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

[sophiecentaur]

That's because torque relates to work,

The 'relationship' you are referring to is just the fact that torque (a Vector Quantity) involves a force and a distance and so does Work (a Scalar Quantity). But the operation in the two cases is very different (a cross and a dot product) and that makes them very distinct. I think this must be distracting / confusing you. I couldn't find this point being made directly in the thread so far so I though I'd make it explicitly.

 

[sophiecentaur]

Let me phrase it a more intuitive way.

I read this way up the thread and I think this must be your problem. Intuition, unless it comes from a valid experience or valid reasoning, can often be totally wrong. Your intuition can make you feel good about right and wrong ideas. It has to be backed up with rigour. I notice that your posts do contain some Maths (that's good) but you seem to have a problem what that Maths is telling you when there is some conflict between the two and you then go back to requiring intuition to be correct.

 

[ynbo0813]

The direction is radial from the mass to the pivot, but the magnitude depends on how fast the mass is moving because the pivot must provide the centripetal force.

Ok so I think I may have got it, but I'm not sure. Please correct me if I'm wrong.

Going back to my diagrams in #41:

In addition to the force illustrated, there is the centripetal force. The centripetal force is bigger closer to the pivot point (because F=(mv^2)/r). The vectors for the forces should then look like this (apologies for the poor quality - I'm just using paint):
(1)

(2)

In (2) the net force is more in the direction of the tangent circular motion. This is why the net force, and therefore the net tangential acceleration, is greater in (2).

This might also provide the intuition I was looking for: why the force is 'magnified', so to speak, by its distance to the pivot point.

However, I suspect I may be completely off the mark here, so please correct me

 

[ynbo0813]

I read this way up the thread and I think this must be your problem. Intuition, unless it comes from a valid experience or valid reasoning, can often be totally wrong. Your intuition can make you feel good about right and wrong ideas. It has to be backed up with rigour. I notice that your posts do contain some Maths (that's good) but you seem to have a problem what that Maths is telling you when there is some conflict between the two and you then go back to requiring intuition to be correct.

Fair point. I thought the maths backed up my intuition (since the moment of inertia is derived from torque, a greater torque should result in a greater moment of inertia). My maths was wrong :(

 

[kuruman]

The centripetal force is bigger closer to the pivot point (because F=(mv^2)/r).

That would be correct if the two points at different radii had the same speed. They do not because the point at a larger radius needs to move faster in order to complete one revolution in the same time as the point closer to the center. You can see what's going on better if you substitute $v=\omega r$ and write the centripetal force as $F_c=m\omega^2 r$ where $\omega$ is the common angular speed of the two points. It is clear that the point at the larger radius has the larger centripetal force.

The acceleration as a vector can be written as $\vec a=\{\alpha r, \omega^2 r\}$. You can see that both the tangential and centripetal components depend directly on the distance from the center. This means that if you wanted to draw arrows at different radii to represent the total acceleration, they should be parallel and an arrow that is twice as far from the center should be twice as long.

 

[ynbo0813]

That would be correct if tthe wo points at different radii had the same speed. They do not because the point at a larger radius needs to move faster in order to complete one revolution in the same time as the point closer to the center. You can see what's going on better if you substitute $v=\omega r$ and write the centripetal force as $F_c=m\omega^2 r$ where $\omega$ is the common angular speed of the two points. It is clear that the point at the larger radius has the larger centripetal force.

The acceleration as a vector can be written as $\vec a=\{\alpha r, \omega^2 r\}$. You can see that both the tangential and centripetal components depend directly on the distance from the center. This means that if you wanted to draw arrows at different radii to represent the total acceleration, they should be parallel and an arrow that is twice as far from the center should be twice as long.

Ok, let me try again. So, I have been rereading the whole thread and I think it’s all coming together. Hopefully I’m at the home stretch. Thanks for sticking through.

Suppose you have a 10m see-saw where the weight on the left is 10 kg and is 2.5m from the left end of the see-saw, and the weight on the right is 5kg and at 10m from the left. The pivot is at 5m.

The see-saw is in balance. This can also be calculated using torques, but I’m going to go a different way.

