Question

Consider the Atwood machine of Figure \(4.15,\) but suppose that the pulley has radius \(R\) and moment of inertia \(I\). (a) Write down the total energy of the two masses and the pulley in terms of the coordinate \(x\) and \(\dot{x}\). (Remember that the kinetic energy of a spinning wheel is \(\frac{1}{2} I \omega^{2}\).) (b) Show (what is true for any conservative one-dimensional system) that you can obtain the equation of motion for the coordinate \(x\) by differentiating the equation \(E=\) const. Check that the equation of motion is the same as you would obtain by applying Newton's second law separately to the two masses and the pulley, and then eliminating the two unknown tensions from the three resulting equations.

Short answer

The equation of motion is \(\ddot{x} = \frac{(m_1 - m_2) g}{m_1 + m_2 + \frac{I}{R^2}}\).

Step-by-step solution

Understand the System

An Atwood machine consists of two masses, say \( m_1 \) and \( m_2 \), connected by a string over a pulley. In this problem, the pulley has a radius \( R \) and a moment of inertia \( I \). The motion is described by the displacement \( x \) of one of the masses.

Write Down Kinetic Energy

The kinetic energy \( T \) of the system comes from three parts: the translational kinetic energies of the two masses and the rotational kinetic energy of the pulley. The kinetic energy of mass \( m_1 \) is \( \frac{1}{2} m_1 \dot{x}^2 \), for mass \( m_2 \) it is \( \frac{1}{2} m_2 \dot{x}^2 \), and for the pulley, it's \( \frac{1}{2} I \omega^2 \) where \( \omega = \frac{\dot{x}}{R} \). Therefore, the total kinetic energy is \[ T = \frac{1}{2} m_1 \dot{x}^2 + \frac{1}{2} m_2 \dot{x}^2 + \frac{1}{2} I \left(\frac{\dot{x}}{R}\right)^2. \]

Write Down Potential Energy

The potential energy \( U \) of the system is due to gravity acting on the masses. If we consider the height of mass \( m_1 \) to be \( x \) and mass \( m_2 \) to be \(-x\), the potential energy is given by \[ U = m_1 g x - m_2 g x. \]

Total Energy Expression

The total energy \( E \) of the system, which is a sum of the kinetic and potential energies, can be written as \[ E = T + U = \frac{1}{2} m_1 \dot{x}^2 + \frac{1}{2} m_2 \dot{x}^2 + \frac{1}{2} I \left(\frac{\dot{x}}{R}\right)^2 + m_1 g x - m_2 g x. \] This can be simplified further if needed.

Differentiate Energy to Find Equation of Motion

Using the principle that energy \( E \, = \, \text{const} \), differentiate the expression for \( E \) with respect to time \( t \). The derivative of potential energy gives force terms and the derivative of kinetic energy gives acceleration terms. This leads to the equation \[ (m_1 + m_2) \ddot{x} + \frac{I}{R^2} \ddot{x} = (m_1 - m_2)g. \] Rearranging gives the equation of motion, \[ \ddot{x} = \frac{(m_1 - m_2) g}{m_1 + m_2 + \frac{I}{R^2}}. \]

Verify with Newton's Second Law

For each mass and the pulley, apply Newton's second law: for \( m_1 \), \( T_1 - m_1 g = m_1 a \); for \( m_2 \), \( T_2 - m_2 g = m_2 a \); and for the pulley, \( T_1 R - T_2 R = I \alpha \) (with \( \alpha = \frac{a}{R} \)). Solve these equations simultaneously to eliminate \( T_1 \) and \( T_2 \) and confirm the equation for \( \ddot{x} \) matches the one derived from energy conservation.

Key concepts

Moment of Inertia

The moment of inertia is a fundamental concept in rotational dynamics. It is often referred to as the rotational equivalent of mass in linear motion. In the context of an Atwood machine, especially when the pulley has a non-negligible moment of inertia, this concept becomes crucial to understanding the dynamics of the system.

The moment of inertia, usually denoted as \( I \), depends on the mass distribution of an object about the axis of rotation. For a simple solid disk, for instance, this is calculated as \( I = \frac{1}{2} m R^2 \), where \( m \) is the mass of the disk and \( R \) is the radius.

In an Atwood machine setup, consider a pulley with radius \( R \) and moment of inertia \( I \). When one mass falls and another rises, the rotation of the pulley must be taken into account. This not only adds complexity to the system but also contributes to the total kinetic energy, as shown in the problem where the rotational kinetic energy is expressed as \( \frac{1}{2} I \omega^2 \), with \( \omega \) being the angular velocity tied to the linear velocity by \( \omega = \frac{\dot{x}}{R} \).

This means the moment of inertia directly influences how the rotational motion of the pulley is coupled to the linear motion of the masses, impacting the system's overall dynamics and energy distribution.

Kinetic Energy

Kinetic energy is the energy associated with motion, and it plays a vital role in understanding the dynamics of the Atwood machine. In this system, both translational and rotational kinetic energies are considered.

• The translational kinetic energy arises from the motion of the two masses connected through a string on the Atwood machine. For each mass, say \( m_1 \) and \( m_2 \), the translational kinetic energy can be expressed as \( \frac{1}{2} m_1 \dot{x}^2 \) and \( \frac{1}{2} m_2 \dot{x}^2 \), respectively.
• The rotational kinetic energy, on the other hand, is due to the spinning of the pulley. For a pulley with moment of inertia \( I \) and angular velocity \( \omega \), this energy is given by \( \frac{1}{2} I \omega^2 \).

The angular velocity \( \omega \) relates to the linear velocity through \( \omega = \frac{\dot{x}}{R} \). Thus, the rotational kinetic energy of the pulley can be rewritten in terms of \( \dot{x} \), integrating the rotational effects into the larger system behavior.

By summing the translational and rotational components, the total kinetic energy \( T \) of the Atwood machine is quantified as shown in the step-by-step solution, playing a critical part in forming the total energy equation and subsequent motion analysis.

Potential Energy

Potential energy is the stored energy of a system due to its position or configuration. For the Atwood machine, this energy arises because of gravitational forces acting on the two masses.

When considering gravitational potential energy, the height of the masses above a reference point determines their energy. For mass \( m_1 \) positioned at height \( x \), its potential energy is \( m_1 g x \). Conversely, for mass \( m_2 \) at a height of \(-x\), its potential energy becomes \(-m_2 g x\).

• The potential energy difference generated as the masses move reflects how energy shifts between kinetic and potential forms in the system.
• It's vital for determining the total mechanical energy of the system \( E = T + U \), which remains constant for a conservative system, barring external forces.

The expression for potential energy showcases how gravitational forces influence the energy state of the Atwood machine. Understanding this concept is key to analyzing the equilibrium and dynamics of the system's motion, particularly when integrated with kinetic energy and moment of inertia insights to complete the picture.