Torque calculation

[clay973]

If we have a rod with an equal weight at both ends and a pivot point in the middle (see attached pic), how do you calculate the amount of weight required to spin it to say 10RPM, and how long it will take to reach the 10RPM from standstill (I believe its called breakaway torque)?

[pequaide]

F = ma

[clay973]

Assuming the rod weights are say 1kg and the mass is say 500g.

What is 'a' (in F=ma) after 1 second, or 5 seconds?

[bilbo]

Hi Clay 973

Have you looked at the simple wheel idea on this thread as your weight configuration appears to be doing something similar to what my rack configuration is doing?

[clay973]

Hi Bilbo, yes i've seen your other thread, but I am really just looking for the formulas that would be relevant for the calculations I mentioned.

[bilbo]

Hi Clay 973

To be honest, those formulae would really help me too as I have limited knowledge of this stuff, unlike a lot of the extremely knowledgeable posters on here..

[jim_mich]

This is a complex engineering problem and you have omitted one portion of required information. The amount of weight required depends upon how fast you want to accomplish the task. Break away torque, IIRC, is the torque required to initially overcome friction and get the device moving. This is usually more friction than present once it is in motion. Then there is air resistance/friction once in motion, which increases as the velocity increases.

But if all friction is ignored then a very tiny amount of weight hanging on the rope will eventually bring the whole device up to any desired speed. A heavier weight will do the job faster.

Also do not forget that you are not only accelerating the rod and weight, but also accelerating the dropping weight. So you need to know the radius of the pully so as to know how much torque is being applied to the device and to know how fast the weight falls relative to the rotation speed of the device.

So, my point is, you have not fully defined the situation, which means there is no definitive answer to your question.

And any calculations that ignore friction (which is the way usually presented) will not mirror real life results since friction ia always present. And air friction is very hard to calculate as it depends on aerodynamics. And if you want to be very precise, aerodynamics depends on air density, which requires knowledge of the air temperature and pressure at the time of the experiment. Of course these have little effect at low speeds, but their effect increases as the speed of the device increases.

The bottom line is, engineering calculations will give you a feel for what the results will be, but only testing will give accurate results in the real world. There are always subtle little things that come into play. In the above situation, we forgot to allow for the weight of the rope upon which the weight hangs. And how stiff the rope as it unwinds. and friction is always unknown since it depends upon lubrication, which depends on the lubricant, which depends upon ...

I'm sorry, you asked a simple question, and I gave back a long complex answer. Me bad!

[clay973]

Thanks for the detailed reply Jim. Lets assume for theory that there is no friction, the rod, rope and pulley are weightless and the following dimensions:
Weights on rod = 1kg each
Rod weights are 1m from the center.
Pulley has a radius of 10cm
Falling mass = 500g

Is there a formula that will tell me how much time and how far the weight would need to fall to reach 50RPM?

[Furcurequs]

Hello Clay973,

Here are some formulas for a very similar problem as posted before by our forum member "Wubbly":

http://www.besslerwheel.com/forum/download.php?id=11164

...from this post and thread:

http://www.besslerwheel.com/forum/viewt ... 746#101746

You would want to replace the moment of inertia (I) of the flywheel in that problem with that of your rod - which is equal to one twelfth of its mass times the square of its length (I=M*L^2/12).

You can see some common moments of inertia here:

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

I have a bit of a headache right now or I would try to solve it for you.

I hope this helps, though.

Dwayne

[Tarsier79]

Here is someone that has built the same device you propose: http://www.besslerwheel.com/forum/viewt ... 924#102924

I have also cobbled together something similar. Are you attempting to "Create energy" through the manipulation of momentum/inertia? or do you have something else in mind?

[pequaide]

It is my opinion that one kilogram that has a leverage advantage of 4 will act like 4 kilograms. If then you place 1 newton of force upon the 4 kilograms you will get 1 = 4a or ¼ meter per second per second acceleration. It makes absolutely no difference that the accelerated mass is moving in a circle or on the end of a lever.

If this is a hanging mass that is accelerating a vertical balanced rim mass wheel then you must include the drive mass in the total mass m.(as Jim stated)

Jim described that start up resistance in the bearing is greater than rolling resistance. I start the dropping mass and if it continues moving but does not accelerate then you are roughly at rolling resistance.

