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

[etotheipi]

Levers do not perform more work: work in = work out

I thought I might add that this holds for ideal simple machines, with neither mass nor frictional losses.

Otherwise we might need to take into account e.g. the rotational kinetic energy of the lever itself, and the work done on the lever is not necessarily equal to the work done on the load.

 

[kuruman]

For example, take two cars accelerating at the same rate, but one has smaller wheels. If I'm not mistaken, the cars will both be accelerating at the same rate, but the smaller wheels will be spinning faster (i.e. it will take them quicker to make a full circle). But in that case, the force is closer to the axis! The force of the friction from the road is closer to the axis because the wheels are smaller. This is why I still don't understand, intuitively at least, how torque arises from Newton's laws. I'm guessing I made a misstep somewhere, just trying to find it.

Instead of trying to understand intuitively, try doing some calculations since intuition can be deceptive. Shown below is a free body diagram of the accelerating wheel of a car. The torque ττ generated at the axle is replaced by an equivalent force $F=τ/R$. The mass of the wheel is mm and its moment of inertia about its center of mass is $I_{cm}$; its moment of inertia about the point of contact with the road is $I=I_{cm}+mR^2$.

We write two equations for Newton's second law. Here acmacm is the acceleration of the center of mass and αα is the angular acceleration about the center of mass. The force of static friction is $f$.
For the translational motion
$F−f=ma_{cm}$.
For the rotational motion we calculate torques about the point of contact
$FR=(I_{cm}+mR^2)\alpha$.
Since the wheel rolls without slipping,$$a_{cm}=\alpha R=\frac{FR^2}{(I_{cm}+mR^2)}.$$Put this back in the first equation to find the force of friction$$f=F−ma_{cm}=\frac{I_{cm}}{(I_{cm}+mR^2)}F.$$Note that the last two equations can be combined to give a relation between the force of friction and the acceleration,$$f=\frac{I_{cm}}{R^2}a_{cm}.$$The moment of inertia of the wheel can always be written as $I_{cm}=q~m~R^2~~~~~ (0<q≤1).$ For example $q=\frac{1}{2}$ for a cylinder and $q=1$ for a hoop. Thus,$$f=q~ m ~a_{cm}.$$This last equation says it all. If two wheels have the same acmacm and qq, the force of friction will depend on their mass and has nothing to do with the radius. The torque about the CM due to friction can be written as$$\tau=fR=q~ m ~a_{cm}R=q mR^2 α=I_{cm}α.$$Everything hangs together perfectly well and the gods of physics are smiling upon us benignly.

So please, if you are going to raise objections to what we say with specific examples, provide calculations substantiating your objections instead of using vague intuitive arguments.

 

[etotheipi]

The torque τ generated at the axle is replaced by an equivalent force $F=τ/RF$.

I wondered whether this is still a treatment of the wheel of an accelerating car or a new example, since you say we can replace the torque with an equivalent force but I'm not sure if it's still the same scenario. For the case of the wheel of an accelerating car, the friction force points in the forward direction to accelerate the car which also has the upshot of opposing the torque produced by the axle.

In this example, I imagine a wheel with a frictionless axle such that the axle exerts a forward contact force on the wheel. And in this case the static friction would indeed point in the backward direction. Is this what you were envisioning?

Apologies if I misunderstand!

 

[kuruman]

For the case of the wheel of an accelerating car, the friction force points in the forward direction to accelerate the car which also has the upshot of opposing the torque produced by the axle.

First the equivalent force was wrongly posted as $F=\tau/RF$, it should be $F=\tau/R$. My apologies for the typo.

The wheel shown in my drawing rolls without slipping to the right and has CM acceleration $a_{cm}$. Undeniably the net force on the CM is $F_{net}=ma_{cm}$. Can we identify this net force with the force of static friction $f$?
If "yes" then it must point to the right. But if it points to the right, the torque about the CM is out of the screen (counterclockwise) which contradicts the assumption of forward linear acceleration which implies clockwise angular acceleration. Therefore the answer must be that the force of friction is cannot be identified as the net force on the CM. This means that there must be an additional force, call it the engine force, $F$, what I call the "equivalent" force.

