Question

A mass \(m\) is suspended from a massless string, the other end of which is wrapped several times around a horizontal cylinder of radius \(R\) and moment of inertia \(I\), which is free to rotate about a fixed horizontal axle. Using a suitable coordinate, set up the Lagrangian and the Lagrange equation of motion, and find the acceleration of the mass \(m\). [The kinetic energy of the rotating cylinder is \(\frac{1}{2} I \omega^{2} .\)]

Short answer

The acceleration of mass \(m\) is \(\ddot{y} = \frac{mg}{m + \frac{I}{R^2}}\).

Step-by-step solution

Define the Coordinates

Let's choose the vertical position of the mass, denoted by \( y \), as the generalized coordinate. The mass moves downward as it unwinds the string, so \( y \) increases as the mass descends.

Express Velocities in Terms of y

The velocity of the mass is \( v_m = \frac{dy}{dt} \). The angular velocity of the cylinder \( \omega \) is related to the linear velocity of the string by \( v_m = R\omega \), which gives us \( \omega = \frac{\dot{y}}{R} \).

Write the Lagrangian

The Lagrangian \( L \) is given by the kinetic energy minus the potential energy, \( L = T - V \). Kinetic energy \( T \) has two components: the translational kinetic energy of the mass \( \frac{1}{2} m \dot{y}^2 \) and the rotational kinetic energy of the cylinder \( \frac{1}{2} I \omega^2 \). Potential energy \( V = mgy \), where \( g \) is the acceleration due to gravity. Thus,\[ L = \frac{1}{2} m \dot{y}^2 + \frac{1}{2} I \left(\frac{\dot{y}}{R}\right)^2 - mgy. \]

Derive the Lagrange Equation of Motion

The Lagrange equation is given by:\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right) - \frac{\partial L}{\partial y} = 0. \]Calculate \( \frac{\partial L}{\partial \dot{y}} = m \dot{y} + \frac{I \dot{y}}{R^2} \) and \( \frac{\partial L}{\partial y} = -mg \). Differentiate \( \frac{\partial L}{\partial \dot{y}} \) with respect to time to get \( (m + \frac{I}{R^2}) \ddot{y} \). Substituting these into the Lagrange equation gives:\[ (m + \frac{I}{R^2}) \ddot{y} = mg \].

Solve for the Acceleration \(\ddot{y}\)

Rearrange the equation from Step 4 to solve for \( \ddot{y} \):\[ \ddot{y} = \frac{mg}{m + \frac{I}{R^2}}. \]

Key concepts

Lagrange Equation of Motion

The Lagrange Equation of Motion is a foundational principle in Lagrangian Mechanics, providing a method for finding the equations of motion for a system. It is particularly useful in systems involving complex constraints or where forces are not easily defined. The Lagrange equation is expressed as:\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0, \]where \(q\) represents the generalized coordinates, variables that describe the configuration of the system. In our exercise, the vertical position \(y\) of the mass acts as this generalized coordinate.
This equation involves derivatives of the Lagrangian \(L\), which is the difference between the kinetic energy \(T\) and the potential energy \(V\):

• \(T\) accounts for both translational and rotational kinetic energy.
• \(V\) represents the potential energy due to gravity.

By substituting the expressions for \(\frac{\partial L}{\partial \dot{y}}\) and \(\frac{\partial L}{\partial y}\) from the original solution, we apply the equation to derive the equation of motion of the system. This allows us to ultimately determine how the system behaves over time, finding the acceleration of the mass.

Kinetic Energy

Kinetic Energy is a crucial concept that differentiates the types of movements in a dynamic system. In our problem, it consists of two forms:

• Translational kinetic energy, which is the kinetic energy due to the motion of the mass itself, given by the formula \( \frac{1}{2} m \dot{y}^2 \).
• Rotational kinetic energy, referring to the energy of the rotating cylinder, expressed as \( \frac{1}{2} I \omega^2 \).

These energies add up to form the total kinetic energy of the system and are used in the Lagrangian to calculate the system's behavior.
In this calculation, it's crucial to note how both translational and rotational components are related. The angular velocity \( \omega \) of the cylinder is connected to the linear speed \( \dot{y} \) of the string and mass. Hence, \( \omega = \frac{\dot{y}}{R} \) is substituted into the kinetic energy expression to create a unified relationship in the Lagrangian. This link between translational and rotational movements helps us describe the entire system's dynamics comprehensively.

Rotational Dynamics

Rotational Dynamics deals with the motion of objects that rotate about an axis. In our exercise, the cylinder exhibits rotational dynamics as it spins due to the string unwinding.
The moment of inertia \(I\) plays a vital role in this process, defined as the measure of an object's resistance to changes in its rotational motion. The larger the moment of inertia, the harder it is to alter the rotation. For our cylinder, its rotational kinetic energy is deduced using this moment of inertia.
Angular velocity \(\omega\) is another key term, representing how fast the object rotates. It's linked to linear velocity through the cylinder's radius \(R\), vital in expressing the system's dynamics as the string causes rotation. The broader the radius, the greater the rotational effect for the same linear speed. The exercise shows how understanding these parameters aids in deriving the equation of motion and predicting how the system will react when forced by the descending mass. This intertwined understanding of rotational and linear elements is what makes Lagrangian mechanics such a powerful method for solving complex motion problems.