Lagrangian and conservation of energy

If Lagrangian of the motion is

$$\mathcal{L}=\frac{1}{2}m\left(a^2\dot\phi^2+a^2\dot\theta^2\sin^2\phi\right)+mga\cos\phi,$$

how can I show that total mechanical energy is conserved? I've read this:

If the time $t$, does not appear [explicitly] in Lagrangian $\mathcal{L}$, then the Hamiltonian $\mathcal{H}$ is conserved. This is the energy conservation unless the potential energy depends on velocity.

Potential energy of this motion doesn't depend on velocity. Also, $t$ does not appear explicitly in Lagrangian. Is this enough to say that total mechanical energy is conserved?

Answer

The reason for the asseveration

If time $t$, does not appear in Lagrangian $\mathcal{L}$, then the Hamiltonian $\mathcal{H}$ is conserved. This is the energy conservation unless the potential energy depends on velocity.

is that, from the definition of the hamiltonian as the Legendre transformation, $$\mathcal{H}\equiv\sum_i\dot{q}_i\frac{\partial\mathcal{L}}{\partial\dot{q}_i}-\mathcal{L}\hspace{1in}(\dagger)$$ and knowing that for any function in phase space, $F=F(q_i,p_i,t)$, $$\frac{dF}{dt}=\frac{\partial{F}}{\partial{q}_1}\frac{d{q_1}}{d{t}}+\ldots+\frac{\partial{F}}{\partial{q}_n}\frac{dq_n}{dt}+\frac{\partial{F}}{\partial{p}_1}\frac{d{p_1}}{d{t}}+\ldots+\frac{\partial{F}}{\partial{p}_n}\frac{dp_n}{dt}+\frac{\partial{F}}{\partial{t}}\\=\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\dot{q}_j+\frac{\partial{F}}{\partial{p}_j}\dot{p}_j\right)+\frac{\partial{F}}{\partial{t}}\\=\left\{F,\mathcal{H}\right\}+\frac{\partial{F}}{\partial{t}}$$ where $\left\{F,\mathcal{H}\right\}$ is the Poisson bracket of $F$ and $\mathcal{H}$, defined as $$\left\{F,\mathcal{H}\right\}\equiv\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\dot{q}_j+\frac{\partial{F}}{\partial{p}_j}\dot{p}_j\right)=\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\frac{\partial{\mathcal{H}}}{\partial{p}_j}-\frac{\partial{F}}{\partial{p}_j}\frac{\partial{\mathcal{H}}}{\partial{q}_j}\right)$$ if $\mathcal{L}=\mathcal{L}(q_i,p_i)$, i.e. the Lagrangian does not depend explicitly on time, which in turn means, from the definition $\mathcal{L}\equiv{T}-V$, that kinetic energy $T$ and potential $V$ does not depend explicitly on time, then $\frac{d\mathcal{H}}{dt}=\left\{\mathcal{H},\mathcal{H}\right\}=0$. Now, a constant of motion is precisely some function $F$ of phase space that is independent of time, i.e. such that $\frac{dF}{dt}=0$, so in this case the hamiltonian would be conserved. Now, from the definition $(\dagger)$, you may verify that the hamiltonian equals the energy, $$\mathcal{H}\equiv{T}+V$$ only if $V=V(q_i)$ alone. So if that is the case, then energy would be conserved.

So identify that in your Lagrangian and get your conclusions, anyway you can always verify it this way for your particular case.