[44-StarGazer (OP)]

Hello!

Non-scientist and Newbie to this forum, with what's probably an elementary engineering-type question, regarding torque. As a reference, I'm a life-long science buff, but I haven't had a course in physics since high school (I live in the USA)... more than fifty years ago.

As an example of the problem I'm attempting to resolve....

Assume a completely friction-free "see-saw" type of apparatus with the fulcrum fixed in position at the exact midpoint of a perfectly balanced rod. After placing a two-kilogram weight on each side of the rod, moderately close to but also equidistant from the fulcrum, the rod remains balanced.

The rod is then rotated about its fulcrum, and the amount of torque required to do so is calculated and measured.

One of the 2Kg weights is then removed, and a 1kg weight is added to that same side of the rod, at exactly twice the distance from the fulcrum point as was the original 2Kg weight... causing the system to once more be in balance.

The rod is again rotated about the fulcrum, and the amount of torque required to do so is again calculated and measured.

My contention -- based on my general and admittedly limited understanding of the the physics and math involved -- is that even though the weighted rod's actual center of gravity has been shifted... in both cases, the torque required to rotate the rod about its fixed fulcrum would be identical.

But I have been told that this is incorrect.

If so, could someone kindly explain -- in layman's terms, and/or using basic mathematics -- what it is that I'm misunderstanding?

I'm truly happy to have discovered this forum, by the way... and I thank you, in advance, for your kind indulgence!

[Colin2B]

We are off to a good start here because you clearly understand the concept of torque and leverage, that the effective force depends on the distance from the fulcrum.
OK, let's think about a flywheel. Start with a bicycle wheel, I think you will agree that if you have 1kg of lead and place it close to the hub it will be easier to spin the wheel compared to if you place the same weight out at the rim. The wheel will also spin longer. The principle here is inertia.
When we push a weight in a straight line we recognise it has inertia, F=ma, greater the force, greater the acceleration, greater the mass the less easy to accelerate.
When pushing a weight in a circle the force has to be replaced by torque to take into account the leverage. But the mass also has to be effectively increased to take account of this leverage effect. In a rotating system the 'torque equivalent' of mass is the moment of inertia I, which is the actual mass x radius squared I=mr².
So the distance we place the mass from the fulcrum has a much greater effect than the effect of placing the force at the same distance. This is why when designing a flywheel we try to put the mass as far away from the pivot point as possible to get greater Inertia.
Hope that helps.

[44-StarGazer (OP)]

Thank you so much! Your explanation-by-example is brilliantly clear, and makes perfect sense. But again, it's been a while... so let's just make sure that I do understand....

Let's say that the total length of the rod is 100cm, and that in the initial setup, each of the two 2Kg weights are arbitrarily placed 25cm from the fulcrum. In "part two", the 1Kg weight would then be 50cm from the fulcrum.

In the initial setup, the total moment of inertia required to rotate the system would then be 2(2Kg x 25²cm) = 2500

In part two, the total moment of inertia required to rotate the system is (2Kg x 25²cm) + (1Kg x 50²cm) = 3750

So... in part two of my example, even though the rod remains statically "balanced" at its fulcrum point, the result of having moved the center of gravity away from that fulcrum point is a very significant increase in the total amount of torque required to rotate the system.

Remarkable.

And thank you, again. I have a strong feeling that this forum is about to very significantly impact my life!

[Colin2B]

Yes, you've got the idea.
The only comments I would make is that in standard units (SI) we use meters rather than centimetres. Your answer is still correct, but you would need to specify the units you have used to avoid confusion, so better to use meters.
Also, I understand what you mean when you say centre of gravity, but strictly speaking the CoG (also centre of mass) is at the fulcrum in this system. The equivalent to centre of mass in a rotating system can be referred to as the radius of gyration which is the effective distance of an equivalent single mass from the pivot point.

I will leave you to work out why the moment of inertia is related to r² rather than just r. A little post Chrismas amusement []