Mass Pulley and torque -- Confusion in this Atwood machine problem

[mancity]

Homework Statement

Consider the triple Atwood’s machine shown in Fig. 4.21. What is the acceleration of the
rightmost mass?

Relevant Equations

torque=I alpha=rF => F_rot = 1/2 mr * alpha

Say you had a pulley with a mass hanging off of it, like in the picture. What I don't understand is, what idealizations are being made such that there is no net torque being generated? My confusion is that we have a massive pulley, hence there will be rotational inertia. But it is pretty clear that in this diagram there is no net torque that resists the acceleration of the pulley. Thanks

 

[kuruman]

What I don't understand is, what idealizations are being made such that there is no net torque being generated? My confusion is that we have a massive pulley, hence there will be rotational inertia.

The diagram is drawn under the assumption that the pulley has negligible moment of inertia. That's because it shows that the pulley changes the direction of the tension but not the magnitude. I am perplexed by the tension 4T shown in the diagram on the left. If that were the case, then the acceleration $a_b$ should be zero. Is that your addition?

 

[Lnewqban]

Your relevant equations show r and rotating m, which are not shown in the book's diagram.
The book's text specifies massless pulleys.
Do you have a reason to try to combine the equations and the ideal case described in the book?

 

[haruspex]

I am perplexed by the tension 4T shown in the diagram on the left. If that were the case, then the acceleration $a_b$ should be zero. Is that your addition?

Why? The pulleys are clearly meant to be massless, and the text supports that: "net force must be zero".

 

[kuruman]

Why? The pulleys are clearly meant to be massless, and the text supports that: "net force must be zero".

Which text is this? I saw

which states that the acceleration of the rightmost mass is $a_b=5g/17$, not zero. The tension in the lower rope in the FBD on the right should be labeled $mg$, not $4T.$ Then the equation $$mg-4T=ma_b$$ would make sense as Newton's second law for a pulley of mass $m$ but zero moment of inertia about its center.

 

[haruspex]

Which text is this? I saw

View attachment 351273
View attachment 351275which states that the acceleration of the rightmost mass is $a_b=5g/17$, not zero. The tension in the lower rope in the FBD on the right should be labeled $mg$, not $4T.$ Then the equation $$mg-4T=ma_b$$ would make sense as Newton's second law for a pulley of mass $m$ but zero moment of inertia about its center.

In the fourth line after "4.12":
“Finally, the tension in the bottom string is 4T because the net force on the bottom pulley must be zero."
The pulley is massless. $a_b$ is the acceleration of that pulley and of the mass $m$ suspended from it. Force balance on that mass gives $ma_b=mg-4T$.

 

[mancity]

Your relevant equations show r and rotating m, which are not shown in the book's diagram.
The book's text specifies massless pulleys.
Do you have a reason to try to combine the equations and the ideal case described in the book?

Sorry. I guess my confusion is that a massless pulley with a mass m hanging off of it does not make the pulley massive/ have a moment of inertia?

 

[haruspex]

Sorry. I guess my confusion is that a massless pulley with a mass m hanging off of it does not make the pulley massive/ have a moment of inertia?

It is different from a massive pulley in two key respects relevant here:

1. It has no moment of inertia
2. The net force on the pulley itself is zero.

The net force on the pulley+mass combination is the same as if the mass m were integrated into the pulley.

 

[kuruman]

Finally, the tension in the bottom string is 4T because the net force on the bottom pulley must be zero."

This sounds like a zen koan to me: "What is the acceleration of a massless pulley when the net force on it is zero?"

Spoiler

Whatever you want it to be because $0=0*a$.

I think it is better to model an ideal pulley with a mass hanging from its center as a pulley with mass but no moment of inertia. I think that is the "idealization" which confused the OP in the first place.

 

[haruspex]

This sounds like a zen koan to me: "What is the acceleration of a massless pulley when the net force on it is zero?"

Spoiler

Whatever you want it to be because $0=0*a$.

That would be so, but it is not the question here, which is, what is the net force on a massless object? As ever, the way to deal with idealisation is as the limit of realistic situations. Taking the limit of $F=ma$ as $m\rightarrow 0$, either $F\rightarrow 0$ or $a\rightarrow\infty$. Since the acceleration is limited by $g$, $F=0$.

I think it is better to model an ideal pulley with a mass hanging from its center as a pulley with mass but no moment of inertia. I think that is the "idealization" which confused the OP in the first place.

Fine, but in the model adopted the tension is 4T.

 

[Lnewqban]

Sorry. I guess my confusion is that a massless pulley with a mass m hanging off of it does not make the pulley massive/ have a moment of inertia?

No.
In this case, the only function of the pulleys is to change the direction of the tension in the string and let it, and the masses, move freely.

There is a difference in the nature of linear and rotational inertias.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mi

 

[kuruman]

Fine, but in the model adopted the tension is 4T.

I think that the textbook solution does not clearly specify what the system is when writing Newton's second law equation from the FBD.

The figure on the right is the FBD of a two-component system consisting of the massless pulley and the hanging mass $m$. The net force on this system is $F_{\text{net}}=(m+0)g-(2T+2T)$. The circled force $4T$ is internal between the two components of the system and has no place in the FBD.

One could also draw a FBD of the hanging mass only and provide an argument why the tension in the string connecting to the pulley must be $4T$ that's better than "Finally, the tension in the bottom string is 4T because the net force on the bottom pulley must be zero". Pedagogically, this statement weakens the generalization that zero net force implies zero acceleration by providing an asterisk with the contents of your post #10.

We may have to agree to disagree on this one.