# Mathematical model of pendulum

• limofunder

## Homework Statement

W
We know the model of an ideal pendulum at rest is given by
$$L \ddot{\theta} + g sin \theta =0, t\geq0$$
$$\dot{\theta}(0)=0$$
$$\theta(0)=\theta_0$$
where $$\theta(t)$$ is the pendulum angle at time t, L is the length of the pendulum, and g is gravity.
Now, consider the total energy per unit mass of the pendulum given by
$$E(\theta,\dot{\theta}) = \frac{1}{2}L \dot{\theta}^2 - g cos \theta$$
Show that this equation is constant along the previously given solutions (the first 3 equations initially given, let's call them EQ [1] ), $$\forall t\geq0$$. Use this result to show that the solution of [1] should satisfy
$$\frac{1}{2}L\dot{\theta}^2-g cos \theta + g cos \theta_0 = 0, \forall t\geq0$$
(lets call this EQN [2])

## Homework Equations

we know that if E(theta, thetadot) is a constant, then
$$\frac{d E(\theta,\dot{\theta})}{dt} = 0$$

## The Attempt at a Solution

so, taking the derivative of the energy, we have:

$$\frac{d E(\theta,\dot{\theta})}{dt} = \frac{d}{dt}[\frac{1}{2} L \dot{\theta}^2 - g cos \theta]$$

$$= L \ddot{\theta} +\dot{\theta}g sin\theta$$
however, this does not seem to equal [1], as $$\dot{\theta}gsin\theta$$ never appears in the solution. what am I doing wrong?
or do we not take the time derivative of the second term in the energy equation (the sin(theta) term), thus giving
$$\frac{d E(\theta,\dot{\theta})}{dt} =L \ddot{\theta} +g sin\theta$$ which is equal to zero by definition of [1], thus if the derivative of the energy equation is equal to zero, then by definition of constant, energy is constant!
now, i am in confusion as how to use this result to prove EQN [2]?

Last edited:
Hi limofunder!

(have a theta: θ )
$$\frac{d E(\theta,\dot{\theta})}{dt} = \frac{d}{dt}[\frac{1}{2} L \dot{\theta}^2 - g cos \theta]$$

$$= L \ddot{\theta} +\dot{\theta}g sin\theta$$

hmm … d/dt (Lθ') = Lθ''

but d/dt (Lθ'2) = … ?

$$\frac{d E(\theta,\dot{\theta})}{dt} = \frac{d}{dt}[\frac{1}{2} L \dot{\theta}^2 - g cos \theta]$$

$$= L \ddot{\theta} +\dot{\theta}g sin\theta$$

$$\frac{d}{dt}\left(\frac{1}{2} L \dot{\theta}^2\right)=L\dot{\theta}\ddot{\theta}$$

EDIT: Tinytim beat me to it

ok, so then that gives way to
dE/dt = θ' (Lθ''+g sinθ)
the second term being by definition of [1] >= 0 for all time.
Thank you,
so that still begs the question, I do not see how this forces
[1] to satisfy
1/2 Lθ' - g cosθ + g cosθ0

ok, so then that gives way to
dE/dt = θ' (Lθ''+g sinθ)
the second term being by definition of [1] >= 0 for all time.
Thank you,
so that still begs the question, I do not see how this forces
[1] to satisfy
1/2 Lθ' - g cosθ + g cosθ0

Because E = constant is a solution, so just call the constant gcosθ0