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?

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?

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?

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?

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

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

For background, see http://en.wikipedia.org/wiki/Pendulum_(mathematics)#Simple_gravity_pendulum. 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?

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?

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?

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?