Pulleys with Strings Having Mass

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In a system with a pulley of mass m1 and two hanging masses m2 and m3 (where m3 > m2), the acceleration is given by (m3 - m2)g/(1/2m1 + m2 + m3). The presence of mass in the pulley causes different tensions on either side of the string, suggesting the string cannot be massless without leading to infinite acceleration. However, the final acceleration formula does not include the mass of the string, raising questions about its assumed masslessness. The discussion clarifies that while tensions differ at the pulley, they remain constant along the length of a massless string. This distinction is crucial for understanding the dynamics of the system.
Jzhang27143
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Suppose there is a pulley (a disc) of mass m1 and a string passes over the pulley with masses m2 and m3 hanging on both ends of the string with m3 > m2. I know that the acceleration should be (m3 - m2)g/(1/2m1 + m2 + m3) and I know how to get there.

However, since the pulley rotates and has mass, the tensions are different on the two ends of the string. That would mean that the string is not massless because there would be an infinite acceleration if the string was massless. However, the mass of the string appears nowhere in the final answer. Why is this so?
 
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Jzhang27143 said:
should be (m3 - m2)g/(1/2m1 + m2 + m3) and I know how to get there.
I(m1), since m1 is going to be rotating.
Jzhang27143 said:
infinite acceleration if the string was massless.
The problem statement has attached masses to the ends of the string.
 
I don't think I made myself clear. An identical problem is here: http://physics.bu.edu/~duffy/semester1/c14_atwood2.html. My question is how we can assume the string is massless because it is clear that the tension of the two sides of the string is different (the pulley would not rotate if the tensions were the same.) Since the tensions are different, there is a net force somewhere on the string and that would yield infinite acceleration on the string if the string was massless.
 
Jzhang27143 said:
Since the tensions are different, there is a net force somewhere on the string
Why do you think that?
 
Jzhang27143 said:
Suppose there is a pulley (a disc) of mass m1 and a string passes over the pulley with masses m2 and m3 hanging on both ends of the string with m3 > m2. I know that the acceleration should be (m3 - m2)g/(1/2m1 + m2 + m3) and I know how to get there.

However, since the pulley rotates and has mass, the tensions are different on the two ends of the string. That would mean that the string is not massless because there would be an infinite acceleration if the string was massless. However, the mass of the string appears nowhere in the final answer. Why is this so?
The tensions are different in the two hanging parts of the string, but do not change along the length, if the string is massless. The tensions are the same at both ends of a string.
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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