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

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Taaha
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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.
 
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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.
 
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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?