In Fig. 9-77, two identical containers of sugar are connected by a cord that passes over a frictionless pulley. The cord and pulley have negligible mass, each container and its sugar together have a mass of 500 g, the centers of the containers are separated by 50 mm, and the containers are held fixed at the same height. What is the horizontal distance between the center of container 1 and the center of mass of the two-container system (a) initially and (b) after 20 g of sugar is transferred from container 1 to container 2? After the transfer and after the containers are released, (c) in what direction and (d) at what acceleration magnitude does the center of mass move?

1. Mass of each container with sugar, $m_1=m_2=500\mathrm{g}$.
2. The separation between containers, $d$is 50 mm .

You use the concept of the position of the center of mass and acceleration of the center of mass. You can find the position of the center of mass initially from the given separation distance. Then using Newton’s second law, we can write an equation for net force for both containers; by solving them simultaneously, you can find acceleration. Using that acceleration in the equation of acceleration of the center of mass, you get the answer.

The separation between two containers is given. From the figure, we can see that this separation gives the diameter of the pulley, and the cord is connected at the centers of the containers. Both containers areat the samedistance from the pulley, so the center of mass of them will be at half of the distance between them which is 25 mm .

So, the center of mass is at 25 mm from the center of container 1.

Center of mass of two container systems after 20 g sugar is transferred from container 1 to container 2:

When 20g of sugar from the left container is transferred to the right container, the mass of the left container will be $m_1=480\mathrm{g}$ , and the mass of the right container will be $m_2=520\mathrm{g}$ , and the distance of the left container from the center point is $x_1=-25\mathrm{mm}$ and the right container is $x_2=+25\mathrm{mm}$.

The position of the center of mass of the system can be calculated as,

$x_{com}=\frac{\left(m_1{x}_1+m_2{x}_2\right)}{m_1+m_2}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} x_{com}=\frac{\left(\left(480\right)\left(-25\right)+\left(520\right)\left(25\right)\right)}{\left(480\right)+\left(520\right)} \\ =1.0\mathrm{mm} \end{gathered}$$

So, the center of mass moves 1.0 mm toward the right from the initial position of the center of mass, that will be $25+1=26\mathrm{mm}$ .

Therefore, the center of mass is at $x_{com}=26\mathrm{mm}$ from container 1.

After the sugar is transferred and when the system is released, the center of mass remains closer to container 2. As container 2 is heavier, it will move downward, so the center of mass will move downward.

Thus, After the sugar is transferred and after containers are released, the direction of the center of mass is downward.

First, you find the acceleration of the system. Tension in the cord is the same everywhere, and it is in the upward direction. You can consider the acceleration positivealong the upwarddirection of motion and negativealong the downwarddirection of motion. You can apply Newton’s second law. You can write the equation for net force on each container and can solve for acceleration.

For container 1 , we can write,

$m_{1}a=T-m_{1}g$

For container 2 , we can write,

$-m_{2}a=-m_{2}g+T$

Rearranging this equation for T, we get,

$T=m_{2}g-m_{2}a$

Plugging this value in equation (1), we get,

$m_{1}a=m_{2}g-m_{2}a-m_{1}g$

We can write,

$$\begin{gathered} m_{1}a+m_{2}a=m_{2}g-m_{1}g \\ \left(m_1+m_2\right)a=\left(m_2-m_1\right)g \\ a=\frac{m_2-m_1}{m_1+m_2}g \end{gathered}$$ (2)

Now we can write the equation for the center of mass of acceleration; we get,

$a_{com}=\frac{\left(m_1{a}_1+m_2\left(-a_2\right)\right)}{m_1+m_2}$

As the acceleration is the same for both the containers, we can write,

$$\begin{gathered} a_{com}=\frac{\left(m_1{a}_1+m_2\left(-a_2\right)\right)}{m_1+m_2} \\ =-\frac{\left(m_2-m_1\right)}{m_1+m_2}a \end{gathered}$$

Using the value of a from equation (2) in the above equation, we get,

$$\begin{gathered} a_{com}=\frac{\left(m_2-m_1\right)}{m_1+m_2}\times \frac{m_2-m_1}{m_1+m_2}g \\ =\frac{{\left(m_2-m_1\right)}^2}{{\left(m_1-m_2\right)}^2}g \end{gathered}$$

Plugging the values we get,

$$\begin{gathered} a_{com}=\frac{{\left(520-480\right)}^2}{{\left(480+520\right)}^2}\left(9.8\right) \\ =1.6\times {10}^{-2}m/s^2 \end{gathered}$$.

Thus, after the transfer and after containers are released, the magnitude of the acceleration of the center of mass, $a_{com}$is $1.6\times {10}^{-2}\mathrm{m}/{\mathrm{s}}^2$.