Where does this derivation of the equation of motion for a simple pendulum go wrong?

Question

I am attempting to derive the following equation of motion for a simple pendulum:

$$\theta''(t) = - \frac{g}{l} \sin(\theta).$$

For background, see this Wikipedia article. I understand this page has multiple derivations, but I don't like any of them so I am attempting my own.

My attempt does not quite work. I am likely missing something stupid. What is it?

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The strategy is to obtain two expressions for the acceleration $a(t)$ of the pendulum bob and set them equal to one another. The first we obtain via Newton's second, the second is obtained by writing down an equation for the position $x(t)$ and deriving it twice.

First, let $u(\alpha) = (\sin(\alpha), -\cos(\alpha))$ for any $\alpha$. Note that $u'(\alpha) = (\cos(\alpha), \sin(\alpha))$ is $u(\alpha)$ rotated $- \pi / 2$.

The position function can be written:

$$x(t) = u(\theta(t))$$

where $\theta$ is an unknown function of time describing the angle of the pendulum at each instant. Using the product rule and the chain rule, we obtain:

$$x'(t) = \theta'(t) u'(\theta(t))$$

$$a(t) = x''(t) = \theta''(t) u'(\theta(t)) + \theta'(t)^2 u''(\theta(t))$$

Now, the net force on the pendulum bob is

$$F = -mg \sin(\theta(t)) \ u'(\theta(t))$$

since the tension on the string and the component of gravitational force parallel to the position vector negate one another.

By Newton's second,

$$a(t) = -g \sin(\theta(t)) u'(\theta(t))$$

Setting the two equations equal to one another:

$$\theta''(t) u'(\theta(t)) + \theta'(t)^2 u''(\theta(t)) = -g \sin(\theta(t)) u'(\theta(t))$$

If we "forget about" the second term on the left, we obtain the desired relation. Note also that $u''(\alpha) = - u(\alpha)$, so this is some component of acceleration that always normal to the arc that the pendulum travels in. Have I forgotten some reason why it shouldn't exist? What step above is incorrect?

Answer 1

The extra term on the left-hand side indicates that there is a force directed along the string that you failed to account for. This is the centripetal acceleration, that is the force that keeps the bob moving in a circular path.

Your error occurs when you say that "the tension on the string and the component of gravitational force parallel to the position vector negate one another". If this were the case then the bob would move in a straight line. Thus we see that the tension force must be greater than the force of gravity perpendicular to the motion of the bob. The difference between tension and gravity is the centripetal force.

Answer 2

The statement "the tension on the string and the component of gravitational force parallel to the position vector negate one another" is false.

The tension is greater than or equal to the gravitational force parallel to the position vector because it provides the centripetal force needed in order for your pendulum to go in a circle.

So there should be an additional term on the RHS.

To obtain your desired expression though, you will have to decompose your vector equation of motion into components. Note that $u''$ and $u'$ are orthogonal, thus, dotting your equation with $u'$ will extract $\theta''(t) = - (g/l) \theta(t)$, while dotting with $u''$ will extract the radial force equation.

Also note that you are missing factors of $l$.

Answer 3

I think you can just get this from conservation of energy, no?

$$K = P$$ $$\longrightarrow \frac{1}{2}mv^2 = mgl\cos\theta$$ where $\theta$ is the angle from the horizontal (i.e. pendulum is held at $\theta=0$ before allowed to drop), and $v$ is the instantaneous tangential velocity of the pendulum mass.

Then we have $v$ = $|\dot{\textbf{r}}|$, where $$ \dot{\textbf{r}} = \frac{d}{dt} \left(\hat{x}l\cos\theta + \hat{y}\sin\theta\right) = l\dot{\theta} \left(-\hat{x}l\sin\theta + \hat{y}\cos\theta\right) $$ so that $$ v^2 = l^2\dot{\theta}^2 $$

Now differentiating both sides: $$ \frac{d}{dt} \left( \frac{1}{2}ml^2\dot{\theta}^2 \right) = \frac{d}{dt} \left( mgl\cos\theta \right) $$ $$ \longrightarrow ml^2\dot{\theta}\cdot \ddot{\theta} = - mgl\sin\theta\cdot{\dot{\theta}} $$

Dividing both sides by $ml^2\dot{\theta}$ gives $$ \ddot{\theta} = - \frac{g}{l}\sin\theta $$

Answer 4

Well It won't let me comment, so I have to answer Mr mcFreid here.

As you will notice, I discovered an error in my derivation; several times in fact so I deleted that false derivation.

But even though my method was erroneous; it was correct in at least some statements.

One statement that was, and still is correct, is that my S = L.sinA is absolutely correct. I even said that S was the horizontal displacement ; it was NOT the arc length.

So the OP's equation, and the supporting statement in my physics hand book do seem to be correct. That's why I removed my mistake ridden derivation.

I assume that somebody else has already, or soon will, show the OP a better derivation than the three in wiki, that he said he didn't like. At least I tried to give him one.

And your most recent comment was NOT posted before I had already removed my erroneous derivation; perhaps you didn't notice that