is analogous to linear kinetic energy.
KE_linear = (1/2)*m*v^2, where m=mass and v=velocity
KE_rotational = (1/2)*I*w^2, where I=moment of inertia and w=angular velocity

"I" depends on the shape of the mass under rotation. In the case of a uniform rod rotating from the center, I = (m*L^2)/12

"w" (actually, omega, but Tex is frustrating and I'm lazy) = V/R, where V = tangential velocity and R = radius

Throw this all in the blender and we get...

KE_rot = (1/2)*((1000kg*1000m^2)/12)*((1000m/s)/500m)^2

= (1/2)*(83,333,333)*(4) = 166,666,666 Joules, assuming I didn't mangle the units or fat-finger the calculator.

So, how do we get that many Joules into your unit with your power source? Well, 1 Watt = 1 Joule / sec. Therefore

166,666,666J = 260W * X seconds.

X = 166,666,666/260 = 641,025 seconds

...or about 7.4 days. Quite a bit shorter than I expected, actually. Again, you would be wise to have someone critique this math before you ask the ESA for funding.

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Also, as a follow-up, I wanted to say that I'm only poking fun at you because of the disparity between the sophistication of your diagrams (which are impressive) and the very basic questions you're asking. I just chuckle at the idea of a guy sitting in the corner of NASA or the ESA posting to online forums for a solution every time he's given a task to do...

Anyway, what graphics program are you using to make your diagrams?

 

http://www.sciforums.com/showpost.php?p=2702056&postcount=81

 

RJBeery
I will run this calculations again myself to be sure that I understood everything correctly (because actually I need to use other values, the ones that I wrote here was just only for example)

Yes, this rod is uniform one-cylinder with constant cross-section. So, “I” would be equal to 1 000 kg (rod’s mass on both sides!)*1 000 meters (rod’s total lenght)^2/12=83333333.3

tangential velocity in my case is equal to 1 000 m/sec, radius-500 meters (rod’s half length!), so w=2

Yes, everything is correct, I received the same result

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and I would like to specify three things:
1. This formula is for energy required for accelerating the rod from zero to some certain velocity (in my case this velocity is equal to 1000 m/sec)? I need to calculate exactly this and NOT the energy required for the rod rotating with some constant velocity

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2. We calculated the energy required for acceleration the rod-it is equal to 166 666 666 Joules and we connected this to power (260 Watts) and time (7.4 days). Does it mean that the rod with any mass (imagine that the mass of the rod was million times more!) could be accelerated to any speed (I do not mean relativistic speeds of course) by means of any power, any area of solar panels? Does not this rotation require some minimal power (area of solar panels) that would be simply necessary for making the heavy rod to begin motion/rotation?
3. We discussed the rod with equally distributed mass and length (500 kg mass and 500 m length at one side and the same on the other side), but what if the length of the rod at one side was less but the mass the same?

I have got nothing to do with NASA, ESA or any space agencies

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1. My method is simply relying on the concept of the conservation of energy. The KE of a nonmoving rod is zero, and the KE of a spinning rod is X. The only way for that rod to acquire X KE energy is from the power source. Note, this ignoring effects like frictional drag; in theory, once the rod is up to speed no more power would be required to keep it spinning, in practice we know that isn't true but the amount of continued power needed would be considerably less than the amount needed to accelerate the unit.

2. Yes, any rod of any mass and length could acquire any speed with any power source in theory. When you refer to "minimal power" you're talking about having enough torque to overcome the static friction of the motionless state. We are ignoring friction effects here, but you could (again in theory) overcome any amount of static friction barrier with the proper GEARING (think of a 10,000,000-speed bicycle).

3. If by "the mass was the same" you actually mean that one side is shorter and fatter (i.e. a counter-balance) but the rotor is still connected to the rod's center of gravity then you can treat the moment of inertia as if the entire unit continues to be a uniform rod. Check out http://en.wikipedia.org/wiki/List_of_moments_of_inertia for the formulas of moments of inertia for various shapes.

 

RJBeery

Yes, the rod is balanced at both sides like this:

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From those variants these two are the best suitable:

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But no completely, since the first one implies the rod with the same shape/length on both sides and second one-without any mass/length on the second side.
So, the formula given by you: KE_rotational = (1/2)*I*w^2 should be somehow altered?

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Eagle, the formula for KE should definitely remain as written. The formula for "I" might need to be changed from (m*L^2)/12 but my off-the-cuff assessment is that it does not as long as you take L to be twice that of the length of the blue half of the rod on the right side of your diagram.

Can anyone else verify this?

 

Eagle, a quick search did not yield an answer. Basically you're going to want to know whether two disparate objects, perfectly counter-balanced on an axis of rotation, have a combined MOI equivalent to having the same object on each side of the axis of rotation. I'm sure someone here knows the answer for sure but that someone is not me. Sorry!

 

Okay, to determine "I" for the whole system, You just add the 'I's for each arm, using the second formula:

\(I = \frac{mL^2}{3}\)

to determine the 'I' for each arm. (assuming both arms remain thin)

So, for example, if you reduce the length of one arm by 1/2 while maintaining the same mass, you will reduce its moment of inertia by 4, and reduce the entire rod's moment of inertia to 5/8 of what it was before.

However, if you do this, the rod will no longer be balanced, as the center of gravity will shift to a position closer to the end of the longer rod. So if you want to keep the same pivot point and keep the rod balanced, you have to add mass to the shorter arm.

 

RJBeery

Thank you very much for your contribution

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Janus58

So, if we reduce one side’s length 1 000 times (mass of both sides will be the same!) its “I” will be reduced 1 000 000 times and whole rod’s “I” will be 1 000 001/2 000 000 of what it was before, right?

Wait a minute, the masses to both sides will be the same

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so as I THOUGHT the center of gravity will coincide to electro motor’s shaft, is not it so? If it IS NOT so because the fact that longer side’s end will rotate with more linear/circular speed (my intuition indicates me that this might be reason) then what can be done?

Well, I need the rotating rod (with one part shorter (this is necessary) and probably (?) with same masses) to be balanced during the rotation. I thought that if it was not balanced then it would cause the electro motor to somersault or something like this. But as you say the mass of one side still should be more and only in this case the rod will be balanced due to rotation, right?

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Treat each arm separately and find its CoG. You can then treat each arm as if its total mass is located at that point, when it comes to finding the CoG of the whole rod.

If the arms are of equal length and mass their indivdual CoGs will be equal distances from the shaft and the CoG of the whole rod will be at the shaft.

If I shorten one of the arms while keeping the same mass, then the new CoG of this arm will be closer to the shaft, Since the CoG of the rod will be halfway between the two arm CoGs the CoG of the rod will shft to a point on the long arm.

It is the same as if you have two equally massed people each sitting the same distance away from the pivot of a seesaw. They will balance.

If however, one of them moves closer to the pivot, the seesaw will unbalance toward she other, even though the mass on both sides of the seesaw didn't change.

 

Eagle, I don't think you mean the masses are the same. The shorter counter-balance will need to be MORE massive to compensate for its shortness (kind of like that guy at the gym!

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)

I've run some numbers and it appears that what you're going to have to do is design your counter-balance, then calculate each side's MOI separately and add them together. I no longer believe it suffices to consider both sides to be of equal shapes when seeking the combined MOI of a system with disparate shapes on each side of the axis of rotation.

 

A quarter round counter weight can help some by putting a higher percentage of the "short arm" further away from the pivot and shifting the CoG outward.

Something like this:

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