Unstable equilibrium in a pendulum Consider a pendulum with a bob and a massless, rigid, hinged rod attached to the bob. The bob is at rest at the bottom most position. Neglecting friction, is it possible to impart such a velocity (parallel to the horizontal) to the bob so as to make it stay upright in an unstable equilibrium. I know that if I impart $\sqrt{4rg}$ velocity to the bob, where $r$ is the length of the rod, the velocity at the top most point will be zero. I asked my physics teacher that if this will cause the pendulum to stay in an unstable equilibrium at the top most position, but he said no, his argument being that the inertia of the bob will break the equilibrium. But my argument was that at the top most position, the velocity of the bob will be zero, so the inertia will actually support the equilibrium. Who is right here?
 A: If you initially give to the bob a velocity $\sqrt{4rg}$, it will actually take an infinite time for the bob to reach the top! A little lesser velocity will cause the bob to stop earlier and come back toward the initial point, while a little greater one will take the bob over the top (the motion will continue, with increasing velocity, to the other side).
Regarding metastable equilibrium: however small a disturbance (thermal motion, ...), the top position is an unstable equilibrium. Remember that the notion of equilibrium is a local one: lowest point of the circle is stable, because a small displacement will cause the bob to come back; top position in an unstable one, so that when the bob leaves that spot it will tend to go further and further away (in the present case the bob is bonded by the rod, but the same concept applies: it just won't turn backward to the top position). From a practical (i.e. realistic) point of view, it's hard to keep the bob on the top spot!
A little math to see why it takes an infinite time to go from the bottom to the top position, when the initial velocity is just enough to get the bob there.
Let $\theta$ be the angle determining the bob position: if initially $\theta=0$, the top position is at $\theta=\pi$. Linear velocity $v=r\dot{\theta}$, so that for $$v_0=\sqrt{4gr} \implies \dot{\theta}_0= \frac{v_0}{r}=\sqrt{\frac{4g}{r}}$$ Conservation of energy may be written as:
$$E=\frac{1}{2} mr^2\dot{\theta}^2 + mgr(1-\cos(\theta))$$
This way, potential energy is 0 at the initial position, while kinetic energy is $$\frac{1}{2}mr^2\big(\frac{4g}{r}\big)=2mgr$$
Energy is a constant of motion, so at every instant of time $E=2mgr$.
It follows that:
$$\frac{2mgr}{\frac{1}{2}mr^2}=\dot{\theta}^2 + \frac{mgr}{\frac{1}{2}mr^2}(1-\cos(\theta))$$
so that the equation for $\dot{\theta}$ [using $1-\cos(x)=2\sin^2(x/2)$] is
$$\dot{\theta} = \sqrt{\frac{4g}{r}}\sqrt{1-\sin^2(\theta/2)}$$
This can be solved (Corben & Stehle, pag. 51) by introducing
$$y=\sin(\theta/2) \implies \dot{y}=\frac{1}{2}\cos(\theta/2)\dot{\theta}$$
now
$$\cos(\theta/2)=\sqrt{1-y^2} \Rightarrow \dot{y}=\frac{1}{2}\sqrt{1-y^2}\dot{\theta} \Rightarrow \dot{\theta} = \frac{2\dot{y}}{\sqrt{1-y^2}}$$
leading to
$$\frac{2\dot{y}}{\sqrt{1-y^2}}=\sqrt{\frac{4g}{r}}\sqrt{1-y^2}$$
As $\sqrt{\frac{4g}{r}} = \frac{v_0}{r}$, we finally get to
$$\dot{y}=\frac{1}{2}\big(\frac{v_0}{r}\big)(1-y^2)$$
which is easily solved remembering that
$$\frac{d}{dx}\tanh(x)=1-\tanh^2(x)$$
so that
$$y(t)-y(t_0) = \tanh\Big(\frac{v_0}{2r} (t-t_0)\Big)$$
For $t_0=0 \to \theta=0$ and for $t=t^* \to \theta=\pi$. In terms of $\theta$, the solutions reads
$$\sin(\pi/2)-\sin(0) = \tanh\Big(\frac{v_0}{2r} (t^*-0)\Big) \implies 1 = \tanh\Big(\frac{v_0}{2r} t^*\Big)$$
As $\tanh(x)$ tends to 1 as $x\rightarrow +\infty$, it follows that it takes the bob an infinite amount of time to get to the top position with the given initial velocity. 
A: Since it is a rigid rod, you are probably right. If the rigid rod is replaced with a string, then your teacher would be right, as the velocity at the top must be non zero in order for the string to remain tight and not collapse before reaching the top. 
In real life scenarios, however, it is nearly impossible to maintain an unstable equilibrium.
A: Maybe in a perfect world with no thermal or atmospheric disturbance (yes simulation) this could happen. Zeno holds true regarding infinite time, but for all practical purposes the bob would appear to get there in a reasonably short time.
In the real world control systems engineers accomplish this feat all the time using feedback to overcome the small disturbances mentioned above. Once the bob reaches 12 o'clock, the controls dither the pivot point back and forth to stabilize it. 
In fact random dither without feedback control can be used to stabilize the inverted pendulum. See an amazing demonstration at Harvard here:
http://youtu.be/5oGYCxkgnHQ
