Finding the angle of a pendulum's period with unusual units:

Inertia_Squared

Inertia_Squared

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I'm trying to find a way to prove with the information given (all said information is on image attached) the value of the angle a pendulum makes from a full period i.e. the 2x angle that it is released from (friction is not being accounted for here), and although I think I may have done it, I'm not sure how valid it is because I'm using different types of units, which is typically a big no-no. I attempted to rectify this by converting the arc-length into a length (I had two values for the arc-length; the average acceleration and the period in seconds) however I'm not sure if that is a valid way to go about it. I've formatted my working as a textbook-style question but I would like to be perfectly clear that its entirely possible for this to have no real solution, although I'm pretty sure that even if this method doesn't work there would be another that does.

Long story short I'm looking for someone to check my working and to see if it's correct, and if not maybe show me a method I could use to solve it? Either way, Thanks in advance!

P.S. Compared to what I'm doing at school (Y11 Ext 1 Maths) I know this is rather simple, however, I'm using this as a double-check an idea I have for an assignment. The current Q I'm on is to make an appropriate graphical model of the length and period of a pendulum, but since its one of the earlier and easier bits, I wanted to go the extra mile and do a real in-depth analysis of its behaviour (its holidays and I have some time to kill so I thought why not). Since its extra work and won't be marked on it I'm 99% sure its be fine to post this here; however I do believe that doing that little bit extra would help me to get a better understanding of the nature of the pendulum, which is why I want to check it here, as doing extra for a better understanding is great, but I don't want to do it wrong and have it end up being harmful in the end!

20200713_154041.jpg
 
Hi,
What are B and C? There is a very neat way of computing the angle in terms of the length of the cord and g if you apply Newton's second law. The simple pendulum is a very classical problem.
 
yoscar04 said:
Hi,
What are B and C? There is a very neat way of computing the angle in terms of the length of the cord and g if you apply Newton's second law. The simple pendulum is a very classical problem.
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B and C represent the points where the bob of the pendulum is still and not accelerating; Their values in terms of the bob's properties are B = 0s, 0m/s, 0m/s2 and C = 1.01s, 0m/s, 0m/s2. Also, the line between them and A is the length of the pendulum; and the arc that is made from B to C passing through point O (its poorly drawn but O is actually the midpoint of the arc) is the path that the bob takes in one full period.

Edit: to clarify their values of time - 0s and 1.01s - is referring to the time it takes for the bob to reach those positions, i.e. the initial position and the final position of one period respectively.
 
I am not sure I follow your reasoning. At B, it is true that the velocity is zero (also at C) since it is a extreme point, but the acceleration at B is not 0, neither at C. If the acceleration there at B and C were 0 it would stay there for ever.
 
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yoscar04 said:
I am not sure I follow your reasoning. At B, it is true that the velocity is zero (also at C) since it is a extreme point, but the acceleration at B is not 0, neither at C.
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I'm trying to find a good way to put this because I'm aware that the way I'm putting it is a little confusing. But imagine for B you take the point in time exactly as the pendulum is released, it is so exact that its velocity i still 0 and thus its instantaneous acceleration is also 0, as for C, take the point in time just as it reaches an extreme point, in that one perfect point it is not speeding up or slowing down, is what I mean. It is still accelerating, but at those two points, the gradient of its change in velocity is null.
 
Inertia_Squared said:
I'm trying to find a good way to put this because I'm aware that the way I'm putting it is a little confusing. But imagine for B you take the point in time exactly as the pendulum is released, it is so exact that its velocity i still 0 and thus its instantaneous acceleration is also 0, as for C, take the point in time just as it reaches an extreme point, in that one perfect point it is not speeding up or slowing down, is what I mean. It is still accelerating, but at those two points, the gradient of its change in velocity is null.
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The acceleration is not 0 at B and C.
 
The points B and C mark the two points in space where the velocity of the pendulum is 0 and is currently not changing, graphically speaking the tangent of its change in velocity at that point is 1 i.e. not changing. It is still accelerating but at that point in time, the velocity is not changing.
 
If I'm not making any sense don't worry about it, the velocity for B and C 0 and their acceleration is their angular acceleration, which is unknown, but their acceleration due to gravity is 9.8.
 
Inertia_Squared said:
If I'm not making any sense don't worry about it, the velocity for B and C 0 and their acceleration is their angular acceleration, which is unknown, but their acceleration due to gravity is 9.8.
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The acceleration (tangential) at B is equal to gsin[MATH]\theta[/MATH].
 
yeah, I'm just getting myself a little confused, I've been doing maths almost all day and my brain is saying the wrong things :p

I agree that acceleration for B is equal to gsinθ, but theta is unknown. I'm trying to find the angle BAC and I was wondering if the method I used to go about calculating it was valid. If you want to calculate the angle using the acceleration at B I'm pretty sure it's possible but the only thing that is known about the point is that the velocity is 0 and the time is 0 (since its the initial velocity of a single period) where at C the velocity is also 0and the time taken to swing from B to C is 1.01 seconds. The only reason I didn't do it this way is that I thought you needed to know the acceleration to find the angle, but to find the angle don't you need to know the acceleration?
 
Inertia_Squared said:
yeah, I'm just getting myself a little confused, I've been doing maths almost all day and my brain is saying the wrong things :p

I agree that acceleration for B is equal to gsinθ, but theta is unknown. I'm trying to find the angle BAC and I was wondering if the method I used to go about calculating it was valid. If you want to calculate the angle using the acceleration at B I'm pretty sure it's possible but the only thing that is known about the point is that the velocity is 0 and the time is 0 (since its the initial velocity of a single period) where at C the velocity is also 0and the time taken to swing from B to C is 1.01 seconds. The only reason I didn't do it this way is that I thought you needed to know the acceleration to find the angle, but to find the angle don't you need to know the acceleration?
Click to expand...
In order to find [MATH]\theta[/MATH] you will need to solve the corresponding differential equation F=ma. Are you taking a physics course?Do you know how to solve second order differential equations with constant coefficients? You will need to assume very small values for \theta in order to be able to solve analytically for [MATH]\theta[/MATH]. If [MATH]\theta[/MATH] is not small enough the problem gets a bit more cumbersome.
 
To summarize and hopefully this will help you:
1) Make a diagram of the pendulum with the forces involved.
2) Write Newton's second law and connect the displacement with the angle of the pendulum and the length of the cord.
3) Solve for the equation of motion, for [MATH]\theta[/MATH]. Remember to use the fact that [MATH]\theta<<1[/MATH] (in radians)
 
okay, I've got a bit of a better idea of what I'm doing, I've got dinner and need to spend some time with the family but I think I'm on track now for when I pick this back up, thanks for the help!
 
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