Prove that Angular Acceleration of Uniform Circular Motion is Zero Using Lagrangian Method

What is The Lagrangian Method?

The Lagrangian method is a method that you can use in classical mechanics, especially if you want to observe an object moving in terms of an energy function. Lagrangian methods are different from the Newtonian method, because the Newtonian method observes an object by decomposing the force  components. You can implement the Lagrangian methods in most Physics formulas, from the simplest to the most complex, because all moving objects have energy. 

Commonly, in Lagrangian method, T described as kinetic energy and V described as potential energy. The Lagrangian itself is symbolized by L and have general solution as:

$\boxed{L = T - V}$

Than we use the second step Lagrangian equations used to identify specific solutions to classical mechanics problems, which expressed by:

$\boxed{\frac{d}{dt} (\frac{\partial L}{\partial \dot{q_k}}) - \frac{\partial L}{\partial q_k}= 0}$

Whereas, V is not a function of velocity and is applied in a conservative system, and L will be derived partially respect to all variables that contain $q_{k}$.

Application of Lagrangian Methods on Uniform Circular Motion

Problem: "Suppose that an object with mass m moving in a plane under the attractive force $\frac{\mu m}{r^2}$ directed to the origin of polar coordinates (r, $\theta$)".

Illustration 1. Circluar Motion Components

We can use Lagrangian method to solve that problem. The problem is about uniform circular motion, because the problem not showing the $\alpha$ or angular acceleration. In that motion, there is an attractive force directed toward the center of the polar coordinates. So, we use the r and $\theta$ as the variables of motion. This attractive force in circular motion is called the centripetal force and will later be used to find the value of the potential energy V. Let we solve this problem together, step by step.

First, determine the kinetic energy of the moving particle

All moving objects have kinetic energy. The particle with mass m has kinetic energy in the form of:

$T = \frac{1}{2} m \cdot v^2$ ...... (1.a.) 

Because the object moves in circular motion, we should express the (1.a) equation into a new equation that contain varibles such as $\theta$ or r. By the fact, the best step is we recalling $v = \omega \cdot r$, so that:

$T =\frac{1}{2} m \cdot \omega^2 r^2$ ...... (1.b.)

You should notice that $\omega$ represents the change in angle $\theta$ with respect to time t or $\frac{d\theta}{dt}$. Then, we can rewrite the (1.b) equation, as follow:

$T = \frac{1}{2} m \: \dot{\theta}^2 r^2$ ...... (1.c.)

Second, determine the potential energy from the centripetal force

We already know that centripetal force is an attractive force that is always directed toward the center of circular motion. From the illustration, the centripetal force can be expressed as $\frac{\mu m}{r^2}$. Then, we can recall a concept that connects potential energy and force, namely divergence. We apply it as follows:

$F=-\nabla V$ 

$F=- \frac{dV}{dx}$ ............ (2.a)

Because the particle is moving around the area from radius r to the center of circular motion, we can transform the dx at (2.a) equation into dr:

$F = - \frac{dV}{dr}$

$-F = \frac{dV}{dr}$

$-F dr = dV $ ............. (2.b)

We already know that $F = \frac{\mu m}{r^2}$, so that:

$dV = -(\frac{\mu m}{r^2})\: dr$

$\int^{V}_{0}{dV} = \int^{r}_{0}{-\frac{\mu m}{r^2} dr}$

$V = - (- \frac{\mu m}{r})$

$V = \frac{\mu m}{r}$ ........ (2.c)

Third, determine the general solution in the form of Lagrangian equation

Congratulation because in this step we already 70% of our topic. Previously, we determine kinetic energy T as (1.c) equation and potential energy V as (2.c) equation. Now, we substitute it at the general solution of Lagrangian equation as follows.

$L = T - V$

$L = \frac{1}{2} m \: \dot{\theta}^2 r^2 - \frac{\mu m}{r}$  .......... (3)

Fourth, determine $\frac{d}{dt} (\frac{\partial L}{\partial\dot{q_k}})$ for $\theta$ and r

• Derive the L respect to the $\dot{q_k}$ for $\dot{\theta}$

$\frac{d}{dt}(\frac{\partial L}{\partial\dot{\theta}})= \frac{d}{dt}(\frac{\partial (\frac{1}{2} m \cdot \dot{\theta}^2 r^2 - \frac{\mu m}{r})}{\partial \dot{\theta}})$

$\frac{d}{dt}(\frac{\partial L}{\partial\dot{\theta}})= \frac{d}{dt}(m \dot{\theta}r^2)$

$\frac{d}{dt}(\frac{\partial L}{\partial\dot{\theta}})= m \ddot{\theta}r^2$ ...... (4.a)

• Derive the L respect to the $\dot{q_k}$ for $\dot{r}$

$\frac{d}{dt}(\frac{\partial L}{\partial\dot{r}})=\frac{d}{dt}(\frac{\partial (\frac{1}{2} m \cdot \dot{\theta}^2 r^2 - \frac{\mu m}{r})}{\partial \dot{r}})$

Because the general solution for L not containing $\dot{r}$, so that we derive L respect to the $\dot{r}$ as:

$\frac{d}{dt}(\frac{\partial L}{\partial\dot{r}}) = 0 $ ...... (4.b)

Fifth, determine $\frac{\partial L}{\partial q_k}$ for $\theta$ and r 

• Derive the L respect to the $q_{k}$ for $\theta$

$\frac{\partial L}{\partial \theta} = \frac{\partial (\frac{1}{2} m \cdot \dot{\theta}^2 r^2 - \frac{\mu m}{r})}{\partial \theta}$ 

Dikarenakan dalam persamaan L tidak memuat variabel yang dapat diturunkan terhadap $\theta$, sehingga:

$\frac{\partial L}{\partial \theta} = 0$ ...... (5.a)

• Derive the L respect to the $q_{k}$ for r

$\frac{\partial L}{\partial r} = \frac{\partial (\frac{1}{2} m \cdot \dot{\theta}^2 r^2 - \frac{\mu m}{r})}{\partial r}$

$\frac{\partial L}{\partial r} = m \dot{\theta}^2 r + \frac{m \mu}{r^2}$ ...... (5.b)

Last, determine the solution in $\theta$ function

Determine the solution in the $\theta$ function is the last step for proving that the angular acceleration of uniform circular motion is zero.

$\frac{d}{dt} (\frac{\partial L}{\partial \dot{\theta}}) - \frac{\partial L}{\partial \theta}= 0$ ................ (6.a)

If we subtitute (4.a) and (5.a) into (6.a)

$m \ddot{\theta} r^2 - 0 = 0$

$m \ddot{\theta} r^2 = 0$

$\ddot{\theta} = 0$ ...... (The angular acceleration equation in uniform circular motion)

Just to be clear, $\ddot{\theta}$ is another form of $\alpha$ (angular acceleration). In this case, with the Lagrangian method for the $\theta$ function, we successfully proved the angular acceleration in uniform circular motion. Interestingly, suppose we use the Lagrangian method for the r function. In that case, we can also prove Kepler's Third Law with the assumption that the planet moves in uniform circular motion. We will post about that topic in the next post. 

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