Moment of inertia where mass and torque are at a different positions

[Delta2]

If you replace torque and MI by Force and Mass, you will see how nonsensical that statement appears. You seem to be determined to fit the Science to your personal intuition. That ain't going to work. It has to be the other way round.

I told OP 2-3 times that the moment of inertia doesn't depend where we apply the torque and how big/small is the torque, but I refrain myself of doing it over and over again. OP chooses perplexed examples (like that of cars with smaller/bigger wheels, or that with the plank and the point mass) where a change in torque comes simultaneously with a change in moment of inertia and that gives OP the false impression that the moment of inertia depends on the torque.

 

[kuruman]

Hope I got it right

You did not. What is the force on each mass? Also, please use units when you write down numbers.

 

[ynbo0813]

If you replace torque and MI by Force and Mass, you will see how nonsensical that statement appears. You seem to be determined to fit the Science to your personal intuition. That ain't going to work. It has to be the other way round.

My original understanding was that τ=Iα should only apply where the torque and mass are at the same position - unless the moment of inertia changes. My mistake was that it was F=ma, not τ=Iα, which doesn't directly apply when the force and mass at different positions. You can repeat ad nauseam that my conclusion is wrong, but it’s far more helpful if you can explain why by pointing out the false premises. Actually #2 did so, but in a hand-waving way so I didn’t really get what they were saying. It took me a while to pinpoint my original mistake, but I think I got there.

I think your accusation is unfair. My original question was mathematical one, not based on intuition, although I did discuss intuition in the thread (perhaps more than I should have).

 

[ynbo0813]

I told OP 2-3 times that the moment of inertia doesn't depend where we apply the torque and how big/small is the torque, but I refrain myself of doing it over and over again.

I'm afraid telling isn't enough. You need to explain. I admit I was a bit stubborn, and I apologise for that, but those who explained to me why I was wrong and what my missteps were got further than those who just asserted I was wrong.

I also had a lot of other misunderstandings on the subject, which came clear as we went along.

 

[Delta2]

I'm afraid telling isn't enough. You need to explain. I admit I was a bit stubborn, and I apologise for that, but those who explained to me why I was wrong and what my missteps were got further than those who just asserted I was wrong.

I also had a lot of other misunderstandings on the subject, which came clear as we went along.

I tried to explain, I gave the general formula the MoI and from that it seems that MoI doesn't depend on Torque , torque doesn't appear anywhere in that formula. It depends on where we put the axis of rotation though and how is the mass density distribution around the axis of rotation. You choose perplexed examples where the changes in the two aforementioned quantities come together with a change in torque (like the example of car with bigger/smaller wheels) and that gives you the false impression that the MoI depends on torque.

@kuruman and others might find your examples interesting and try to explain you in detail where you get wrong, I am afraid I just don't find them so interesting. My area of expertise is not classical mechanics anyway.

And yes you are stubborn especially on trying to fit science to your personal intuition as @sophiecentaur very successfully said.

 

[sophiecentaur]

where a change in torque comes simultaneously with a change in moment of inertia

My mistake was that it was F=ma, not τ=Iα, which doesn't directly apply when the force and mass at different positions.

You need to explain. I admit I was a bit stubborn,

Actually, no one 'needs' to explain. Those are not the terms of PF interchanges. You are the one with the 'need' and all PF can do is to state the facts and, perhaps, explain away popular misconceptions. You will appreciate that your reactions all seem like a last ditch stand to maintain that you are actually right. This is hard to deal with.
My suggestion is, again, to apply the logic that applies with F = ma and to get the causal relationship in the right direction. Let me try this: A given point mass cannot be altered but the MI of an object varies with the axis you choose. The minimum MI is when the axis is through the CM. (If you are familiar with basic statistics, the Standard Deviation is equivalent to the MI) MI about other axes increases (and the SD of a distribution is minimal about the mean and increases from side to side). So we have the same object which is harder to rotate if you want it to rotate about a point that's not the CM because the MI has increased. So MI is a property of the object (as with Mass) but it depends on the reference axis. That makes total sense when you think about it.
If you want to give a certain value of angular acceleration to an object then the Torque required will depend on which axis you want to rotate it about.
I am labouring this point because I think it is what your problem is all about. Cause and Effect...

 

[ynbo0813]

You did not. What is the force on each mass? Also, please use units when you write down numbers.

Here’s my calculation for #56:

1. Net torque is [(20kg)(9.8m/s^2) × (2.5m)] – [(5kg) (9.8m/s^2) × (5m)] = 490Nm -245Nm = 245Nm
2. The common acceleration is α = τ/I = 245Nm/[(20kg)(6.25m)] + (5kg)(25m)] = 245Nm/250kg m^2 = 0.98 rad/s^2
3. The COM is at 1m from the pivot. The tangential acceleration should then be 0.98 m/s^2
4. The force on the COM is (25Kg)(0.98m/s^2) =245N
5. The tangential acceleration of the left mass is (0.98 rad/s^2)(2.5m) = 2.45m/s^2. The force on the left mass is therefore (20kg)(2.45m/s^2) = 49N
6. The tangential acceleration of the right mass is (0.98 rad/s^2)(5m) = 4.9 m/s^2. The force on the right mass is therefore (5kg)(4.9 m/s^2) = 24.5N

I think I have a better grasp on the issues now, and have cleared up my initial misunderstandings. Thanks all for your help and patience! It took a while, but I learned a lot.

 

[kuruman]

Here’s my calculation for #56:

1. Net torque is [(20kg)(9.8m/s^2) × (2.5m)] – [(5kg) (9.8m/s^2) × (5m)] = 490Nm -245Nm = 245Nm
2. The common acceleration is α = τ/I = 245Nm/[(20kg)(6.25m)] + (5kg)(25m)] = 245Nm/250kg m^2 = 0.98 rad/s^2
3. The COM is at 1m from the pivot. The tangential acceleration should then be 0.98 m/s^2
4. The force on the COM is (25Kg)(0.98m/s^2) =245N
5. The tangential acceleration of the left mass is (0.98 rad/s^2)(2.5m) = 2.45m/s^2. The force on the left mass is therefore (20kg)(2.45m/s^2) = 49N
6. The tangential acceleration of the right mass is (0.98 rad/s^2)(5m) = 4.9 m/s^2. The force on the right mass is therefore (5kg)(4.9 m/s^2) = 24.5N

I think I have a better grasp on the issues now, and have cleared up my initial misunderstandings. Thanks all for your help and patience! It took a while, but I learned a lot.

All that is correct. i am happy that you have a better grasp now.

 

[Adesh]

What I can see is that OP has not replied to anyone in the thread after posting his/her question.

 

[sophiecentaur]

I think I have a better grasp on the issues now, and have cleared up my initial misunderstandings. Thanks all for your help and patience! It took a while, but I learned a lot.

That's great; well done.
Now you are convinced that the 'theory' gives a 'convincing' answer, the next step is to see it working in symbolic (algebraic) form. That half page of numbers would all change if you used a slightly different set of conditions, moreover, when you use numbers, they lose their identity when you multiply them together and a final numerical answer doesn't tell you how it arrived. The algebra stays the same for all the numbers - you know what I mean.
So now you should look at the formulae and try to see the 'forms' without the numbers - like where there are squared values or divisions etc. etc. and see how that all goes to give the answer. You should find that your grasp improves further.