Motion of bead on a rod Why does the bead that is free to move on on a frictionless rod, move outward when the rod is rotated with constant angular velocity about one of its end? 
So, due to change in direction there is centripetal acceleration ($\omega^2 r$). The other acceleration is due to coriolis force ($2 \omega v_r$, where $v_r$ is radial velocity), which is tangentially  directed. 
So, if the force is tangential, then why does the bead move outward?
 A: In the frame co-rotating with the rod, there is a centrifugal force $$F_{cf}=m\omega^2 r \tag 1$$ acting on the bead which moves it outward along the rod. The same happens in centrifuges and in your laundry machine when the water is forced out of your laundry by high-speed rotation.
Following the comment of @suiz, I'll try to solve the problem in the inertial frame using the Lagrange equations. The Lagrange function in polar coordinates with $\phi=\omega t$, where $\omega $ is a constant angular velocity, is given by $$L=\frac {m}{2}(v_r^2+v_{\phi}^2)=\frac {m}{2}(\dot r^2+r^2\omega^2) \tag 2$$ Thus the Lagrange function is independent of $\phi$ and the Lagrange equation is: $$\frac {d}{dt}\frac {\partial L}{\partial \dot r}-\frac {\partial L}{\partial r}=m(\ddot r-r\omega^2)=0 \tag 3$$ giving the homogeneous ordinary second order differential equation for $r(t)$ $$\ddot r-r\omega^2=0 \tag 4$$ with the general solution $$r(t)=A\exp(\omega t)+B\exp(-\omega t) \tag 4$$ Assuming the initial conditions $$r(t=0)=r_0; \dot r(t=0)=0 \tag 5$$ the solution for the time dependent radius is $$r(t)=r_0\cosh(\omega t) \tag 6$$ yielding the radial velocity $$v_r=r_0 \omega \sinh(\omega t) \tag 7$$ and the "radial acceleration"* $$\ddot r=r_0 \omega^2 \cosh(\omega t)=r\omega^2 \tag 8$$ Thus, it can be seen that the frictionless bead flies away radially eventually with exponentially increasing radial velocity. It is interesting that due to the exponential increase, at the end of a long rod the magnitude of the radial velocity $v_r$ tends towards the rotational velocity $v_{\phi}=\omega r$ so that the bead would fly away practically at 45° to the momentary direction of the radius. 
The "radial acceleration" eq. (8) is exactly the centrifugal acceleration in the rotating frame see eq. (1). [In the rotating frame, eq. (4) and its solution follows directly by using Newton's law with the centrifugal force (1).] It is also noteworthy that in the present problem of the frictionless bead, there is no centripetal force present. 
It is instructive to inspect the kinetic energy of the bead $$T=\frac {m}{2}(v_r^2+v_{\phi}^2)=\frac {m}{2}r_0^2\omega^2 [\sinh(\omega t)^2+\cosh(\omega t)^2=\frac {m}{2}r_0^2\omega^2 \cosh(2\omega t) \tag 9$$ which increases strongly with $t$ and eventually becomes exponential.
Although this derivation for the inertial system using the Lagrange function is completely transparent, it would be interesting if somebody could come up with an intuitive physical explanation for the strong increase in (radial) velocity and total kinetic energy of the bead without using the concept of a centrifugal force.
[*] In quotation marks, because this is the second time derivative of the generalized coordinate $r$ and not the radial vector component of the acceleration in polar coordinates, which is $\ddot r-r\omega^2$, and thus according to eq. (4) equal to zero, as @pgml has correctly pointed out in his answer below.   
A: There is no force directed along the rod ("radial"). And at each instant the radial component of the acceleration is zero. But the radial component of the acceleration is not the derivative of the radial component of the velocity: denoting acceleration by $\boldsymbol{a}$, velocity by $\boldsymbol{v}$, and radial by the subscript $\text{r}$,
$$\boldsymbol{a}_{\text{r}} = \left(\frac{\mathrm{d}\boldsymbol{v}}{\mathrm{d}t}\right)_{\text{r}} \neq \frac{\mathrm{d}(\boldsymbol{v}_{\text{r}})}{\mathrm{d}t}.$$
In other words, we must first take the time derivative and then project along the rod, not vice versa. The two operations don't commute because we are considering rotating projection directions. For this reason we must also be very careful and distinguish "the radial component of the acceleration" from "the second time derivative of the radial coordinate", because they're different.
Let's see this in detail and work out the solution.
Consider a coordinate system $xy$ on the plane of the rotating rod and fixed in an inertial frame, with the rod's pivot at $(0,0)$.
The position of the bead can be written as
$$r(t)\,(\cos\omega t, \sin\omega t),$$
where $r(t)$ is the distance from the pivot and we're assuming that the rod is on the $x$ axis at $t=0$.
The velocity of the bead is, denoting time derivatives with a superposed dot,
$$\boldsymbol{v}(t) = \dot{r}(t) \, (\cos\omega t, \sin\omega t) + \omega r(t) \, (-\sin\omega t, \cos\omega t).$$
The first summand is the instantaneuos radial component, the second is the azimuthal (normal to the rod) component. In the inertial frame the bead has a spiral motion. The velocity, tangent to its trajectory, is inclined with respect to the rod.
