A long horizontal rod has a bead which can slide along its length and is initially placed at a distance L from one end A of the rod Question: A long horizontal rod has a bead which can slide along its length and is initially placed at a distance $L$ from one end $A$ of the rod.The rod starts from rest in angular motion about $A$ with a constant angular acceleration $α$.
If the coefficient of friction between the rod and the bead is $μ$ and gravity is neglected then what is the time after which the bead starts slipping?
Online Solution:

*

*The bead will start sliding when centripetal force is equal to limiting friction.
$$mω²L = μN\quad\dots(1)$$

*We know that here normal reaction is provided by tangential force. $$mLα = N$$

*Substituting value of $N$ in eq.$(1)$
Now, at time $t, \,\omega = 0 + αt \implies w = αt$.
Now,
$\displaystyle mω²L = μmLα\\
 m(αt)²L = μmLα\\
 (αt)² = μ\\
\therefore t = √(μ/α)$
My questions:
My issue with this solution is that why is the centripetal force equated with the frictional force? Aren't both forces directed radially inward? Intuitively the ball would be moving outwards but why exactly?
 A: 
Aren't both forces directed radially inward?

It seems that you have confused centripetal force for a "new  kind" of force , that acts towards the center , whenever there's a circular motion. This is a common misconception.
Whenever an object undergoes circular motion, it must have an acceleration=$v^2/r=\omega^2r$ towards the center (this is called the centripetal acceleration). Which is equivalent to saying, it experiences a force= $mv^2/r$ towards the center. Now the important part is, this is not an extra force acting on the object.
The components of the existing forces on the object, towards the center, always add up to $mv^2/r$. In your example, friction is the only force that acts towards the center. Thus, the value of the frictional force has to be = $mv^2/r$. The appropriate way to say this is that the frictional force is the centripetal force.
Here's another example: consider a satellite(mass $m$) in circular orbit (of radius $r$) around the earth (mass $M$). Here, the only force acting on the satellite towards the center is the gravitational force. Thus, the gravitational force has to be equal to $mv^2/r$, i.e ,$GMm/r^2=mv^2/r$.Thus, $v=\sqrt{GM/r}$.Here, we will say that the Gravitional force is the centripetal force.
Summary till now:Centripetal force is not an"extra" force on the object. Whenever theres  circular motion,the central components of all the EXISTING forces on the object always add up to $mv^2/r$.This is what the centripetal force is: simply the sum of all the central components of the existing forces.

Intuitively the ball would be moving outwards but why exactly?

Lets analyse your problem a bit more thoroughly, and through a different perspective:

*

*First: We need to understand that the Rod is what's undergoing circular motion:
i.e: each point on the rod is moving in a circle. The radius of the circle is different for different points on the rod: if the length of the rod is x, then the center rotates in a circle of radius =$x/2$. The end point of the rod rotates in a circle of radius $x$.

*Each point on the rod has an acceleration= $\omega^2r$ towards the center and $\alpha r$ tangentially, where r is the radius of the circle in which the point rotates.

*Let the point on the rod, which is in contact with the bead be P. As per the question, $r=l$.

*The Contact Forces (friction and $N$)always ensure (as long as friction is static) that the bead and P have identical motion: i.e they do not move relative to each other: i,e  acceleration of bead=acceleration of point P.

*Let the central direction be denoted by $\hat{i}$, and the tangential direction by $\hat{j}$. Then , $\vec{a_{point-p}}=\omega ^2l\hat{i}+\alpha r\hat{j}$.

6.Point no.4 tells us that the acceleration of the bead must be identical to this.Thus, we conclude that the friction acts towards the center, (since the normal acts towards the tangent), so that we have, $\vec{a_{bead}}=(f/m)\hat{i}+(N/m){j}$, which we can equate to $\vec{a_{point-p}}$, to yield $f=m\omega^2l$ and $N=ml\alpha$.
Note: The result in point 6 can be obtained directly through point 1 and 4. 1 and 4 together imply that the bead is also undergoing circular motion.However,points 7 and 8 show that analyzing the motion of the bead and point P separately turns out to be quite useful.
7.There will come a time when the value of $m\omega^2l$ will exceed $\mu N$.($t=\sqrt{\mu/\alpha}$)
$\vec{a_{bead}}=(f/m)\hat{i}+(N/m){j}$, and we still have $N/m=l\alpha$,but now, we cant have $f/m = \omega^2l$, since that would imply $f>\mu N$. Clearly, once $m\omega^2l$ exceeds $\mu N$, it is no longer possible to have the bead and point moving together. Which means $f$ will be kinetic from this point onwards. $f$ will have a fixed value=$\mu N$.


*After this point, lets compute the acceleration of the bead relative to point P:
$\vec{a_{rel}}=\vec{a_{bead}}-\vec{a_{p}} = [(f/m)\hat{i}+(N/m){j}]-[\omega ^2l\hat{i}+\alpha r\hat{j}]=(\dfrac{f-m\omega^2/l}{m})\hat{i}.$. Since $f=\mu N$ and $m\omega^2l > \mu N$, We can conclude that $\vec{a_{rel}}$ is along the $\hat{-i}$ direction.Thus, relative to point p, the bead accelerates outwards.

A: Either you stay in Earth frame and say that the bead rotates only up to the time that frictional force on it can provide sufficient centripetal force i.e. only up to when friction can turn the velocity vector that is continuously increasing. Or you go in the rotating frame and apply a (pseudo) centrifugal force on the bead that the frictional force can only balance up to when the bead is not too fast. The mantra is that the centrifugal force is to be balanced whereas the centripetal force is to be caused by another force that turns the velocity vector.
A: Your confusion is thinking of the frictional force and centripetal force as two different forces.  In this case the frictional force is the centripetal force.  The centripetal force is the force necessary to keep the ball in circular motion at a constant distance $L$ from the center of motion.  That is what the frictional force is doing.
It is analogous to a ball held in circular motion by a string.  In that case the centripetal force is provided by the tension in the string.  There is not a centripetal force and a tensile force.  They are one in the same.
They are one in the same in your case also.
A: In this case the coefficient determines the maximum value of the static friction which provides the centripetal acceleration.
