Pendulum concept nitty gritty....Finding Force and PE relations. Plz Help!

[DrBTS]

Assume we have all the necessary requirements for a pendulum undergoing SHM.

I figured out that the restoring Force to create HM in the x (horizontal) is Frx = mg sin(theta)

I have a three part question:
1.)
Thus, to relate this to PE, we have W=-PE=KE = FD = L mg sin(theta). Is this correct?

Thus, at the bottom of the pendulum when it's at equil. KE = PE, 1/2 m v^2 = L mg sin(theta). Is this correct?

Doing the same for a spring, where F = -k x, W = 1/2 k x^2, because we had to take the average force across the distance...why dont we do the same for the pendulum?


2.)
Work due to gravity converting to KE from PE = mgh . I'm not understanding how the above equation is able to relate to this one.
Is it just that when we are given h, we resort to this equation vs theta we resort to the above eq?

3.)
How does centripetal force come into play here? Since it's centripetal motion F = mv^2 / L
then which force is really at play here and how does it differ with the restoring force?

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[DrBTS]

Work due to gravity converting to KE from PE = mgh . I'm not understanding how the above equation is able to relate to this one.
Is it just that when we are given h, we resort to this equation vs theta we resort to the above eq?

So far I got that the the height H is related with theta by H = L (1 - cos (theta))

if the restoring force Frx = mg sin (theta) and W is then W = L mg sin (theta) and we can set the two equal:

L mg sin(theta) = mg (L (1-cos (theta)), mg cancels out and we are left with

sin(theta) = 1-cos(theta)

 

[FeinMS]

Assume we have all the necessary requirements for a pendulum undergoing SHM.

I figured out that the restoring Force to create HM in the x (horizontal) is Frx = mg sin(theta)

I have a three part question:
1.)
Thus, to relate this to PE, we have W=-PE=KE = FD = L mg sin(theta). Is this correct?

Thus, at the bottom of the pendulum when it's at equil. KE = PE, 1/2 m v^2 = L mg sin(theta). Is this correct?

Doing the same for a spring, where F = -k x, W = 1/2 k x^2, because we had to take the average force across the distance...why dont we do the same for the pendulum?


2.)
Work due to gravity converting to KE from PE = mgh . I'm not understanding how the above equation is able to relate to this one.
Is it just that when we are given h, we resort to this equation vs theta we resort to the above eq?

3.)
How does centripetal force come into play here? Since it's centripetal motion F = mv^2 / L
then which force is really at play here and how does it differ with the restoring force?

1) Force=mg sin(theta) is correct, but it varies along the course of swinging, so you can't simply use F*d=W. You must integrate dW=F*dx. Fortunately enough, there's an equation for potential energy of the pendulum. As you said it's mg (L (1-cos (theta)). That's when the pendulum is at its highest position. At the lowest position, PE will become KE, so KE=mg (L (1-cos (theta)).
2) PE=mgh works here. Note h=L(1-cos(theta)) at the highest position by some complicating geometry. As pendulum swings, h will vary and this will convert PE to KE.
3) When the pendulum is at it's lowest position, it experiences centripetal force. F=mv^2/L. I was wondering if the centripetal force is present at other positions, but it turned out that it is only present at the lowest position. Also, note that the centripetal force is provided by Tension-mg. You will be able to find the tension in the pendulum when it's at the lowest position given L,m and v (or it's amplitude h_max)

 

[DrBTS]

3) When the pendulum is at it's lowest position, it experiences centripetal force. F=mv^2/L. I was wondering if the centripetal force is present at other positions, but it turned out that it is only present at the lowest position. Also, note that the centripetal force is provided by Tension-mg. You will be able to find the tension in the pendulum when it's at the lowest position given L,m and v (or it's amplitude h_max)

I think I should note that this explanation as well as the explanation for the use of mg l sin theta is incorrect so we might wanna keep thinking about this... The centripetal force is the total force responsible as long as the mass on the pendulum is swinging ( of course at the rest and theta0 positions v = 0) in a circular motion we employ the Centripetal Force as the force of that object.

It holds that we must use mgh = 1/2mv^2 because the restoring force is not proportional to the theta displacement, which confuses me because the height term is still in terms of cos(theta) at L(1-cos theta).

 

[DrBTS]

1) Force=mg sin(theta) is correct, but it varies along the course of swinging, so you can't simply use F*d=W. You must integrate dW=F*dx. Fortunately enough, there's an equation for potential energy of the pendulum. As you said it's mg (L (1-cos (theta)). That's when the pendulum is at its highest position. At the lowest position, PE will become KE, so KE=mg (L (1-cos (theta)).

My question still remains why you can't use that same theta as you use in mg (L (1-cos(theta))) and put it into the restoring force F= mgcos (theta) and then just multiply that by the length to the get the max PE and then set that equal to KE?

 

[FeinMS]

My question still remains why you can't use that same theta as you use in mg (L (1-cos(theta))) and put it into the restoring force F= mgcos (theta) and then just multiply that by the length to the get the max PE and then set that equal to KE?

