Limiting case for an Atwood's machine

[brotherbobby]

TL;DR

In Atwood's machine with masses $m_1,m_2\; (m_2 >m_1)$ , the acceleration and tension in the rope are $a = \dfrac{m_2-m_1}{m_1+m_2}g,\quad T=\dfrac{2m_1 m_2}{m_1+m_2}g$. If $m_2\gg m_1$, the tension in the rope $T=2m_1g$. In this case the system is freely falling. Shouldn't the tension in the rope be 0?

System : The Atwood's machine is well-known, and drawn alongside, where $m_2 > m_1$. The (common) acceleration of the masses $a = \dfrac{m_2-m_1}{m_1+m_2}g$ and the tension in the (massless) rope is $T=\dfrac{2m_1 m_2}{m_1+m_2}g$.

Question : Make sense of the above result(s) in the event (i) $m_2=m_1$ and (ii) $m_2\gg m_1$.

Response : (i) When the masses become equal $\left( m_1=m_2=m \right)$, $\boxed{a = 0\quad,\quad T = mg}\quad\color{green}{\large\checkmark}$. These match what we expect.

(ii) When the mass $m_2\gg m_1$, the acceleration of the system $\boxed{a = g}$, which we expect. The system is in free fall. However, the tension in the rope in this limit comes out to be $\displaystyle{T = \lim_{m_2\rightarrow\infty} \dfrac{2m_1g}{\frac{m_1}{m_2}+1}=2m_1g}\;{\color{red}{\Large\times}}$.

Surely this is wrong. In free fall, shouldn't the tension in the rope be zero? Both masses "fall" with acceleration g.

Request : A help or a hint as to how to calculate the tension in the rope in this limit. It should come out to zero, or so I suspect.

 

[jbriggs444]

TL;DR: In Atwood's machine with masses $m_1,m_2\; (m_2 >m_1)$ , the acceleration and tension in the rope are $a = \dfrac{m_2-m_1}{m_1+m_2}g\; T=\dfrac{2m_1 m_2}{m_1+m_2}g$. If $m_2\gg m_1$, the tension in the rope $T=2m_1g$. In this case the system is freely falling. Shouldn't the tension in the rope be 0?

In that situation, mass $m_2$ is approximately freely falling while mass $m_1$ is approximately accelerating upward at $g$. Of course, the tension will be approximately $T \approx 2m_1g$.

Meanwhile, the mass $m_2$ is only approximately free falling and has a much larger mass. You seem to want to approximate the tension as the product of a very large mass ($m_2$) by an acceleration difference from free fall that is very small. You are approximating very small multiplied by very large as zero times something you do not care to calculate. That failure to calculate is an error.

 

[brotherbobby]

In that situation, mass $m_2$ is approximately freely falling while mass $m_1$ is approximately accelerating upward at $2g$. Of course, the tension will be approximately $T \approx 2m_1g$.

I failed to understand. Mass $m_2\; (\rightarrow\infty)$ is accelerating downward at $g$. How is mass $m_1$ accelerating upward with acceleration $2g$? Aren't them both going down and going up with acceleration $g$?

 

[jbriggs444]

I failed to understand. Mass $m_2\; (\rightarrow\infty)$ is accelerating downward at $g$. How is mass $m_1$ accelerating upward with acceleration $2g$? Aren't them both going down and going up with acceleration $g$?

Right. That was a braino in my original. Corrected now. Apparently you read it before I fixed it.

If you go up with an acceleration of $g$ while weighing $mg$ then the required tension force is $2mg$.

 

[brotherbobby]

I failed to understand. Mass $m_2\; (\rightarrow\infty)$ is accelerating downward at $g$. How is mass $m_1$ accelerating upward with acceleration $2g$? Aren't them both going down and going up with acceleration $g$?

Oh am sorry; you made the correction. However, I find myself struggling with something else. How can the tension be $T = 2m_1g$?

Imagine the situation where both masses are going "down" with acceleration g. The system would be freely falling. The rope connecting them will be slack, the tension 0.

Isn't this scenario the same as that happening in the problem?

 

[jbriggs444]

Oh am sorry; you made the correction. However, I find myself struggling with something else. How can the tension be $T = 2m_1g$?

Imagine the situation where both masses are going "down" with acceleration g. The system would be freely falling. The rope connecting them will be slack, the tension 0.

Isn't this scenario the same as that happening in the problem?

Accelerating upward at 1g is different from freely falling downward at 1g. One requires an external force. The other does not.

 

[brotherbobby]

Accelerating upward at 1g is different from freely falling downward at 1g. One requires an external force. The other does not.

Gravity is an external force. I suppose what you mean is that if $m_1$ were to accelerate "up" with $g$, we'd need an agent to apply a force of amount $2m_1g$ on it upwards. We need no such agent to let it go "down" with acceleration $g$.
Thank you. I suppose the matter is settled unless there's something else of interest here that I am unaware of.

Yes, by the way there is, though I have solved it. How about the motion of the center of mass of the system? If you calculate the "net external force" on the system and the acceleration of the center of mass, you'd find that those values satisfy the equation $\displaystyle{\Sigma F_E = M a_C}$. Just note that the net external force on the system is not the obvious $(m_1+m_2)g$. Therein lies the interesting aspect of this question, which most students would ignore, since books don't talk about it.

Spoiler: Hint

Is there a force acting upon the system from "above"?

 

[jbriggs444]

Gravity is an external force. I suppose what you mean is that if $m_1$ were to accelerate "up" with $g$, we'd need an agent to apply a force of amount $2m_1g$ on it upwards. We need no such agent to let it go "down" with acceleration $g$.

Right. That external agent is, of course, the tension in the rope.

Thank you. I suppose the matter is settled unless there's something else of interest here that I am unaware of.

Yes, by the way there is, though I have solved it. How about the motion of the center of mass of the system? If you calculate the "net external force" on the system and the acceleration of the center of mass, you'd find that those values satisfy the equation $\displaystyle{\Sigma F_E = M a_C}$. Just note that the net external force on the system is not the obvious $(m_1+m_2)g$. Therein lies the interesting aspect of this question, which most students would ignore, since books don't talk about it.

Right. The center of mass of the two masses is accelerating downward. So the supporting force from the pulley must not be enough to match $(m_1 + m_2) g$.

By inspection, the required supporting force must be equal to $2t$.

 

[brotherbobby]

Right. That external agent is, of course, the tension in the rope.

Right. The center of mass of the two masses is accelerating downward. So the supporting force from the pulley must not be enough to match $(m_1 + m_2) g$.

By inspection, the required supporting force must be equal to $2t$.

If you put the value of the tension $2T = \dfrac{4m_1m_2}{m_1+m_2}g$ and subtract it from the force due to gravity $(m_1+m_2)g$, it would come out to be $\dfrac{(m_2-m_1)^2}{(m_1+m_2)^2}$, which is the total mass $M(=m_1+m_2)$ times the acceleration of the center of mass, $a_C=\dfrac{(m_1-m_2)a}{m_1+m_2}$. Some algebra is required to show that both sides are the same : $\Sigma F_E\; (= Mg-2T) = Ma_C$

 

[Herman Trivilino]

In free fall, shouldn't the tension in the rope be zero?

Only in the limiting case where $m_1$ is zero.

In other words, you have an ideal pulley and you hang two objects such that the mass of one of them is much much larger than the mass of the other one. But to get a tension of zero you would have to completely remove the object with the smaller mass.