Objects and are connected by a string of negligible mass and suspended vertically over a pulley of negligible mass, creating an atwood’s machine, as shown in the figure. the objects are initially at rest, and the mass of object is greater than the mass of object . as object falls, how does the kinetic energy of the center of mass of the two-object system change? justify your selection. all frictional forces are considered to be negligible.

Question: Objects and are connected by a string of negligible mass and suspended vertically over a pulley of negligible mass, creating an atwood’s machine, as shown in the figure. the objects are initially at rest, and the mass of object is greater than the mass of object . as object falls, how does the kinetic energy of the center of mass of the two-object system change? justify your selection. all frictional forces are considered to be negligible.

In this blog post, we will explore the concept of kinetic energy of the center of mass of a system of two objects. We will use an example of an Atwood's machine, which is a device that consists of two objects of different masses attached by a string and suspended over a pulley. We will assume that the string and the pulley have negligible mass and that there is no friction in the system.

The figure below shows an Atwood's machine with two objects, A and B, where A has a greater mass than B. The objects are initially at rest, but when released, A will fall and B will rise due to the gravitational force acting on them.


The question we want to answer is: how does the kinetic energy of the center of mass of the two-object system change as A falls and B rises?

To answer this question, we need to recall the definition of kinetic energy and the center of mass. The kinetic energy of an object is given by:

$$K = \frac{1}{2}mv^2$$

where m is the mass and v is the speed of the object. The center of mass of a system of objects is the point where the total mass of the system can be considered to be concentrated. The position of the center of mass is given by:

$$\vec{r}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$$

where m_i and r_i are the mass and position vector of each object in the system. The speed of the center of mass is given by:

$$\vec{v}_{cm} = \frac{d\vec{r}_{cm}}{dt}$$

The kinetic energy of the center of mass is then given by:

$$K_{cm} = \frac{1}{2}M\vec{v}_{cm}^2$$

where M is the total mass of the system.

Now, let us apply these formulas to our example. We can choose a coordinate system where the positive y-axis points upward and the origin is at the center of the pulley. We can also label the lengths of the string segments as L_A and L_B, as shown in the figure.


The position vectors of A and B are then given by:

$$\vec{r}_A = -L_A \hat{j}$$

$$\vec{r}_B = L_B \hat{j}$$

The position vector of the center of mass is then given by:

$$\vec{r}_{cm} = \frac{m_A \vec{r}_A + m_B \vec{r}_B}{m_A + m_B}$$

Simplifying, we get:

$$\vec{r}_{cm} = \frac{(m_A - m_B)(L_A - L_B)}{2(m_A + m_B)} \hat{j}$$

The speed vector of A and B are then given by:

$$\vec{v}_A = -\frac{dL_A}{dt} \hat{j}$$

$$\vec{v}_B = \frac{dL_B}{dt} \hat{j}$$

The speed vector of the center of mass is then given by:

$$\vec{v}_{cm} = \frac{d\vec{r}_{cm}}{dt}$$

Using the chain rule, we get:

$$\vec{v}_{cm} = \frac{(m_A - m_B)(-dL_A/dt - dL_B/dt)}{2(m_A + m_B)} \hat{j}$$

Simplifying, we get:

$$\vec{v}_{cm} = -\frac{(m_A - m_B)v}{2(m_A + m_B)} \hat{j}$$

where v is the common speed of A and B (since the string is inextensible).

The kinetic energy of A and B are then given by:

$$K_A = \frac{1}{2}m_A v^2$$

$$K_B = \frac{1}{2}m_B v^2$$

The kinetic energy of the center of mass is then given by:

$$K_{cm} = \frac{1}{2}(m_A + m_B)\left(-\frac{(m_A - m_B)v}{2(m_A + m_B)}\right)^2$$

Simplifying, we get:

$$K_{cm} = \frac{(m_A - m_B)^2 v^2}{8(m_A + m_B)}$$

Now, we can see that the kinetic energy of the center of mass depends only on the difference in mass and the common speed of the objects. As A falls and B rises, the speed of the objects increases, and so does the kinetic energy of the center of mass. Therefore, we can conclude that **the kinetic energy of the center of mass of the two-object system increases as object A falls**.

We can also justify this result by using the conservation of energy principle. The total mechanical energy of the system is given by the sum of the kinetic and potential energies of the objects:

$$E = K_A + K_B + U_A + U_B$$

where U is the gravitational potential energy. Since there is no friction or external force in the system, the total mechanical energy is conserved:

$$E = \text{constant}$$

The potential energy of A and B are given by:

$$U_A = -m_A g L_A$$

$$U_B = m_B g L_B$$

where g is the gravitational acceleration. Substituting these expressions into the conservation of energy equation, we get:

$$\frac{1}{2}m_A v^2 + \frac{1}{2}m_B v^2 - m_A g L_A + m_B g L_B = \text{constant}$$

Rearranging, we get:

$$(m_A + m_B) v^2 - 2 g (m_A L_A - m_B L_B) = \text{constant}$$

Dividing by 4(m_A + m_B), we get:

$$\frac{(m_A - m_B)^2 v^2}{8(m_A + m_B)} - \frac{g (m_A - m_B)(L_A - L_B)}{2(m_A + m_B)} = \text{constant}$$

Comparing this equation with the expression for the kinetic energy of the center of mass, we can see that:

$$K_{cm} - \frac{g (m_A - m_B)(L_A - L_B)}{2(m_A + m_B)} = \text{constant}$$

Therefore, if K_cm increases, then the term in parentheses must decrease. This means that as A falls and B rises, L_A - L_B decreases, which makes sense since the length of the string is constant. This also shows that **the kinetic energy of the center of mass of the two-object system increases as object A falls**.

We hope you enjoyed this blog post and learned something new about kinetic energy and center of mass. If you have any questions or comments, please feel free to leave them below. Thank you for reading!