It seems to me that what you can measure and see is what you have. You can measure and see four kilograms at a radius of one; balancing one kg at four, and it accelerates smoothly. Only when people insert their immutable beliefs (moment of inertia, angular momentum, energy conservation) does it become so difficult.

[jim_mich]

Weights on rod = 1kg each
Rod weights are 1m from the center.
Pulley has a radius of 10cm
Falling mass = 500g

Is there a formula that will tell me how much time and how far the weight would need to fall to reach 50RPM?

Radius of Gyration formula:
RoG = (a²×W+b²×w)÷(a×W+b×w)

W = 2000 g (two 1_kg weights at ends of rod)
a = 100 cm (radial distance of weight 'W')
w = 500 g (dropping weight)
b = 10 cm (radius of pulley)

Filling in the values:
RoG = (100²×2000+10²×500)÷(100×2000+10×500)

Radius of Gyration result:
RoG = 97.804878 cm

In other words, the arrangement will accelerate and rotate as if it were a single 2500g weight located at 97.804878 cm radius, sans gravity.

The 500 kg dropping weight is at 10 cm while the RoG is at 97.8 cm, so the torque force is leveraged.
It will be 97.804878÷10 = 9.7804878_g of force acting on 2500 g of mass.

The desired speed is 50 RPM. The RoG is 97.804878.
Note that Radius of Gyration value must be used as the radius of the moving weight, else the calculations WILL NOT BE CORRECT.
I've seen some members just grab any old radius that they come up with from who know where.
If you want accurate calculations concerning rotating objects then you MUST reduced everything that is rotating down to equivalant Radius of Gyration values.

So the circumference at RoG is 2×pi×97.804878 = 614.526172727 cm.
The desired velocity is 614.526172727 × 50 RPM = 30726.3086364 cm/min.
Or 30726.3086364 ÷ 60_seconds = 512.10514394 cm/s velocity desired.
Or 5.1210514394 m/s velocity desired.

We want to know the time needed to accelerate 2500 grams up to a Radius of Gyration velocity of 5.1210514394 m/s using a force of 9.7804878 grams where the gravitational acceleration is 9.80665_m/s².

Time = (M × V) ÷ (Ga × F)
Time = (2500_g × 5.121_m/s) ÷ (9.80665_m/s² × 9.7804878_g)
Time = 133.48 seconds to accelerate the mechanism up to a speed of 50 RPM.

Final velocity of dropping weight = 50_RPM × 10_cm × 2 × pi = 3141_cm/min = 52.36_cm/s
Distance dropped = (V×T)÷2
Distance dropped = 52.36_cm/s × 133.48_s ÷ 2 = 3494.5_cm or about 114.65 feet.


Hopefully I've made no mistakes. I make no guarantee. I've checked and double checked and believe this is the correct answer.

[clay973]

Thanks guys, some great responses!

[pequaide]

The two kilograms has a leverage advantage of 10 so it will act like 20 kilograms at the pulley radius.

The .500 kilograms exerts 4.905 newtons of force. So you have 4.905 = 20.5 kg * a; or .239 m/sec/sec.

The final velocity is 50 RPM / 60 sec = .833 RPS * 3.1415 * 2r = .523 m/sec

d = ½ v² / a; so d =.5 * .523* .523 / .239 = .573 m

d = ½ at², so it will take = 2.19 seconds

I have a wheel that I am now playing with where the wheel itself probably has a mass of about 1.8 kilograms. I have 3.8 kilograms on the circumference as in an Atwood's. The bearings are used and at times you can feel them grinding. The rolling resistance is probably 40 grams (force) or so. The drive mass is at .5 radius and the drive mass is 150 grams. Subtracting the rolling resistance the actual drive mass is about 110 grams.

This wheel will make one rotation in 5.5 seconds. This is a final velocity of .319 m/sec for the drive mass. And the drive mass will drop about .88 meters.

The drive mass of the real experiment is about the same (.110 kg at .5 and .500 kg at .1) as your thought experiment but the driven mass is at least double and you have a real bearing for the 5.5 seconds.

It seems that the calculation of 2.19 seconds for your experiment is consistent with real data.

[Trevor Lyn Whatford]

Hi Pequaide,

in this set up the weights are in counterbalance, "thus no leverage advantage" only more air drag friction on the T bar!

Regards Trevor