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How does conservation of mechanical energy in a pendulum work?

I understand that gravitational potential energy is converted, as an object falls, into kinetic energy, and is otherwise given by $mgh$, for an object at a height of $h$. $ \ $ So, if the object falls a distance of $x$, from rest, then it gains $mgx$ kinetic energy at the height of $ \ h-x$.

But the bob of a pendulum isn't simply falling. Rather it follows a circular-arc trajectory, so the resultant force causing its motion is different than the weight of the bob. Yet, the potential energy in a bob at a height of $h$, is also $mgh$. So, as the bob reaches a height of $ \ h-x \ $, in its circular-arc motion, it will also have a kinetic energy of $mgx. \ \ $

This then must imply that the bob will be moving at the same speed as the object described above, when it reaches the height of $ \ h-x \ $, but in a very different direction. Is this accurate?

I'm having a hard time understanding how this makes sense intuitively.

On one hand, the bob has the same speed as the object even though a different resultant force than its weight got it to that speed. On another hand, it's not even moving in the same direction. But, still, both phenomena are referred to as gravitational potential energy, and no distinction seems to be made between the two when solving problems.

Why is this (assuming I got everything right in the above)?

Stephen
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    Well there other forces at play here other than gravity, specifically the tension of the string which is responsible for the circular motion. The extra force doesn't do any work (i.e. give/take energy from the bob) because it acts perpendicular to the bob's motion. – Jared Apr 25 '18 at 01:10
  • @Jared Of course. I wasn't taking work into consideration at all! (I'm really not a physicist...) – Stephen Apr 25 '18 at 01:15
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    In energy terms the gravitational field is conservative: the potential energy "assigned" to an object with finite mass located at a particular point in a gravitational field is completely independent of the path taken by the object to arrive at that point. (N.B. clarifying "assigned", the actual numerical value of the potential energy at a point will depend on how we simplify a particular problem to make it tractable and on the associated zero potential energy datum we choose to use. This arbitrary constant offset cancels when calculating a potential difference of course.) – James Arathoon Apr 25 '18 at 01:56
  • @JamesArathoon We need to be careful here. It's true that, even in the constant approximation, gravity is conservative. However this is not always true when external forces are applied. Certainly "conservation of energy" doesn't apply when I throw a ball up and catch it (where the final kinetic energy is 0). The key in this situation is the direction that the external force is applied (always perpendicular to the motion). – Jared Apr 26 '18 at 04:46
  • @Jared I find this specific concept rather strange. So essentially, a force can act whichever way we want on an object, as long as in the end its direction is perpendicular to the motion, that force will be a conservative force? Because it does no work. – Stephen Apr 26 '18 at 06:28
  • @Jared: conservation of energy always applies. Obviously we can devise ways of wasting arbitrary amounts of energy to reach a particular gravitational potential above the earths surface, but that does not mean that the conservation of energy is falsified. In your example throwing a ball vertically straight up and catching it at the highest point it can reach, will leave us with just gravitational potential energy and no kinetic energy. Even the conservation of momentum still applies, it just that the earth is too big a mass to measure what is happening. – James Arathoon Apr 26 '18 at 09:32
  • @Stephen Work is given by: $\int \vec{F}\circ d\vec{l}$. When $\vec{F}$ is perpendicular to the particle's motion (i.e. the direction of $\vec{l}$), that dot product is $0$ and thus $\vec{F}$ does no work. This has nothing to do with a conservative force. A conservative force is any that can be written as the gradient of a scalar: $\vec{F} = \vec{\nabla} \Lambda$. We recognize this scalar function as the potential energy. Conservative forces have the property that the difference between the potential energy at any two points does not depend on the path taken between those two points. – Jared Apr 26 '18 at 21:57
  • @Stephen For further reading, I suggest reading up on the mathematics behind a conservative force: Gradient Theorem. Furthermore a conservative force field, $\vec{F}$, has the property that its curl is zero: $\vec{\nabla}\times\vec{F} = 0$ – Jared Apr 27 '18 at 05:03
  • @Jared Yes, I have actually studied some of this, but a few months ago. Much of it should come back to me after my upcoming revision for my applied exam! I'm better with pure concepts... – Stephen Apr 27 '18 at 08:17

1 Answers1

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This is a surprisingly difficult problem to solve, actually. Looking at the figure, we can surmise that:

$$ \frac{dv}{dt} = \ddot{x} = g\sin(\theta) $$

It's natural to switch to angular coordinates here ($x = r\theta, \dot{x} = r\dot{\theta}, \ddot{x} = r\ddot{\theta}$):

$$ r\ddot{\theta} = g\sin(\theta) $$

Unfortunately this is a non-linear differential equation and my recollection (and a cursory search supports this) is that there is no analytical solution.

We suspect however, that there should be a conserved quantity here: energy. The kinetic energy is given by $\frac{1}{2}mr^2\dot{\theta}^2$ and the potential energy is given by (above the lowest point) $mgr\left(\cos(\theta) - 1\right)$. Therefore we expect the following to be true:

$$ mr\left(\frac{1}{2}r\dot{\theta}^2 + g\cos(\theta) - g\right) = C $$

Which really means that the following is constant:

$$ \frac{1}{2}r\dot{\theta}^2 + g\cos(\theta) = C $$

You can verify, through inspection that, sure enough, if you differentiate the above equation with respect to time, you get:

$$ r\dot{\theta}\ddot{\theta} - g\sin(\theta)\dot{\theta} = 0 \leadsto \dot{\theta}\left(r\ddot{\theta} - g\sin(\theta)\right) = 0 $$

Finally we recover the original acceleration equation:

$$ r\ddot{\theta} = g\sin(\theta)\text{, q.e.d.} $$

Comments

I'll be the first admit this isn't a complete answer. 1) I worked backwards from what we already assumed (conservation of energy) and 2) at no point did I show what that value of C should be or how it can be used to calculate the actual speeds.

Super late edit

It's worth pointing out that our differentiation of energy gave us a weird (yet valid) solution which was $\dot{\theta} = 0$, corresponding to the case that we hold the bob at any angle we want--and never ever! move it!

Note that this solution isn't possible from the original assumptions because this extraneous solution requires introducing another force (one that cancels out gravity exactly). In the original, $r\ddot{\theta} = g\sin\left(\theta\right)$, certainly $\theta = 0$ (and thus $\dot{\theta} = 0$) is a solution but $\theta = A$ (a constant) does not satisfy this equation.

Jared
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