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Let's say I've an Atwood's machine in which I have two point masses which are different. Now, when I accelerate the pulley, the two masses accelerate at a different rate which can be shown by calculation. The proof is given in Kleppner's book of mechanics, where he showed the acceleration to be \begin{equation}\frac{(2A+g)M2-M1g}{M1+M2}\end{equation} for mass 1 and likewise for M2.

Note that it's different from the general Atwood's machine because the pulley itself is being accelerated upwards. In the ideal case, we assume that the rope cannot be stretched. However, in this case, as the two masses are accelerating at different rates, the rope must stretch to support the two masses, right? So, are we giving up our assumption of the rope being inexpandible here or is something wrong with my concept?

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2 Answers 2

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There is no flaw in the question. It's all about frames of reference.

If you watch the motions of both the objects from the pulley's frame of reference then the 2 objects will have the same acceleration magnitudes (in opposite directions). But if you observe the motion from the ground's frame of reference, the acceleration:

1) Of bigger mass: A-a

2) Of smaller mass: A+a

If A has been directed upwards.' a' is the acceleration of the masses from pulley's frame of reference. You may calculate that using a pseudo force.

This is called relative motion. I would encourage you to watch a few videos on relative motion and then try to solve this question. It is a very easy one.

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  • $\begingroup$ So, from an inertial frame, will I observe that the rope is being stretched? $\endgroup$ Jan 27, 2019 at 11:41
  • $\begingroup$ Nope. It's just the illusion due to frame of reference. $\endgroup$ Jan 27, 2019 at 12:11
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We can check if the rope stretches by calculating its total length during the acceleration. Since the acceleration is constant for the masses and the pulley, this makes things much simpler. Let's further simplify by assuming all parts of the machine start at rest and the hanging masses start at the same height. The length of the rope is $L$ and we'll set the starting height of the pulley is zero, so the masses start out hanging down at a height of $-L/2$. So, the equations of motion for every part is: $$y_P(t) = \frac{1}{2}At^2$$ $$y_{m_1} = \frac{1}{2}\left(\frac{(2A+g)M_2-M_1g}{M_1+M_2}\right)t^2 - L/2$$ $$y_{m_2} = \frac{1}{2}\left(\frac{(2A+g)M_1-M_2g}{M_2+M_1}\right)t^2 - L/2$$ In these equations, $y_P$ is the position of the pulley and $y_{m_1}$ and $y_{m_2}$ are the positions of mass 1 and mass 2. The total length of the rope is the sum of the distances from the pulley to the masses. \begin{align} L(t) &= \left[y_P(t) - y_{m_1}(t)\right] + \left[y_P(t) - y_{m_2}(t)\right] \\ &= 2y_P(t) - y_{m_1}(t) - y_{m_2}(t) \end{align} If you substitute in the expressions for the heights of the pulley and masses, you will find that the length of the rope is a constant $L.$

Furthermore, this works in any frame of reference, even non-inertial ones. If an observer is moving with some random motion given by $y_O(t)$, this observer will see the parts of the machine moving in this coordinate system as $y_P(t) - y_O(t),$ $y_{m_1}(t) - y_O(t),$ and $y_{m_2}(t) - y_O(t).$ Looking at the expression for $L(t)$, we can see that the function $y_O(t)$ will be canceled out, leaving the original answer $L.$

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