Theta in h=L(1-cos(theta)) is the angle the pendulum makes with the vertical at a given point. We plug in the value of the maximum angle theta to find the maximum value of h (amplitude) At this point, the pendulum will have only PE.
I don't understand why you would use F*L to find PE though. PE in pendulum is provided by the difference in height between the highest and lowest point, so mgh.
We can definitely use F=mgsin(theta) to find the force at a given point, but we can't use one value of theta to find work because force changes as angle theta changes. Look at the picture below, see the force (acc vector times mass) changing at every point? So we must use integration W=int(F*dx) if you choose to find work by your method.
Let me know if I misunderstood your question.

 

[DrBTS]

Awesome gif, csadmin, thanks for your patience.

So by your logic theta is changing, and thus the work is also changing....but that also holds true for h= L(1-costheta).
To clarify further: I'm saying that we don't need to take into account the fact that theta changes, because we really only need the maximal value of theta displacement because at this value it will give us the parameters to calculate PE. It's by this logic that we have some angle theta that corresponds with height that we put into mgh. My above question is asking: if it is the case that we have angle theta that corresponds to some maximal height parameter, then this same angle theta, at that instant which has the maximal force required to restore pendulum to equilibrium. The equation for this restoring force is F = mg sin(theta) and the work is simply the length this force acts on, W = L mg (sin (theta)).

Aside question: This L mg (sin (theta)) is the torque, correct?

I know that we can calculate the instantaneous maximal work to restore back to equilibrium, because this was a question and subsequent answer choice (L mg (sin (theta))) in TPR. But all questions corresponding to finding the theta proportionalities or the maximal velocities only use the mgh or mg L (1-cos(theta)) equilibrium....and when I plug in with torque its never correct ....so thats what I'm trying to figure out 🙂

I think the crux of the my question might be in figuring out what the difference is between torque (torque = work) at theta and mgh = W ?

All of these doubts came from a few problems, which i attached, that demonstrate them asking 1. for torque, where i used mgl sin (theta) 2. for Work done by gravity where it had to be mgh and 7. proportionality of two different starting maximal thetas, where mgh/mgh and thus 1-cos(theta)/1-cos(theta*) but you couldn't do sin (theta)/ sin (theta).

 

[FeinMS]

Awesome gif, csadmin, thanks for your patience.

So by your logic theta is changing, and thus the work is also changing....but that also holds true for h= L(1-costheta).
To clarify further: I'm saying that we don't need to take into account the fact that theta changes, because we really only need the maximal value of theta displacement because at this value it will give us the parameters to calculate PE. It's by this logic that we have some angle theta that corresponds with height that we put into mgh. My above question is asking: if it is the case that we have angle theta that corresponds to some maximal height parameter, then this same angle theta, at that instant which has the maximal force required to restore pendulum to equilibrium. The equation for this restoring force is F = mg sin(theta) and the work is simply the length this force acts on, W = L mg (sin (theta)).

Aside question: This L mg (sin (theta)) is the torque, correct?

I know that we can calculate the instantaneous maximal work to restore back to equilibrium, because this was a question and subsequent answer choice (L mg (sin (theta))) in TPR. But all questions corresponding to finding the theta proportionalities or the maximal velocities only use the mgh or mg L (1-cos(theta)) equilibrium....and when I plug in with torque its never correct ....so thats what I'm trying to figure out 🙂

I think the crux of the my question might be in figuring out what the difference is between torque (torque = work) at theta and mgh = W ?

All of these doubts came from a few problems, which i attached, that demonstrate them asking 1. for torque, where i used mgl sin (theta) 2. for Work done by gravity where it had to be mgh and 7. proportionality of two different starting maximal thetas, where mgh/mgh and thus 1-cos(theta)/1-cos(theta*) but you couldn't do sin (theta)/ sin (theta).
View attachment 24546

View attachment 24547View attachment 24549

Ok, the reason why you can't use Lmgsin(theta) for PE at the amplitude is the following. If you want to find the work by integration (that is not by mgL(1-cos theta)) you need to integrate from the equilibrium to the amplitude. As you said earlier, W= delta(KE)= -delta(PE). You're arguing that Force exerting at the amplitude is mgsin(theta) which is correct, but it cannot be used to find PE at that point. Further, the distance over which this force exerts on is really small (not L,) once you move by a small angle, the force will change.
Let's say I'm on top of an inclined plane, and I'm pushing the box down. I apply a force of 10N for fraction of a second over fraction of centimeters. Can you find the potential energy of the box? No, not enough information is given. This is a similar situation. To find the PE of the box, I have to integrate how much force over what distance I pushed the box up to bring it up to that position. (analogous to integrating force*dx from equilibrium to the amplitude.)
Yes it seems like Lmgsin(theta) is torque because L is perpendicular to mgsin(theta) and the pendulum is attached to the ceiling. By the reason I explained above, this torque doesn't equal work or PE.

 

[DrBTS]

Dude thank you.