Question

The center of a long frictionless rod is pivoted at the origin, and the rod is forced to rotate in a horizontal plane with constant angular velocity $\omega$. Write down the Lagrangian for a bead threaded on the rod, using $r$ as your generalized coordinate, where $r,\varphi$ are the polar coordinates of the bead. (Notice that $\varphi$ is not an independent variable since it is fixed by the rotation of the rod to be $\varphi =\omega t$.) Solve Lagrange's equation for $r(t).$ What happens if the bead is initially at rest at the origin? If it is released from any point $r_{\mathrm{o}}>0,$ show that $r(t)$ eventually grows exponentially. Explain your results in terms of the centrifugal force $m{\omega }^{2}r$.

Short answer

The Lagrangian is $L=\frac{1}{2}m({\dot{r}}^2+r^2{\omega }^2)$, and $r(t)=r_0{e}^{\omega t}$ grows exponentially due to centrifugal force.

Step-by-step solution

Define the system parameters

We are considering a bead of mass, $m$, moving on a rod rotating with an angular velocity $\omega$. The position of the bead is described by the polar coordinates $(r,\varphi )$ where $\varphi =\omega t$.

Write down the expression for kinetic energy

Since the bead moves in circular motion, the kinetic energy, $T$, is given by:$$T=\frac{1}{2}m({\dot{r}}^2+r^2{\omega }^2)$$Here, $\dot{r}$ is the radial velocity and $r^2{\omega }^2$ is the tangential component due to rotation about the pivot.

Write down the expression for potential energy

The rod is frictionless and there are no external fields; thus, the potential energy, $V$, is zero: $$V=0$$

Construct the Lagrangian

The Lagrangian, $L$, is the difference between the kinetic and potential energy:$$L=T-V=\frac{1}{2}m({\dot{r}}^2+r^2{\omega }^2)$$

Derive the Euler-Lagrange equation

For one coordinate $r$, the Euler-Lagrange equation is:$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{r}}\right)-\frac{\partial L}{\partial r}=0$$Here, $\frac{\partial L}{\partial \dot{r}}=m\dot{r}$ and $\frac{\partial L}{\partial r}=mr{\omega }^2$.

Solve the differential equation for $r(t)$

Plugging into the Euler-Lagrange equation, we obtain:$$\frac{d}{dt}(m\dot{r})-mr{\omega }^2=0$$Which reduces to:$$\ddot{r}=r{\omega }^2$$This is a second-order linear differential equation with constant coefficients.

Solve the differential equation

The solution to $\ddot{r}=r{\omega }^2$ is given by:$$r(t)=A{e}^{\omega t}+B{e}^{-\omega t}$$where $A$ and $B$ are constants determined by the initial conditions.

Apply the initial conditions

For a bead initially at rest at the origin, $r(0)=0$ and $\dot{r}(0)=0$, implying $A+B=0$ and $A-B=0$. Thus, $A=B=0$, so $r(t)=0$, which means the bead stays at the origin.

Analyze the condition $r_0>0$

For initial conditions where $r(0)=r_0>0$ and $\dot{r}(0)=0$, we get $r(t)=r_0{e}^{\omega t}$. This shows that $r(t)$ grows exponentially with time due to the term $A{e}^{\omega t}$.

Explain results with centrifugal force

The centrifugal force on the bead is $m{\omega }^{2}r$, which effectively tries to "fling" the bead outwards along the rod. This results in exponential growth of $r(t)$ when $r_0>0$.

Key concepts

Euler-Lagrange Equation

The Euler-Lagrange equation is a fundamental equation in Lagrangian mechanics, a formulation of classical mechanics introduced by the mathematician Joseph-Louis Lagrange. It provides a powerful way to derive the equations of motion for a system. The equation is derived from the principle of least action, which suggests that the path taken by a physical system between two states is the one for which the action integral is minimized.
The action integral is defined as the time integral of the Lagrangian function. The Lagrangian, denoted as $L$, is typically the difference between the kinetic and potential energies of the system: $L=T-V$.

• The Euler-Lagrange equation itself is $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)-\frac{\partial L}{\partial q}=0$, where $q$ is the generalized coordinate and $\dot{q}$ is its time derivative.
• Solving this equation yields the equations of motion for the system.

In the context of the exercise, the Euler-Lagrange equation is used to find $r(t)$, the radial position of the bead along the rod. The specific form $\frac{d}{dt}(m\dot{r})-mr{\omega }^2=0$ is derived considering the rotational motion of the rod and the bead.

Kinetic Energy

Kinetic energy is an essential component in dynamics and plays a significant role in the formulation of the Lagrangian. It represents the energy that an object possesses due to its motion. In mechanical systems, kinetic energy can be translational (due to linear motion) or rotational (due to rotational motion).
For a particle of mass $m$, the kinetic energy $T$ is given by $T=\frac{1}{2}m{v}^2$, where $v$ is the particle's velocity. For a system undergoing circular motion, such as the bead on the rotating rod in the exercise, the velocity $v$ incorporates two components: radial $\dot{r}$ and tangential $r\omega$. Thus, the kinetic energy expression is decomposed into:

• Radial kinetic energy: $\frac{1}{2}m{\dot{r}}^2$.
• Tangential kinetic energy: $\frac{1}{2}m{r}^2{\omega }^2$.

This comprehensive kinetic energy accounts for both the motion of the bead along the rod and its circular path around the pivot. When calculating the Lagrangian of the system $L=T-V$, this total kinetic energy plays a decisive part, especially given that there is no potential energy $V$ due to a lack of external forces.

Centrifugal Force

Centrifugal force is an apparent force that acts outward on a body moving in a circular path. It is perceived in a rotating reference frame where the body appears to be pulled away from the center of rotation. This pseudo-force is essential to consider because it has noticeable effects on objects in rotational systems.
In the context of the rotating rod, centrifugal force attempts to "fling" the bead outward as it accelerates along the rod’s length. Mathematically, this force is expressed as $F_{cent}=m{\omega }^{2}r$, where $m$ is the mass of the bead, $\omega$ is the angular velocity, and $r$ is the radial distance from the rotation axis.

• The presence of centrifugal force in the equation $\ddot{r}=r{\omega }^2$ reveals how it contributes to the exponential growth of $r(t)$, highlighting its pivotal role in driving the bead outward.
• Initially, if the bead is located at $r=0$, there is no centrifugal force acting on it, and it remains stationary.
• If released from a position $r_0>0$, the bead feels the centrifugal pull, causing $r(t)$ to expand exponentially.

Understanding this concept is crucial for solving and interpreting motion in rotating systems like the one described in the exercise.