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

In summary, the conversation revolves around finding the time it takes for a pendulum to fall from the horizontal position, which involves solving a difficult differential equation. The speaker mentions that typically, this problem is simplified by using the small angle approximation, but in this case, the full range of motion needs to be considered. The differential equation is not separable, making it challenging to solve. The speaker suggests using the Euler method for approximation, but it would not provide an explicit solution. There is also a clarification that the angle is actually π/2 radians when the pendulum is at its lowest point.
  • #1
Taaha
3
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
  • #2
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 scottdave
  • #3
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.
 
  • #4
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.
 
  • #5
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
 
  • #6
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?
 
  • #7
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?
 

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

What is a pendulum and how does it work?

A pendulum is a weight attached to a rod or string that is able to swing back and forth. The motion of a pendulum is governed by the laws of physics, specifically gravity and inertia. As the pendulum swings, it converts potential energy into kinetic energy and back again, creating a regular and repeatable motion.

What is the relationship between the length of a pendulum and its period?

The period of a pendulum, or the time it takes for one full swing, is directly proportional to the length of the pendulum. This means that a longer pendulum will have a longer period and a shorter pendulum will have a shorter period. This relationship is described by the equation T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

How does the angle of a pendulum affect its period?

The angle at which a pendulum is released does not affect its period. This is because the length of the pendulum, which is the determining factor of the period, remains constant regardless of the angle at which it is released. However, the amplitude of the pendulum's swing will decrease as the angle increases, which can affect the accuracy of the period measurement.

Why is the period of a pendulum at an angle θ equal to the period of a pendulum with a length equal to L/cosθ?

This is due to the concept of effective length. When a pendulum is released at an angle, the effective length of the pendulum decreases due to the component of gravity acting in the direction of the swing. The effective length is equal to the actual length of the pendulum multiplied by the cosine of the angle. Therefore, the period of a pendulum at an angle can be calculated using the effective length, which is L/cosθ.

How can the period of a pendulum at any given angle θ be measured accurately?

The period of a pendulum at any given angle θ can be measured accurately by using a stopwatch to record the time for a certain number of swings and then calculating the average period. It is important to ensure that the amplitude of the pendulum's swing remains constant throughout the measurement and to take multiple measurements to reduce errors. Alternatively, the period can also be measured using a photogate timer, which can provide more precise measurements.

Similar threads

  • Introductory Physics Homework Help
Replies
26
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
746
  • Introductory Physics Homework Help
Replies
5
Views
943
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
5K
Back
Top