The acceleration of the bead is, taking the derivative of all terms above and omitting some "$(t)$",
\begin{align}
\boldsymbol{a}(t) &= 
\ddot{r} \, (\cos\omega t, \sin\omega t) +
\omega \dot{r} \, (-\sin\omega t, \cos\omega t) +
\omega \dot{r} \, (-\sin\omega t, \cos\omega t) -
\omega^2 r \, (\cos\omega t, \sin\omega t) 
\\
&= (\ddot{r}-\omega^2 r)\, (\cos\omega t, \sin\omega t) +
2\omega \dot{r} \, (-\sin\omega t, \cos\omega t).
\end{align}
In the first equality we can see that both the radial and the azimuthal components of the velocity contribute to the radial component of the acceleration; and both contribute to its azimuthal component too.
The radial component of the acceleration, $\ddot{r}-\omega^2 r$, has two terms because the velocity is changing not only in direction, but also in magnitude: the term $-\omega^2 r$ reflects the former change (it's the centripetal acceleration that the bead would have if it were glued to the rod); the term $\ddot{r}$, the latter.
Multiplying by the mass of the bead we have
$$m\boldsymbol{a}(t) = m\,(\ddot{r}-\omega^2 r)\, (\cos\omega t, \sin\omega t) +
2m\omega \dot{r} \, (-\sin\omega t, \cos\omega t).$$
By Newton's second law, since we are in an inertial frame, the azimuthal component of the expression above must be equal to the sum of the forces normal to the rod constraining the bead. There are no radial forces, so the radial component must identically vanish, which happens only if
$$\ddot{r}(t)-\omega^2 r(t) = 0.$$
That is, the acceleration cannot have a radial component. But note again that such component is not just $\ddot{r}$. This is the same equation that  freecharly's answer obtains through a Lagrangian.
Just as in freecharly's answer, the general solution of the differential equation above is
$$r(t) = A\exp(\omega t) + B\exp(-\omega t).$$
Let's assume that at $t=0$ the position of the bead is $r(0)=r_0>0$ and the radial component of its velocity is $\dot{r}(0)=0$. Then $A=B=r_0/2$ and
$$r(t) = \frac{1}{2}r_0\,[\exp(\omega t) + \exp(-\omega t)] \equiv r_0\,\cosh\omega t.$$
The velocity is
$$\dot{r}(t) =  \omega r_0\,\sinh\omega t.$$
So the bead is moving inwards for $t<0$ and outwards for $t>0$. This makes sense if we consider a non-inertial, rotating system fixed with the rod: the inertial force is pushing the bead outwards (and sideways), so the only way for it to have zero radial velocity at $t=0$ is to have been thrown towards the centre at some earlier time $t<0$. The inertial force made it decelerate and then reverse direction at $t=0$.
If you assume some other velocity at $t=0$ (but compatible with the fact that $r\geq 0$ at all times, since it isn't clear whether the rod is open or closed at the pivot, or whether it extends in the other direction) you'll notice that in any case $r(t) \to \infty$ as $t\to+\infty$, that is, the bead is eventually always pushed outwards.
A: This may be a bit late but this problem has good insights:
Lets consider the problem in inertial frame by polar coordinates. $$F=m((\ddot{r}-r{\dot{\theta}^2})\hat{r} +(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta})$$
The rod only applies a force in the tangential direction. ($F_\theta=N\hat{\theta}$). The force in the radial direction is $0$. Hence we see that $a_r$ should be zero and there should be no motion along radial direction.
Since we assumed no motion along radial direction, let's take $\dot{r}=0=\ddot{r}$ and see what will happen.
$$\vec{v}=r\dot{\theta}\hat{\theta}$$
and hence by differentiating, acceleration is:$$\vec{a}=-r\dot{\theta}^2\hat{r}+(\dot{r}\dot{\theta}+r\ddot{\theta})\hat{\theta}$$
Since $\dot{r}=0$ and $\ddot{\theta}=0,$ We get the preposterous result:$$\vec{a}=-r\dot{\theta}^2\hat{r}$$
This result implies that entire acceleration is directed radially which contradicts the facts that $F_r=a_r=0$ and $N \neq 0$ . What went wrong?
Well we were right when we said that $F_r=a_r=0$, but wrong when we assumed that $\dot{r}=0=\ddot{r}$. This is the stubborn illusion that we carried from Cartesian coordinate system . In Cartesian coordinate system, since $$F_x=ma_x=m\ddot{x}$$ $$F_x=0 \Rightarrow \ddot{x} =0$$
But in polar coordinates the same intuition breaks down since $$F_r=ma_r =(\ddot{r}-r{\dot{\theta}^2})\neq{m\ddot{r}}$$  And hence despite $F_r=0; \ddot{r}\neq0,r{\dot{\theta}^2}\neq 0$
So we have to accept that motion in radial direction is inevitable and remove the subconscious assumptions that $F_r = m\ddot{r}$ and velocity and force should be in the same direction.
So, radial force is zero, radial acceleration is zero but radial velocity is non zero ($\dot{r} \neq 0$) and the fact it gets away from centre faster is not surprising  ($\ddot{r} \neq 0$).
