Tension in Atwood machine revisited

Question

In their book "Theoretical Mechanics of Particles and Continua", Fetter and Walecka ask to solve the problem (problem 1.3 of Chapter 1) of an Atwood machine in which a massless string is placed over a massless pulley, and each end is wound around and fastened to a vertical hoop; each hoop has a different mass and radius. The first question is about the tension in the string, which, according to them, is different by a factor of 2 from the classical treatment in which the 2 masses are treated as point masses. Hoops have mass $m_1,m_2$ and radius $R_1,R_2$. The situation is depicted as follows:

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They claim that in their situation, the tension is $T=m_1m_2g/(m_1+m_2)$ whereas when masses are treated as point masses, one gets $T=2m_1m_2g/(m_1+m_2)$. I am wondering how to get this factor of 2 difference. The whole point is the qualifier of the string being "fastened" to the hoops! So it sounds like the string is not unwinding ... but perhaps I am misinterpreting the text!

Answer 1

Here is a trial solution using Newtonian mechanics. The notation is shown in the following figure :

Gravity is oriented along the positive x axis.

First let us establish a relation between the 4 parameters $x_1,x_2,\theta_1$ and $\theta_2$. If the pulley rotates by $d\theta$, it contributes to the displacement of $m_1$ by $dx_1=R*d\theta$, where $R$ is the radius of the pulley; if the hoop rotates by $d\theta_1$,it contributes to the displacement of $m_1$ by $dx_1=R_1*d\theta_1$. So the overall displacement of $m_1$ is given by $dx_1=R*d\theta+R_1*d\theta_1$. For $m_2$, taking into account sign conventions, one has a similar expression $dx_2=-R*d\theta-R_2*d\theta_2$. Summing these 2 equations in order to eliminate the pulley parameters lead to :

$$dx_1+dx_2=R_1*d\theta_1-R_2*d\theta_2 \tag{1}\label{eq1}$$

We will then apply the law of angular momentum change for hoop 1, angular momentum $L_1$ being calculated at point A. We have $L_1=I_1*\dot\theta_1$, where $I_1$ is the moment of inertia of hoop 1 about an axis perpendicular to the figure passing by point A; by the parallel axis theorem $I_1=2*m_1*R_1^2$. The only force which provides moment is the weight of hoop 1 since the string tension goes through A; thus we have $I_1*\ddot\theta_1=m_1*g*R_1$, which reduces to :

$$\ddot\theta_1=\frac{g}{2*R_1} \tag{2}\label{eq2}$$

Similarly for hoop2 one gets :

$$\ddot\theta_2=-\frac{g}{2*R_2} \tag{3}\label{eq3}$$

We can now combine equations \eqref{eq1}, \eqref{eq2} and \eqref{eq3} to obtain : $$\ddot x_1+\ddot x_2=R_1*\ddot\theta_1-R_2*\ddot\theta_2=g \tag{4}\label{eq4}$$

Finally we will write Newton's law for the displacements of the 2 hoops : $$m_1*g-T=m_1*\ddot x_1$$ $$m_2*g-T=m_2*\ddot x_2$$ which can be combined to give : $$\ddot x_1+\ddot x_2=2*g-T*(\frac{1}{m_1}+\frac{1}{m_2})$$

Using \eqref{eq4}, one finally gets the desired result :

$$T=\frac{m_1*m_2}{m_1+m_2}*g$$

Answer 2

As already suggested previously (How to use Lagrangian Mechanics, not Newtonian Mechanics, to find tension in a system of two hoops attached to a massless string over a pulley?) the same problem can be solved using Lagrangian Mechanics.

In that case, using the same notation, we can write the Lagrangian of the system :

$$L=\frac{m_1}{2}\dot x_1^2+\frac{m_2}{2}\dot x_2^2+\frac{m_1\cdot R_1^2}{2}\dot \theta_1^2+\frac{m_2\cdot R_2^2}{2}\dot \theta_2^2+m_1\cdot g\cdot x_1+m_2\cdot g\cdot x_2$$ with the previously established constraint written as : $R_2\cdot \dot\theta_2=R_1\cdot \dot\theta_1-\dot x_1-\dot x_2$.

This allows to eliminate $\dot\theta_2$ from the previous expression of the Lagrangian, which can then be expressed with only 3 generalized coordinates : $x_1$, $x_2$, $\theta_1$ and their derivatives :

$$L=\frac{m_1}{2}\dot x_1^2+\frac{m_2}{2}\dot x_2^2+\frac{m_1\cdot R_1^2}{2}\dot \theta_1^2+\frac{m_2}{2}\cdot (R_1\cdot \dot\theta_1-\dot x_1-\dot x_2)^2+m_1\cdot g\cdot x_1+m_2\cdot g\cdot x_2$$

Starting from this expression one can use Euler-Lagrange equations to obtain the following 3 equations for the movement :

$$m_1\cdot \ddot x_1-m_2\cdot (R_1\ddot\theta_1-\ddot x_1-\ddot x_2)=m_1\cdot g$$

$$m_2\cdot \ddot x_2-m_2\cdot (R_1\ddot\theta_1-\ddot x_1-\ddot x_2)=m_2\cdot g$$

$$R_1\cdot \ddot\theta_1=\frac{m_2}{m_1+m_2}\cdot (\ddot x_1+\ddot x_2)$$

The third equation can be used to eliminate $\ddot\theta_1$ from the first two equations which leads to the following set of equations :

$$\begin{align} (m_1+2\cdot m_2)\cdot \ddot x_1+m_2\cdot \ddot x_2&=(m_1+m_2)\cdot g \\ m_1\cdot \ddot x_1+(2\cdot m_1+m_2)\cdot \ddot x_2&=(m_1+m_2)\cdot g \end{align}$$

This simple algebraic system can be easily solved and gives : $$\ddot x_1=\frac{m_1\cdot g}{m_1+m_2}$$ and $$\ddot x_2=\frac{m_2\cdot g}{m_1+m_2}$$

At this point we leave Lagrangian approach to deduce the tension using Newton law for instance for $m_1$ : $$m_1\cdot \ddot x_1=m_1\cdot g+T$$ which gives : $$T=-\frac{m_1\cdot m_2}{m_1+m_2}\cdot g$$ which is the desired result (with the right sign, i.e. T is indeed negative). In fact we could probably stay in the Lagrangian approach using the technique of Lagrange multiplier without eliminating one of the generalized coordinates; the Lagrange multiplier would then be identified as the desired tension.