Now imagine a FBD similar to the one I showed in #37 except that the arrow representing $f$ has reversed direction and points to the right. Reversing the arrow's direction means that one needs to reverse the algebraic sign in front of $f$, which represents the magnitude of static friction, in all the equations where it appears. This will result in$$f=-\frac{I_{cm}}{(I_{cm}+mR^2)}F.$$Because $f$ and $F$ are both magnitudes, if we insist that the force of friction is correctly drawn, we must assume that $F$ must be incorrectly drawn and therefore must point to the left. this last result says that the choice of direction was incorrect and that it should actually point to the left. This is consistent with the definition of $\vec F$ as the force that generates a torque about the CM such that $\vec \tau=\vec R \times \vec F$ where $\vec R$ is from the CM to the point of contact. In #37 I considered the torque about the point of contact. Since friction generates no torque about that point, $F$ must be in the forward direction. Both pictures agree on the direction and magnitude of $a_{cm}$.

(I edited this post to remove material that was not correct.)

 

[ynbo0813]

So please, if you are going to raise objections to what we say with specific examples, provide calculations substantiating your objections instead of using vague intuitive arguments.

I made a specific argument about the fact that the car scenario I described seemed to contradict τ=dL/dt. It was answered in #28. I'm not trying to argue, I specifically asked for someone to point out my misstep - which @etotheipi did.

 

[ynbo0813]

If you fix the moment of inertia then you're right, $\tau = I\alpha$ so a larger torque yields a larger angular acceleration.

If we instead fix the tangential force, then the angular acceleration is greater for smaller radii (even though the torque of the force at the larger radius is larger!).

Let me go back to my original scenario, hopeful this will make things clearer (sorry for all the back and forth - I'm trying to wrap around my head around a number of concepts at the same time I guess):
(1)

(2)

Assume the see-saw thing has no mass or friction. Further, assume that the forces are of the same magnitude in (1) and (2). The tangential accelerations should, therefore, be the same in (1) and (2). Likewise, the moment of inertia is the same between (1) and (2) since the masses are in the same position.

However, the torque of (2) is double that of (1). Therefore, the angular acceleration of (2) should also be double that of (1). Analytically, in (2):

2τ = I(2α)
2rF = m(r^2)2(a/r) = 2rma
F=ma

In other words, in (2) the angular acceleration is double that of (1), but the tangential acceleration is the same.

Here’s my problem. If masses (1) and (2) are in the same position, how can their tangential accelerations be the same but their angular accelerations different?

 

[Lnewqban]

Therefore, the angular acceleration of (2) should also be double that of (1).

What makes you think so?

 

[kuruman]

The tangential accelerations should, therefore, be the same in (1) and (2).

This is not so. When there is angular acceleration, you get the tangential acceleration from $a_t=\alpha r$ instead of asserting what it is using $F_{net}=ma$. That's because the CM is not free to move in a straight line, but is constrained by the pivot to go around in a circle. I showed you how to handle this in post #8. The general strategy is (a) find the angular acceleration of the rigid body about the pivot using $\tau=I\alpha$; (b) find the linear (tangential) acceleration of any point on the rigid body using $a_t=\alpha r$.

Here, the tangential acceleration is $$a_t=\alpha~L=\frac{\tau}{I}L=\frac{Fr}{mL^2}L=\frac{Fr}{mL}$$ where $L$ is the distance of the mass from the pivot. In the top picture (1) $r=L$ so that $a_{1t}=\dfrac{F}{m}$. In the bottom picture (2), $r=2L$ so that $a_{2t}=\dfrac{2F}{m}$. Note that the angular acceleration is twice as much in (2) than in (1) because the torque is twice as much and the moment of inertia is the same. Then for the same position of the mass, the tangential acceleration is also twice as much in (2) than in (1).

 

[ynbo0813]

When there is angular acceleration, you get the tangential acceleration from $a_t=\alpha r$ instead of asserting what it is using $F_{net}=ma$. That's because the CM is not free to move in a straight line, but is constrained by the pivot to go around in a circle.

I understand your argument. I'm just trying to reconcile it with F = ma. Was my mistake that I used F = ma without considering the internal forces which constrain it? It helps if you can pinpoint my error :)

 

[etotheipi]

The wheel shown in my drawing rolls without slipping to the right and has CM acceleration $a_{cm}$. Undeniably the net force on the CM is $F_{net}=ma_{cm}$. Can we identify this net force with the force of static friction $f$?
If "yes" then it must point to the right. But if it points to the right, the torque about the CM is out of the screen (counterclockwise) which contradicts the assumption of forward linear acceleration which implies clockwise angular acceleration. Therefore the answer must be that the force of friction is cannot be identified as the net force on the CM. This means that there must be an additional force, call it the engine force, $F$, what I call the "equivalent" force.

