Pendulum time at any given angel θ, 0<=θ<=π/2

Click For Summary
SUMMARY

The discussion focuses on calculating the time it takes for a pendulum to reach the lowest point (θ=0) from a horizontal position (θ=π/2) under the influence of gravitational acceleration. The key challenge is that the acceleration of the pendulum bob is a function of the angle, specifically expressed as Gcosθ, complicating the integration process. The participants emphasize the limitations of the small angle approximation and the necessity of solving a non-separable differential equation to find an explicit solution. The conversation highlights the importance of correctly defining the angle θ in relation to the pendulum's motion.

PREREQUISITES
  • Understanding of pendulum dynamics and gravitational forces
  • Familiarity with differential equations and their classifications
  • Knowledge of angular motion and its relationship to linear motion
  • Basic grasp of numerical methods, such as the Euler method
NEXT STEPS
  • Study the derivation of the pendulum's differential equation using g sin(θ)
  • Learn about non-separable differential equations and methods to solve them
  • Explore numerical methods for approximating solutions to differential equations
  • Investigate the implications of the small angle approximation in pendulum motion
USEFUL FOR

Students and professionals in physics, mathematics, and engineering who are interested in understanding pendulum dynamics, solving differential equations, and applying numerical methods for motion analysis.

Taaha
Messages
3
Reaction score
0
Thread moved from the technical forums, so no Homework Template is shown
I got stucked with a solution of a problem.
I have a pendulum of r radius which is in G gravitational field. So the working acceleration on the bob is a function of the angle. That is Gcosθ.
  • The mass of the bob = m
  • The radius of the string of the pendulum = r
  • Gravitational force is G. So the magnitude of the component of the force that is working on the bob is Gcosθ.
My question is what time is needed to reach at the bottom where θ=0 rad. I know the acceleration is a function of time and its integral is velocity and its integral is the position. But here the acceleration is a function of position (or angle). So I can't convert the parameter of the function.
 
Physics news on Phys.org
You want to find the time it takes for a pendulum to fall from the horizontal position? Welcome to the world of differential equations, as far as I know this is not a simple one to solve.

Typically when pendulums are introduced as an introductory topic, there is the restriction that the range of angles the pendulum oscillates through is small. That way, we can use the approximation sin(θ) ≈ θ, which turns the difficult differential equation into a (very) simple one.

This simpler differential equation is only accurate for a limited range of motion (because it is based upon this "small angle" approximation).

You are asking about the full range of motion, where we cannot use that approximation, and so you have a solid differential equation on your hands.
 
  • Like
Likes   Reactions: scottdave
Hiero said:
You want to find the time it takes for a pendulum to fall from the horizontal position? Welcome to the world of differential equations, as far as I know this is not a simple one to solve.

Typically when pendulums are introduced as an introductory topic, there is the restriction that the range of angles the pendulum oscillates through is small. That way, we can use the approximation sin(θ) ≈ θ, which turns the difficult differential equation into a (very) simple one.

This simpler differential equation is only accurate for a limited range of motion (because it is based upon this "small angle" approximation).

You are asking about the full range of motion, where we cannot use that approximation, and so you have a solid differential equation on your hands.
I know what you are saying. That the pendulum is restricted and all those you told. But I actually want to know about that differential equation. How can I just reach to this? I have tried many procedure like in a short range dθ suppose the acceleration is constant so time ti = (Vi - Vi - 1)/ acosθ

Total time t = ∑ti
But here velocity is also a function of angle. Here the actual problem is. I can not integrate that function. Maybe the procedure is faulty or there can be a better procedure.
 
Well if we are going to get into it you should first double check which component you have; you said θ = 0 when it is at it's lowest point, and you also say the tangential (working) component of gravity is gcosθ, is this all correct?

Taaha said:
But I actually want to know about that differential equation. How can I just reach to this?
You said you had the angular acceleration as a function of angle, right? That is the differential equation I'm talking about.

Taaha said:
But here velocity is also a function of angle. Here the actual problem is. I can not integrate that function. Maybe the procedure is faulty or there can be a better procedure.
The actual problem is the differential equation is not separable. A 'separable differential equation' (which might be worth looking up) is basically one in which you can just integrate. For example, dx/dt = x is a differential equation analogous to yours, where the velocity is a function of position, instead of time, but we can still integrate if we separate the differentials first:
dx/dt = x
dx/x = dt
∫dx/x = ∫dt
ln(x) = t + C
x = e^(t+C)
The differential equation you wish to solve, unfortunately, cannot be separated like this. I know of no simple method of solving the one you want to solve.

Taaha said:
I have tried many procedure like in a short range dθ suppose the acceleration is constant so time ti = (Vi - Vi - 1)/ acosθ

Total time t = ∑ti
I think it is essentially what you're getting at: https://en.wikipedia.org/wiki/Euler_method
This would work for approximating the solution to the differential equation, given some initial conditions. It wouldn't work for finding an explicit solution, though.
 
Hiero said:
you said θ = 0 when it is at it's lowest point, and you also say the tangential (working) component of gravity is gcosθ, is this all correct?
Sorry it is θ = π/2 rad
 
Taaha said:
Sorry it is θ = π/2 rad
So you've set your angular coordinate up such that the lowest position of the pendulum is at θ = π/2 radians?
Then the two horizontal positions are at θ = {0, π}, right?

Your working component is correct, but does this seem like the most natural way to set up the θ coordinate?
 
Taaha said:
I actually want to know about that differential equation.
As @Hiero suggests, it will be more convenient to make theta the angle to the vertical. The tangential acceleration is then g sin(θ). (Use lowercase g for a constant gravitational field. G is for the gravitational constant.)
How does the tangential acceleration relate to derivatives of θ?
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
2K