Non Rigid Pendulum

Classical Mechanics Level 3

Consider a point mass (marked as red) attached to a spring O A OA of instantaneous length r r . The spring makes an angle − θ -\theta to the horizontal. The equations of motion of the system are:

r ¨ = r θ ˙ 2 − a r − b sin ⁡ θ + c \ddot{r} = r\dot{\theta}^2 - ar - b\sin{\theta} + c θ ¨ = − ( 10 cos ⁡ θ + 2 r ˙ θ ˙ ) r \ddot{\theta} = -\frac{(10\cos{\theta} + 2\dot{r}\dot{\theta})}{r}

Here, a a , b b and c c are positive integers. Find a + b + c \boxed{a + b + c} .

Details:

  • g = 10 m / s 2 g = 10 \ m/s^2
  • The point O O is fixed at the origin of the coordinate system and is marked in blue.
  • Natural length of the spring is 0.3 m.
  • Spring constant = 20 N / m 20 \ N/m
  • Mass = 1 k g 1 \ kg
  • Naming the angle − θ -\theta instead of θ \theta is deliberate.

Bonus: Numerically solve this system of equations over a large period of time, by considering the initial configuration of the system to be at rest, and the mass is initially placed on the positive x-axis such that the spring is stretched. Plot the trajectory of the mass. What is the nature of the motion observed?


The answer is 36.

Karan Chatrath by Karan Chatrath , 27, Netherlands

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2 solutions

Steven Chase
Sep 30, 2019

I assume that θ \theta is the magnitude of the angular displacement below the positive x x axis. I will proceed accordingly.

x = r cos ⁡ θ y = − r sin ⁡ θ x ˙ = − r sin ⁡ θ θ ˙ + r ˙ cos ⁡ θ y ˙ = − r cos ⁡ θ θ ˙ − r ˙ sin ⁡ θ v 2 = x ˙ 2 + y ˙ 2 = r 2 θ ˙ 2 + r ˙ 2 x = r \, \cos \theta \\ y = -r \, \sin \theta \\ \dot{x} = -r \, \sin \theta \, \dot{\theta} + \dot{r} \, \cos \theta \\ \dot{y} = -r \, \cos \theta \, \dot{\theta} - \dot{r} \, \sin \theta \\ v^2 = \dot{x}^2 + \dot{y}^2 = r^2 \, \dot{\theta}^2 + \dot{r}^2

The kinetic energy, gravitational potential energy, and spring potential energy are:

E = 1 2 m v 2 = 1 2 m ( r 2 θ ˙ 2 + r ˙ 2 ) U g = m g y = − m g r sin ⁡ θ U s = 1 2 k ( r − ℓ 0 ) 2 E = \frac{1}{2} m v^2 = \frac{1}{2} m (r^2 \, \dot{\theta}^2 + \dot{r}^2) \\ U_g = m g y = -m g r \, \sin \theta \\ U_s = \frac{1}{2} k (r - \ell_0)^2

Lagrangian:

L = E − U g − U s = 1 2 m ( r 2 θ ˙ 2 + r ˙ 2 ) + m g r sin ⁡ θ − 1 2 k ( r − ℓ 0 ) 2 L = E - U_g - U_s = \frac{1}{2} m (r^2 \, \dot{\theta}^2 + \dot{r}^2) + m g r \, \sin \theta - \frac{1}{2} k (r - \ell_0)^2

Equations of motion:

d d t ∂ L ∂ r ˙ = ∂ L ∂ r d d t ∂ L ∂ θ ˙ = ∂ L ∂ θ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{r}}} = \frac{\partial{L}}{\partial{r}} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{\partial{L}}{\partial{\theta}}

Crunching out the derivatives results in:

r ¨ = r θ ˙ 2 − k m r + g sin ⁡ θ + k ℓ 0 m θ ¨ = g cos ⁡ θ − 2 r ˙ θ ˙ r \ddot{r} = r \dot{\theta}^2 - \frac{k}{m} r + g \sin \theta + \frac{k \ell_0}{m} \\ \ddot{\theta} = \frac{g \cos \theta - 2 \dot{r} \dot{\theta}}{r}

There is a slight sign discrepancy between my results and the expected result, but they are clearly describing the same thing.

Simulating this per the recommendations and making an ( x , y ) (x,y) scatter plot results in a nice "bird's nest" pattern. The first plot is for 10 seconds, and the second plot is for 50 seconds.

2 Helpful 2 Interesting 2 Brilliant 0 Confused

Nice solution! I defined the angle as negative as I wanted to stick to the convention of an anticlockwise rotation as positive. In the end, as you rightly said, it doesn't matter.

Also, I was wondering if you got a chance to look into my latest comment on your problem on the two machine dynamics. Please let me know your thoughts on it.

Karan Chatrath - 1 year, 8 months ago

I have posted a follow-up. I think you will enjoy it

Karan Chatrath - 1 year, 8 months ago
Krishna Karthik
Apr 12, 2020

This is a classic university level dynamics problem; the best way to solve it is by using Lagrangian mechanics, because The vectors and forces in Newtonian mechanics get really tedious as this problem deals with 2 degrees of freedom, θ , r \theta, r

This paper here explains the equations of motion in both models; energy (Lagrangian) and Force vectors (Newtonian). Also, at the end of the paper, it shows the simulation results and the trajectory of the bob over time. Here's the paper:

elastic pendulum dynamics

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