Many thanks for the explanation. I agree with all of your analysis, but it still seems to me like the force $\vec{F}$ is a contact force between the axle and the wheel in your example and the setup is such that the axle can transmit no torque. This is completely valid, and demonstrates rolling also, though not what I was imagining!

I imagined a FBD like this, with the torque from the axle equivalent to a force couple exerted by the axle on the wheel:

Then we can identify the static friction force $\vec{f}$ as the forward force providing $\vec{a}_{CM}$, however we must also have that the clockwise torque of the couple is greater than the anticlockwise torque of the static friction, i.e. $\tau_{couple} > fR$. That provides the clockwise angular acceleration and with the constraint $a_{CM} = R\alpha$ we can then solve the system.

I haven't the faintest clue how cars are actually designed, though I presumed it was a force couple (or equivalent) that the axle exerts on the wheel and not a forward contact force. In reality, it probably does both, and we need to draw a couple in addition to a contact force from the axle!

 

[kuruman]

I understand your argument. I'm just trying to reconcile it with F = ma. Was my mistake that I used F = ma without considering the internal forces which constrain it? It helps if you can pinpoint my error :)

Yes, the net force on the mass is not represented by the arrow that you have drawn. The net force on the mass is the normal force exerted by the plank on the mass. The fact that the plank is approximated as massless does not mean that it accelerates in a straight line when you apply the force. The arrow that you have drawn is one of the forces exerted on the plank. There is also the constraint force at the pivot. That is why you first have to solve for the angular acceleration of the plank and then use it to find the linear acceleration of any point along it. If you choose to calculate torques about the pivot, you do not have have to worry about it.

 

[kuruman]

I understand your argument. I'm just trying to reconcile it with F = ma. Was my mistake that I used F = ma without considering the internal forces which constrain it? It helps if you can pinpoint my error :)

Yes, the net force on the mass is not represented by the arrow that you have drawn. The net force on the mass is the normal force exerted by the plank on the mass. The fact that the plank is approximated as massless does not mean that it accelerates in a straight line when you apply the force. The arrow that you have drawn is one of the forces exerted on the plank. There is also the constraint force at the pivot. That is why you first have to solve for the angular acceleration of the plank and then use it to find the linear acceleration of any point along it. If you choose to calculate torques about the pivot, you do not have have to worry about it.

Many thanks for the explanation. I agree with all of your analysis, but it still seems to me like the force $\vec{F}$ is a contact force between the axle and the wheel in your example and the setup is such that the axle can transmit no torque. This is completely valid, and demonstrates rolling also, though not what I was imagining!

I imagined a FBD like this, with the torque from the axle equivalent to a force couple exerted by the axle on the wheel:

View attachment 263562
Then we can identify the static friction force $\vec{f}$ as the forward force providing $\vec{a}_{CM}$, however we must also have that the clockwise torque of the couple is greater than the anticlockwise torque of the static friction, i.e. $\tau_{couple} > fR$. That provides the clockwise angular acceleration and with the constraint $a_{CM} = R\alpha$ we can then solve the system.

I haven't the faintest clue how cars are actually designed, though I presumed it was a force couple (or equivalent) that the axle exerts on the wheel and not a forward contact force. In reality, it probably does both, and we need to draw a couple in addition to a contact force from the axle!

I have retracted some of what I said. I do not wish to pursue the accelerating wheel in this thread because I don't want to detract the discussion from OP's question. Also, as you may know, there have been two accelerating wheel threads in the past that were eventually shut down by the mentors. If there is to be a third thread, one must proceed with caution and say something hat has not been said before.

 

[etotheipi]

Yes, apologies for the digression! I don't have any other comments on the OP's question further to what you've already said in #46 so I will step back now .

 

[ynbo0813]

There is also the constraint force at the pivot.

Ok that makes sense. Is this force along the lever arm, and in the direction of the pivot point?

 

[kuruman]

Ok that makes sense. Is this force along the lever arm, and in the direction of the pivot point?

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.

 

[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.