The pivot is at the centre of mass. Choosing 0m to be the left end of the see-saw:

(10kg)(2.5m)+(5kg)(10m)/15kg = 5m.

Therefore, the upwards normal force applied by the pivot balances out the downwards forces from the weights.

F=ma only applies at the centre of mass (as I have learned). This is probably the key to all of my many misunderstandings. The above (if correct) would give me an intuition for torque and its relationship to F=ma.

Again, please correct me if I’m wrong.

To bring this closer back to my original question, suppose the weight on the left was now 20kg. The COM would now be 4m from the left. Suppose further that we cannot move the pivot, so the see-saw will fall to the left. What is the force (in Newtons) on the COM? I think the same strategy will help me to correctly calculate #41.

Thanks all for your patience.

 

[etotheipi]

F=ma only applies at the centre of mass (as I have learned). This is probably the key to all of my many misunderstandings. The above (if correct) would give me an intuition for torque and its relationship to F=ma.

I think you have a better grasp, but it might be worth doing some further reading on the mechanics of rigid bodies to gain further insights. And also to have a read through this page on Euler's laws of motion (basically just reformulations of Newton's laws for rigid bodies). What you have said about F=ma is more accurately stated as "the net force on a body (the vector sum of forces no matter where they are applied on the body, i.e. they don't need to be applied at the CM) equals the mass times the acceleration of the centre of mass".

If the plank is rigid, and its centre of mass is always just above the triangular pivot, then the acceleration of the centre of mass of the plank is constrained to be the zero vector. The vector sum of forces on the plank consisting of the sum of the contact force from mass 1, contact force from mass 2, contact force from pivot must then also be zero in equilibrium. You might also analyse either of the masses with the same ideas. Or you could analyse torques on the plank after first choosing an origin for your coordinate system (here it is convenient to use the pivot).

You can also take your system to be the plank + mass 1 + mass 2, and then the only external forces are the weights and the contact forces from the hinge.

 

[kuruman]

To bring this closer back to my original question, suppose the weight on the left was now 20kg. The COM would now be 4m from the left. Suppose further that we cannot move the pivot, so the see-saw will fall to the left. What is the force (in Newtons) on the COM? I think the same strategy will help me to correctly calculate #41.

Once more, the method for handling this question and others like it is
1. Find the net torque about the pivot.
2. Find the angular acceleration $\alpha$ about the pivot.
3. Find the tangential acceleration of the COM using $a_t=\alpha X_{cm}$ where $X_{cm}$ is the distance of the COM from the pivot.
4. The force on the COM is the linear acceleration times the mass of the system $F_{cm}=(m_1+m_2)a_t$. If the COM has speed, you need to also consider the centripetal component as we discussed.

I will let you do it as an exercise to see for yourself how it is done. Please post your work if you want it checked. Note that you cannot do this problem without using torques. That's because you are given two of the external forces acting on the system, the two weights, but you don't know the third force at the pivot.

 

[sophiecentaur]

a greater torque should result in a greater moment of inertia

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.

 

[ynbo0813]

Once more, the method for handling this question and others like it is
1. Find the net torque about the pivot.
2. Find the angular acceleration $\alpha$ about the pivot.
3. Find the tangential acceleration of the COM using $a_t=\alpha X_{cm}$ where $X_{cm}$ is the distance of the COM from the pivot.
4. The force on the COM is the linear acceleration times the mass of the system $F_{cm}=(m_1+m_2)a_t$. If the COM has speed, you need to also consider the centripetal component as we discussed.

I will let you do it as an exercise to see for yourself how it is done. Please post your work if you want it checked. Note that you cannot do this problem without using torques. That's because you are given two of the external forces acting on the system, the two weights, but you don't know the third force at the pivot.

Ok here goes.
1. Torque on the left is 20 × 2.5 = 50. The torque on the right is 5 × 5 = 25. The net torque is 25.

2. α = τ/I. I = (20)(2.5)^2 + (5)(5)^2 = 250.
25/250 = 10 (the common angular acceleration)

3. The tangential acceleration is also 10 since the COM is at 1m from the pivot as shown in #56.

4. The force on the COM is therefore 250N.

Hope I got it right

 

[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.