Rotational motion and angular velocity Why does $N=0$ when w=w critical? 
What does "Mass m falls away from the drum if N ≤ 0, when ω ≤ ωcritical" mean?
I don't understand the relationship between N and W here.

 A: Basically, if we know the concrete is traveling in a circle with frequency $\omega$, we know its trajectory - that is, its position vector as a function of time.  So, by taking the second derivative of that function, we know its acceleration.  For circular motion, that works out to be a vector of magnitude $R\omega^2$, always pointing toward the center of the circle, aka. centripetal.
$a = R\omega^2$
So, the acceleration increases with frequency - at low frequency, there is little centripetal acceleration.
But Newton's law tells us that acceleration is proportional to applied force:
$F_{net} = ma$
At the top of the drum, all forces are downward and so is the acceleration, so if we make downward positive, they are all positive in that equation:
$W + N = ma\\
mg + N = mR\omega^2$
The normal force is the force between the drum and the block - but the drum is not sticky, so it can only push the block down, not pull it up.  Since we made downward positive, that means N cannot be negative.  So the net downward force cannot be less than the weight!
From the equation we can see that as frequency increases, normal force does too.  That's because, when the block circles faster, it has more acceleration, so it requires more force - and the weight does not change, so the normal force must.
But by the same token, as frequency decreases, the block requires less force.  At some point, it only requires its own weight, and so N = 0.  We call that the critical frequency.
$mg = mR\omega_{critical}^2$
And if the frequency drops below that, the block actually requires less force than its own weight to keep moving on the circle.  But the force can't be less than the weight.  Hence, there is too much force on the block, and it is pulled right off of the drum - it "falls away from the drum".
$F_{net} = ma = mg > mR\omega^2\\a > R\omega^2$
=> No more circular motion!
A: In this answer I am referring to top position only.Similar arguments can be made for any position.
Well think it like this.You know that radius of curvature of circular motion is given as:
R= F/(m ω^2)
where F is net force.
Notice if we keep angular speed decreases, radius decreases. However you know that such a thing is not happening here.The reason of this is that net force is decreasing simultaneously.
Now,net force= Weight + Normal
Notice how Weight cannot change . It must be Normal that is changing . In short , as speed decreases , Normal decreases to maintain same radius .
Now you might argue why normal is coming into play anyway. This is to balance centrifugal force from frame of body.
Finally what happens when Normal becomes zero . At N = 0 , Weight can alone support the current radius .However if you want reduce speed even lesser , you need to make Normal opposite to Weight (to make net force even lesser) . This is not possible as Normal  cannot be negative. It violates the very definition of Normal.
SO after this,radius just needs to become a little lesser so that the body falls. At this stage Weight cannot provide centripetal as it is always downwards and not towards centre. In short ,body falls.
P.S. Normal is defined as the force that stops two objects from attaining the same space.
