Question: If I have a Lagrangian (made up equation in this case):

$$L = \frac{1}{2}mr^2\dot\theta + \frac{1}{4}mg\ddot\theta \, ,$$

can I immediately conclude that the total energy is constant because $\partial L / \partial t = 0$, because $L$ does not depend explicitly on $t$? Can I also conclude that the angular momentum is constant because $L$ does not depend explicitly on $\theta$, and that the linear momentum isn't because $L$ depends on $r$?

The notes seem to imply that one can just figure the constants of motion like so without any computations! Is this correct?

If not, what actual computations would I have to do to figure out the constants of motion from a Lagrangian?

Answer: 
If $q\_i$ denotes the generalised coordinates, then note that a time translation $t \to t+\epsilon$ infinitesimally corresponds to $q\_i \to q\_i + \epsilon \frac{d}{dt}q\_i$ and so $\delta q\_i = \dot{q}\_i$. The change in the Lagrangian is,

$$\delta L = \frac{\partial L}{\partial q\_i} \dot q\_i + \frac{\partial L}{\partial \dot q\_i} \ddot q\_i = \frac{d}{dt}L$$

which is the total derivative if $L$ has no explicit time dependence. By Noether's theorem, we have a conserved quantity,

$$Q = \frac{\partial L}{\partial \dot q\_i}\dot q\_i - L $$

which we can recognise as the Legendre transform of the Lagrangian, that is, the Hamiltonian and thus the energy of the system is conserved. Clearly this applies to your Lagrangian.

As you noted, your Lagrangian is also invariant under $\theta \to \theta + \alpha$ for which $\alpha \delta \theta = \alpha$, and $\delta L =0$. We thus have a conserved quantity which we can identify as the conjugate momentum, namely,

$$p\_\theta = \frac{\partial L}{\partial \dot \theta} = \frac12 mr^2.$$

Another way to see this is that the Euler-Lagrange equations read, $\frac{\partial L}{\partial \theta} = \frac{d}{dt}\frac{\partial L}{\partial \dot \theta}$ and since there is no $\theta$ dependence,

$$\frac{d}{dt}\frac{\partial L}{\partial \dot \theta} = \frac{d}{dt} p\_\theta = 0.$$
In your case, the answer is yes, it's that simple. The reason is that all the constants of motion you're dealing with are related to some translational symmetry. Energy corresponds to translations in time, angular momentum to rotations (translations in $\theta$) and linear momentum to translations.

There's a theorem that relates symmetries of the lagrangian and constants motion: **Noether's theorem**. It says that for any symmetry there's an associated constant of motion, which can be calculated using the lagrangian and the symmetry transformation. That's a general way of computing constants of motion.

For time-independent lagrangians, a symmetry (a transformation $x\mapsto x(\lambda)$ such that $L(x(\lambda),\dot{x}(\lambda))=L(x,\dot{x})$, where $x$ denotes collectively all coordinates) implies a constant of motion
\begin{align}
 C=&\sum\_i\frac{\partial L}{\partial \dot{x\_i}}\frac{\partial x\_i}{\partial\lambda}.
\end{align}

A simple derivation (without too much detail) is
\begin{align}
 \frac{dC}{dt}=&\sum\_i\left[\frac{d}{dt}
 \left(\frac{\partial L}{\partial 
 \dot{x\_i}}\right)\frac{\partial x\_i}{\partial\lambda}+
 \frac{\partial L}{\partial \dot{x\_i}}
 \frac{d}{dt}\frac{\partial x\_i}{\partial\lambda}\right] \\
 =& \sum\_i\left[
 \frac{\partial L}{\partial 
 x\_i}\frac{\partial x\_i}{\partial\lambda}+
 \frac{\partial L}{\partial \dot{x\_i}}
 \frac{\partial \dot{x}\_i}{\partial\lambda}\right]
 = \frac{dL}{d\lambda}=0.
\end{align}

Now, for the case of rotations $\theta\mapsto \theta+\lambda$, our conserved quantity is $C\_\theta=\partial L/\partial\dot{\theta}=mr^2/2$, the angular momentum. For a lagrangian invariant under $r\mapsto r+\lambda$, $C\_r$ is the conserved linear momentum. Because in our case this isn't a symmetry, $C\_r$ is not a constant of motion.

For the time symmetry we would need a generalized version of Noether's theorem but since in this case we are specifically interested in the relation between energy and time, observe that the derivative of energy is
\begin{align}
 \frac{dH}{dt}=\frac{d}{dt}(p\dot{q}-L)=
 \frac{d}{dt}\left(\frac{dL}{d\dot{q}}\dot{q}-L\right)=
 \frac{dL}{dq}\dot{q}+\frac{dL}{d\dot{q}}\ddot{q}-\frac{dL}{dt} =
 -\frac{\partial L}{\partial t}
\end{align}
and therefore, $H$ is conserved if $L$ is explicitly time-independent.