Question

Consider a mass \(m\) moving in a frictionless plane that slopes at an angle \(\alpha\) with the horizontal. Write down the Lagrangian in terms of coordinates \(x,\) measured horizontally across the slope, and \(y\), measured down the slope. (Treat the system as two-dimensional, but include the gravitational potential energy.) Find the two Lagrange equations and show that they are what you should have expected.

Short answer

Lagrangian: \(L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - mg y \sin\alpha\). Equations: \(m\dot{x} = \text{constant}\), \(\ddot{y} = -g\sin\alpha\).

Step-by-step solution

Define the coordinates and kinetic energy

We define the coordinates: horizontal position across the slope as \(x\) and down the slope as \(y\). The kinetic energy of the mass \(m\) is given by the expression \(T = \frac{1}{2}m(v_x^2 + v_y^2)\), where \(v_x\) and \(v_y\) are the velocities in the \(x\) and \(y\) directions, respectively.

Express velocities in terms of derivatives of coordinates

The velocities \(v_x\) and \(v_y\) can be expressed using time derivatives of \(x\) and \(y\) as follows: \(v_x = \frac{dx}{dt}\) and \(v_y = \frac{dy}{dt}\). Thus, the kinetic energy can be rewritten as \(T = \frac{1}{2}m\left( \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 \right)\).

Determine the gravitational potential energy

The gravitational potential energy \(V\) is given by \(V = mgh\), where \(h\) is the height. In this system, \(h = y \sin\alpha\), leading to the potential energy \(V = mg y \sin\alpha\).

Write down the Lagrangian

The Lagrangian \(L\) is defined as \(L = T - V\). Using the expressions derived for kinetic and potential energies, the Lagrangian becomes \(L = \frac{1}{2}m\left( \dot{x}^2 + \dot{y}^2 \right) - mg y \sin\alpha\).

Apply Euler-Lagrange equation to find the equations of motion

The Euler-Lagrange equation is \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0\) for each coordinate \(q\). For \(x\), this yields: \[\frac{d}{dt}(m\dot{x}) = 0 \Rightarrow m\dot{x} = C,\] where \(C\) is a constant. For \(y\), we get: \[m\ddot{y} = -mg\sin\alpha \Rightarrow \ddot{y} = -g\sin\alpha.\]

Interpret the results

The equation for \(x\) indicates constant horizontal velocity, typical for motion without external forces. The equation for \(y\) reflects motion under constant gravitational acceleration projected down the slope, consistent with expected mechanics.

Key concepts

Euler-Lagrange equation

In Lagrangian mechanics, the Euler-Lagrange equation is a fundamental tool used to derive the equations of motion for a system. It is a statement about how nature operates optimally, providing the path that minimizes the action (a particular quantity in physics). The equation is expressed as:

• \( \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0 \).

Here, \( L \) represents the Lagrangian, \( q \) is the generalized coordinate, and \( \dot{q} \) is the time derivative of this coordinate (essentially its velocity).
The Lagrangian \( L \) is a function usually given by \( L = T - V \), where \( T \) is the kinetic energy and \( V \) is the potential energy. By plugging these into the Euler-Lagrange equation, we can derive differential equations that describe the system's dynamics.
The Euler-Lagrange equation in our problem demonstrates how forces act on the mass as it moves. For the coordinate \( x \), the equation finds that there is no change in velocity, indicating that horizontal motion is unhindered by forces. In contrast, for the coordinate \( y \), the equation reflects the gravitational pull acting down the slope, illustrating how gravity influences the system through its component along the slope.

kinetic energy

Kinetic energy is a crucial concept in classical mechanics, representing the energy that a body possesses due to its motion. The kinetic energy \( T \) for a moving object with mass \( m \) and velocity \( v \) is expressed as:

• \( T = \frac{1}{2} m v^2 \).

When dealing with multiple dimensions, the velocity \( v \) is broken down into components, such as \( v_x \) and \( v_y \), corresponding to motion along each axis.
In this exercise, the mass moves across a frictionless slope, making it important to consider kinetic energy in both directions. Therefore, we determine the kinetic energy as
\( T = \frac{1}{2} m (v_x^2 + v_y^2) \).

• The velocity components are expressed in terms of derivatives: \( v_x = \frac{dx}{dt} \) and \( v_y = \frac{dy}{dt} \).

By substituting these into our expression for kinetic energy,
we get \( T = \frac{1}{2} m \left( \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 \right) \).
This representation allows us to capture the motion of the mass accurately, considering how it moves both horizontally and down the slope.

gravitational potential energy

Gravitational potential energy (GPE) is associated with an object's position in a gravitational field. It represents the work done against gravity to bring an object to a particular height. The formula for gravitational potential energy is:

• \( V = mgh \),

where \( m \) is the mass of the object, \( g \) is the gravitational acceleration, and \( h \) refers to the height above the reference point.
In this problem, the concept of height is slightly different. Since we are dealing with a slope inclined at an angle \( \alpha \), we calculate the height \( h \) along the slope.

• Here, \( h = y \sin\alpha \), making the potential energy \( V = mg y \sin\alpha \).

This expression reflects how much work would be required to lift the object a vertical height, expressed in the direction of \( y \). By integrating this gravitational potential energy into the Lagrangian equation, we understand the influence of gravity on the system in the plane of motion.