# Why in this example the gravity not considered on the $y$-axis? [closed] In this example 7.6 from the book "Classical Dynamics of Particles and Systems 5th Edition by Thornton and Marion"

"Find the frequency of small oscillations of a simple pendulum placed in a railroad car that has a constant acceleration $a$ in the $x$-direction."

Then he writes the equations:

$$x=v_{0}t+\frac{1}{2}at^{2}+l\sin\theta$$ $$y=-l\cos\theta$$

But I did not understand why gravity is not considered in the y-axis. It would look like the x-axis.

To find the equation of x I started from the second law:

$$F_{x}=m\ddot{x}$$ $$ma=m\ddot{x}$$

integrating and putting the initial conditions I found the example equation.

For the coordinate $y$ I started the same way

$$F_{y}=m\ddot{y}$$ $$-mg=m\ddot{y}$$

which resulted in

$$y=v'_{0}t-\frac{1}{2}gt^{2}-l\cos\theta$$

• Welcome to Physics! Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. – ZeroTheHero May 16 '18 at 14:57
• What is your solution to the problem? – sammy gerbil May 16 '18 at 15:38
• Gravity acts along the $z$-axis not the $y$-axis. – Qmechanic May 16 '18 at 17:21
• @Qmechanic: I have the book in front of me, and Thornton & Marion choose to have the $y$-axis be the vertical axis in this problem. – Michael Seifert May 17 '18 at 12:39
• – sammy gerbil May 18 '18 at 14:01

Gravity is considered in the functions of theta (sine and cosine), considering that the angle theta is the equilibrium angle due to the car's acceleration and the acceleration of gravity.

Gravity will also be considered in part of the potential energy of the system. When finding the frequency of small oscillations of a pendulum, a key component is solving for the Lagrangian, which in turn is composed of both kinetic and potential energy. The potential energy will be of the magnitude of the mass multiplied by the acceleration of gravity (the gravitational force), and then multiplied by the length of the pendulum and its cosine of theta. Do not forget though that this potential energy is negative.

• This makes sense. So $\theta$ is that known function of the simple pendulum solution: $\theta (t) = \theta_{0}\cos(\omega t+\phi)$ where gravity is in $\omega^{2}=g/l$ ? – Amarildo May 16 '18 at 17:05
• To my belief, yes. Though, keep into consideration that when omega squared is equivalent to g/l, that is assuming the train car in the problem is not accelerating. So that is true when a=0. – A. Rudzinski May 19 '18 at 1:06
• And my apologies on responding two days late. But very valid question, considering some users above visualize gravity acting upon the z axis. :) – A. Rudzinski May 19 '18 at 1:08

In problems like this one, we assume that some external agency is moving the train car; in other words, there is some external force that causes the train car to move with a constant acceleration $a$ in the $x$-direction. There is also presumably some kind of normal force from the tracks that ensures that the train car does not accelerate in the $y$-direction. Integrating twice, the $x$-coordinate of the train car is then $x_c = v_0 t + \frac{1}{2} a t^2$, and the $y$-coordinate is a constant (which we can take to be zero.)

The question is then asking what the motion of a pendulum bob fixed to a particular point inside the train car would be. The bob will experience the force of gravity and the force of tension from the string, the latter of which will vary in direction and magnitude as the pendulum swings. Moreover, the string is assumed to be inextensible, so it must be the case that the distance between the bob and the point of suspension is a constant. This determines what the tension force must be, but only as an implicit function of the pendulum's angle and its velocity. Thus, Newton's Second Law for the pendulum bob becomes much harder to write down in an explicit form.

This is where the Lagrangian formalism comes in. By writing down the bob's position relative to the car, we can find its kinetic energy and potential energy quite easily. The Euler-Lagrange equations then give us a set of ODEs for the angle $\theta$, which we can then investigate further to find properties of the motion.