Help with elements acting on simple pendulum!

[TXKnight]

I need some alternative view to grasp this concept better:
I am tired and I think my brain is starting to f**k things up for me,
can someone provide a clarification/critique of this simple concept for me?

forces acting on a simple pendulum

during the cycle of the pendulum you have one force, that of mg (always pointing straight down), that can be resolved in two components.
Component of mg perpendicular to arc is Tension= (mg cos theta) and the other componet -tangent to arc- is restoring force (mg sin theta). At theta 90 (extreme position) I resolve my two forces as mg (tangential-pointing straight down) and tension T=0.
During the travel down to the equilibrium position, the restoring force vector (F=mg sin theta) gets smaller until theta=0, thus Fx=0. No net force is present in the tangential component at equilibrium, but mg cos theta is maxed out, thus greatest tension at equilibrium.
What's up with acceleration? I know that at equilibrium acceleration,is also pointing to center (radial) along the pendulum's rod. So you can say acceleration on x is 0 but acceleration in y component is not.

Is it correct to think that acceleration can be calculated from a=v^2/R (borrowed from UCM)
It follows that:
mg cos theta = mv^2/r
velocity can be obtained from conservation of energy KE=PE

WTF...I may need some sleep i think i've gotten all this wrong...

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[milski]

It's not UCM - the angular velocity is increasing from 90 to bottom and then decreasing again when you go up. The acceleration will be non-constant, pointed somewhere below the attachment point of the pendulum, except when at equilibrium, when it will point straight up. I think the UCM formula for acceleration magnitude will work for the momentary acceleration at that point but that's all - you cannot use it at any other point.


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[TXKnight]

It's not UCM - the angular velocity is increasing from 90 to bottom and then decreasing again when you go up. The acceleration will be non-constant, pointed somewhere below the attachment point of the pendulum, except when at equilibrium, when it will point straight up. I think the UCM formula for acceleration magnitude will work for the momentary acceleration at that point but that's all - you cannot use it at any other point.

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Thanks milski...as you well pointed out this motion is by no means UCM (except for theta=0), Ithanx for the insight...now...what about the other factors of my pendulum layout?

 

[milski]

It would be helpful to know what you are trying to solve. If you are trying to describe the motion of a pendulum, basic algebra won't be enough. The setup to solve that is very similar to what you describe. You have two forces acting on the pendulum - gravity and tension from the string. You can break down gravity in two components, just as you said - one tangent to the motion and one normal to it. Here is the first difference from what you describe - the normal component is pointed away from the center, the tension is towards the center. The magnitudes of the tension and the normal component are not the same! If they were, the pendulum would continue straight ahead in the direction that it was moving at that instant.

With what we've said so far, you cannot compute what the tension (and thus the net force and the acceleration) are going to be. All your logic about the tension having the largest magnitude at the equilibrium point and not having a tangent component are generally correct but you don't know the exact values unless you know the velocity at that point and you won't know the velocity unless you know the acceleration along the "fall" of the pendulum.

The general case can be solved by applying F=ma for the tangental component of the weight. F is known and a can be expressed as a second derivative of the length of the arc that the pendulum is following. That leads to a differential equation which is not easy to solve. For smaller angles of deflection where θ=sinθ=tanθ that equation can be simplified and leads to estimating the motion of the pendulum with a simple harmonic motion. For such small deflections you can write φ=φ0 * cos(ωt+&#948 where φ0 is the angle of max deflection.

TLDR: Describing pendulum motion from kinematics perspective sucks.

Using energies works a lot better! Energy is conserved during the whole motion, so you can write PE+KE=const. From there you need to know V at only one point to be able to calculate V at any other point along the motion of the pendulum. Even better, that will be the exact velocity while the SHM would be only estimation.

Saying that mg cos θ = mv^2/r would imply that the only force acting at the equilibrium point is gravity which is not correct - tension is acting in the opposite way. The resulting force from tension+gravity at that point should satisfy mf=mv^2/r or f=v^2/r

 

[TXKnight]

It would be helpful to know what you are trying to solve. If you are trying to describe the motion of a pendulum, basic algebra won't be enough. The setup to solve that is very similar to what you describe. You have two forces acting on the pendulum - gravity and tension from the string. You can break down gravity in two components, just as you said - one tangent to the motion and one normal to it. Here is the first difference from what you describe - the normal component is pointed away from the center, the tension is towards the center. The magnitudes of the tension and the normal component are not the same! If they were, the pendulum would continue straight ahead in the direction that it was moving at that instant.

With what we've said so far, you cannot compute what the tension (and thus the net force and the acceleration) are going to be. All your logic about the tension having the largest magnitude at the equilibrium point and not having a tangent component are generally correct but you don't know the exact values unless you know the velocity at that point and you won't know the velocity unless you know the acceleration along the "fall" of the pendulum.

The general case can be solved by applying F=ma for the tangental component of the weight. F is known and a can be expressed as a second derivative of the length of the arc that the pendulum is following. That leads to a differential equation which is not easy to solve. For smaller angles of deflection where θ=sinθ=tanθ that equation can be simplified and leads to estimating the motion of the pendulum with a simple harmonic motion. For such small deflections you can write φ=φ0 * cos(ωt+&#948 where φ0 is the angle of max deflection.

TLDR: Describing pendulum motion from kinematics perspective sucks.

Using energies works a lot better! Energy is conserved during the whole motion, so you can write PE+KE=const. From there you need to know V at only one point to be able to calculate V at any other point along the motion of the pendulum. Even better, that will be the exact velocity while the SHM would be only estimation.

Saying that mg cos θ = mv^2/r would imply that the only force acting at the equilibrium point is gravity which is not correct - tension is acting in the opposite way. The resulting force from tension+gravity at that point should satisfy mf=mv^2/r or f=v^2/r

Nice post, yeah i agree that a energy perspective is a lot better. I was thinking about being unable to get the correct values without taking the derivative, thanks for putting it in clearer terms. And no, i don't want (need) to derive/integrate anything anymore ever..hahaha.... Now, this is nice -- φ=φ0 * cos(ωt+&#948 where φ0 is the angle of max deflection--

point well taken, you know your physics man.