# "Shifting" non-inertial frame in Accelerating Atwood Machine

• ln(
In summary, the conversation discusses the problem of determining the acceleration of individual masses in an accelerating Atwood machine. The correct approach is to consider the additional inertial force on any massive object in an accelerated frame, which is equal to the object's mass multiplied by the acceleration and acts as a gravitational force. This means that in an accelerating frame, the system is equivalent to a non-accelerating one with a different gravitational acceleration. This also affects the tension in the string and the forces acting on the masses in the system.
ln(
Hello,

I was referring to this thread: https://www.physicsforums.com/threads/accelerating-atwoods-machine-problem.44305/ to solve a problem on accelerating atwood machines in which, naturally, one attempts to find the acceleration of each of the individual masses.

Although I now understand how to solve the problem thanks to that thread, I have difficulty identifying the issue with my first approach, where I solve for the accelerations as if the atwood machine were NOT accelerating and then adding the acceleration of the entire system to these accelerations. Doing this results in a different answer than the correct approach mentioned in the thread.

What is wrong with this approach? Thanks.

PS: My apologies if this should be in the Homework section. I was unsure where to put it since technically I know how to do the problem; I am just unsure of the general nature of changing our reference frame in this instance.

Here is some food for thought. Imagine the Atwood machine was in free fall. How would the masses move? How does that change when you stop the free fall?

Orodruin said:
Here is some food for thought. Imagine the Atwood machine was in free fall. How would the masses move? How does that change when you stop the free fall?
I'm not sure, since my intuition is clearly wrong. Without freefall, solving the accelerations is easy. With freefall, I clearly cannot just subtract g from the accelerations in the system without freefall.

Will there be any tension in the string in free fall? (Imagine placing an Atwood machine on the International Space Station, which is in free fall)

Orodruin said:
Will there be any tension in the string in free fall? (Imagine placing an Atwood machine on the International Space Station, which is in free fall)
I believe not.

So therefeore the acceleration is?

Orodruin said:
So therefeore the acceleration is?
Just g downwards.

ln( said:
Just g downwards.
Yes, for all masses. This is clearly different from just adding the gravitational acceleration compared to the case when the system is not in free fall. When the system is in free fall, all masses are falling with the same acceleration.

Does that help?

Orodruin said:
Yes, for all masses. This is clearly different from just adding the gravitational acceleration compared to the case when the system is not in free fall. When the system is in free fall, all masses are falling with the same acceleration.

Does that help?
Yes, thank you!

The other issue is that my "logic" here is, to my understanding, essentially what ehild applied in his approach with the accelerations relative to the pulley., with the difference being that he said that ##a_r = -a_r## and ##a_1 = a-a_r## and ##a_2 = a+a_r## during the derivation while I attempt to apply this after the fact.
Although what I did is wrong, and thank you for giving me an intuitive sense of that, I still don't know intuitively WHEN it is appropriate to actually apply this transformation (as in adding the ##a## to the accelerations relative to the pulley itself).

Another possibility is that such a transformation (adding ##a## to the accelerations relative to the pulley) is always valid. If this is indeed the case, that means what I find by pretending there is zero acceleration on the system is not actually the accelerations of the masses with respect to the pulley. So then, how can you find them while accelerating with the pulley itself, i.e. while in the frame of reference of the pulley?

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In an accelerated frame there is an additional inertial force on any massive object. This force is equal to the object’s mass multiplied by the acceleration and opposite in direction to the acceleration. Thus, it acts exactly as a gravitational force and the accelerating system is equivalent to the non-accelerating one, just with what appears to be a different gravitational acceleration.

Orodruin said:
In an accelerated frame there is an additional inertial force on any massive object. This force is equal to the object’s mass multiplied by the acceleration and opposite in direction to the acceleration. Thus, it acts exactly as a gravitational force and the accelerating system is equivalent to the non-accelerating one, just with what appears to be a different gravitational acceleration.
Ah, fantastic.

So does this mean that, given a mass on a table in an elevator, the table will have to support the mass with more force if the elevator is accelerating upwards than if it were on ground?

And say we have a mass of mass ##m_1## connected to one side of a simple atwood machine and ##m_2## to the other and ##m_2 > m_1##. Let this system be accelerating with acceleration ##a'##. I will use the convention of ##a##, the acceleration of the masses relative to the pulley, going in the direction of the overall motion. If I want to write down expression of the forces acting on them with respect to the accelerated frame, is it correct to say that $$m_1a = F_\textrm{tension} - m_1g - m_1a'$$ and $$m_2a = m_2g + m_2a' - F_\textrm{tension}$$ This seems to be correct, at least.

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## 1. What is the concept of "shifting" a non-inertial frame in an Accelerating Atwood Machine?

The concept of "shifting" a non-inertial frame in an Accelerating Atwood Machine refers to the process of changing the perspective from which we observe the system. This is necessary because the Atwood Machine is a non-inertial frame of reference, meaning that it is accelerating and therefore does not follow the laws of classical mechanics.

## 2. Why is it important to understand how to shift a non-inertial frame in an Accelerating Atwood Machine?

Understanding how to shift a non-inertial frame in an Accelerating Atwood Machine is important because it allows us to accurately describe and analyze the motion of objects in this system. Without taking into account the non-inertial frame, our observations and calculations may be incorrect.

## 3. How do we shift a non-inertial frame in an Accelerating Atwood Machine?

To shift a non-inertial frame in an Accelerating Atwood Machine, we use the concept of fictitious forces. These are forces that appear to act on objects in a non-inertial frame, but are actually a result of the frame's acceleration. By including these fictitious forces in our calculations, we can shift our perspective and accurately describe the motion of objects in the system.

## 4. Can the concept of shifting a non-inertial frame be applied to other systems besides the Accelerating Atwood Machine?

Yes, the concept of shifting a non-inertial frame can be applied to any system that is accelerating. This includes systems such as a rotating carousel or a swinging pendulum. In these cases, we must also take into account the fictitious forces that arise due to the frame's acceleration.

## 5. Are there any practical applications of understanding how to shift a non-inertial frame in an Accelerating Atwood Machine?

Yes, understanding how to shift a non-inertial frame in an Accelerating Atwood Machine has practical applications in fields such as engineering and navigation. For example, when designing structures or vehicles that are subject to acceleration, it is important to take into account the non-inertial frame and the resulting fictitious forces. In navigation, understanding how to shift a non-inertial frame allows for more accurate measurements and predictions of motion in systems that experience acceleration.

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