Atwood's fall machine consists of two weight of mass \(m_{1}\) and \(m_{2}\) attached with a massless rope running around a spinning wheel of mass \(M\) and radius \(R\) without slipping. The spinning wheel is attached at its center and rotates around an axis through its center without friction. (a) Find velocity of each of the weights as a function of their vertical positions. (b) Find angular velocity of the spinning wheel as a function of the vertical positions of the weights.

Newton's laws of motion are fundamental principles that describe how objects move.
In the context of Atwood's machine, these laws help us understand how the masses and the pulley interact.

• First Law (Inertia): An object will remain at rest or move in a straight line unless acted upon by an external force.
• Second Law (Force and Acceleration): This is expressed as \textbf{F = ma}, where F is the net force acting on an object, m is the mass, and a is the acceleration.
• Third Law (Action and Reaction): For every action, there is an equal and opposite reaction.

Using the second law, we analyze forces on both masses in Atwood's machine:
For mass 1: \( m_1 a_1 = m_1 g - T \)
For mass 2: \( m_2 a_2 = T - m_2 g \)
Here, T is the tension in the rope and g is the gravitational acceleration.

Torque is a measure of the force causing an object to rotate.
It is essential in analyzing the motion of the pulley in Atwood's machine.
The torque (\textbf{τ}) exerted by a force F at a radius R is given by \ τ = R \times F \
In this problem, we consider the torque acting on a pulley:
If we consider rotational motion, we use \[ \tau = I \frac{d^2 \theta}{dt^2} \]
Here, I stands for the moment of inertia, and \( \frac{d^2 \theta}{dt^2} \) is the angular acceleration.
For the pulley, \( \tau = T R \), where T is the tension in the rope.
Substituting in the moment of inertia for a solid disk, we get \[ I = \frac{1}{2} M R^2 \]
Thus, we apply the torque equation: \[ \frac{1}{2} M R^2 \frac{d^2 \theta}{dt^2} = T R \]

Kinematic equations describe the motion of objects without considering the forces that cause the motion.
They are crucial in determining the velocity and displacement of the masses in Atwood's machine.
The general forms of kinematic equations are:

• \( v = u + at \)
• \( s = ut + \frac{1}{2} at^2 \)
• \( v^2 = u^2 + 2as \)

Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, \(t\) is the time, and \(s\) is the displacement.
In our exercise, from rest (\(u = 0\)), the velocity of the weights can be determined using:
\[ v^2 = 2ax \]
Where \(x\) is the position change of the weights and \(a\) is the common acceleration found by: \[ a = \frac{(m_1 - m_2) g}{m_1 + m_2} \]

Moment of inertia indicates how difficult it is to change the rotational motion of an object.
It plays a key role in Atwood's machine when considering the pulley's rotation.
The general formula for the moment of inertia (I) depends on the mass distribution relative to the axis of rotation.
For a solid disk (like our pulley), the moment of inertia is:
\[ I = \frac{1}{2} M R^2 \]
Where M is the mass of the pulley, and R is its radius.
This value is essential in using the torque equation to find the relationship between tension, the pulley's mass, and its angular acceleration.

Angular velocity is how fast an object rotates and is especially important for the rotating pulley in an Atwood's machine.
It is often denoted by the Greek letter \( \omega \).
The relationship between angular velocity and linear velocity (V) of a point on a rotating object is given by:
\[ \omega = \frac{v}{R} \]
In our problem, once we find the linear velocity of the masses (v), we can relate it to the angular velocity of the pulley:
\[ \omega = \frac{v}{R} \]
Where \( R \) is the radius of the pulley. This gives us the rotational speed corresponding to the linear speed of the masses.