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

  • Thread starter Thread starter mancity
  • Start date Start date
AI Thread Summary
The discussion centers on the confusion regarding the application of torque and tension in an Atwood machine problem involving a massive pulley. Participants clarify that the textbook assumes massless pulleys, which simplifies the analysis by eliminating net torque and moment of inertia considerations. The tension in the system is debated, particularly the assertion that the tension is 4T, which some argue should not be included in the free body diagram (FBD) as it complicates the understanding of net forces. The conversation highlights the distinction between massless and massive pulleys and the implications for force balance and acceleration in the system. Ultimately, the complexities of idealizations in physics problems are emphasized, particularly in relation to Newton's laws.
mancity
Messages
26
Reaction score
2
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
 

Attachments

  • 1.png
    1.png
    3.3 KB · Views: 55
  • Screenshot 2024-09-17 at 5.43.40 PM.png
    Screenshot 2024-09-17 at 5.43.40 PM.png
    42.6 KB · Views: 42
Physics news on Phys.org
mancity said:
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?
 
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?
 
kuruman said:
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".
 
haruspex said:
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

Screen Shot 2024-09-18 at 7.35.38 AM.png

Atwood_4T.png
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.
 
kuruman said:
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##.
 
Lnewqban said:
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?
 
mancity said:
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.
 
haruspex said:
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?"
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.
 
  • #10
kuruman said:
This sounds like a zen koan to me: "What is the acceleration of a massless pulley when the net force on it is zero?"
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##.
kuruman said:
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.
 
  • #11
mancity said:
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
 
  • #12
haruspex said:
Fine, but in the model adopted the tension is 4T.
Atwood_4T.png
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.
 

Similar threads

Back
Top