"Centrifugal" weight spinning causing vertical weight to lift So the problem Im dealing with has a corresponding diagram below where a mass is being spun in a circular path (at an increasing angular velocity) causing the hanging weight to be lifted through the rope. There is a demonstration here by Prof Julius Miller at the 8 minutes and 22 second mark.
Could anyone explain using senior high school terminology why the weight is being lifted?

 A: When the mass $m$ performs a circular motion, the centripetal force is provided by the tension in the rope. Let angular velocity of the mass be $\omega$ and assume the mass performs a horizontal circular motion(which assumes the part of the string is horizontal, but in reality it is slightly inclined*). Then tension of the rope can be found using $T=mr\omega^2$. When $\omega$ increases, $T$ also increases. For simplicity, if we consider the string is massless, the tension throughout the string is the same. That means, the mass $M$ is under that tension by the rope. That tension acts upwards while weight $Mg$ acts downwards. When $T$ overcomes $mg$ there is a net vertical force on the mass $M$ helping it to accelerate upwards.

*Actually you can find this angle easily, then the corresponding equation will be $T\sin\theta=mr\omega^2$, whereas $\theta$ is the angle between the string and the vertical. From vertical equilibrium of $m$ you can get another equation $T\cos\theta=mg$.
A: weight spinning causing vertical weight to lift
In the horizontal plane there is only one force acting on the rotation mass and that force is the horizontal component of the tension on the string  acting radially inwards.
The equation of motion of the mass on the ned of the string is
$T_{\rm horizontal} = m \dfrac {v^2}{r}$ where $m$ is the mass on the end of the string, $v$ its speed and $r$ the radius of the circle.
Now assume that $T_{\rm horizontal}$ stays approximately constant, $\approx Mg$, where $M$ is the mass of the hanging weight, what will happen if the speed of the rotating mass is increased?
$T_{\rm horizontal}$, and $m$ cannot change so the only way for the equation of motion to be satisfied is for $r$ to increase, $T_{\rm horizontal} = m \dfrac {v^2\uparrow}{r\uparrow}$.
To increase the radius the hanging mass must rise up.
The intermediate phase of the radius of motion of the rotating mass increasing and the hanging mass accelerating upwards is complex.
The hanging mass is subjected to a net upward force and the rotating mass as well as rotating faster also has a net radial acceleration with the radius of the motion increasing.
Think of the following as infinitesimal steps.
The rotating mass is made to move faster and to keep rotating mass at the same radius the tension in the string must increase.
An increase in the tension of the string will result on a net force on the hanging mass and the hanging mass will move upwards.
The radius of the motion of the rotating mass increases so it needs a small tension to maintain its motion.
The net force on the hanging mass decreases such that the hanging mass is no longer moving upwards.
The rotating mass is made to move even faster . . . . . . . . .
A: Imagine the weight has some initial angle like 20° instead of 90°. When you start to spin the weight, it's angular velocity cause it to have a radial force wich is perpendicular to the center (90°) and actually pushes it outwards, but you could split that outwards force to two forces (since force is just a vector): a force in the direction of the rope, and a force perpendicular to the first force, which causes the object to rise and increase it's angle.
If you're struggling to understand how radial force works, just Google it. I'm sure there are plenty of intuitive explanations.
A: The spinning thing is changing its velocity. Velocity is a vector - both direction and magnitude. If that vector doesnt stay pointed in the same direction with the same magnitude, then it is changing. Changing velocity is called acceleration. So the spinning thing is accelerating. At any split second, its velocity is tangent to the circle of travel. But its acceleration (change in that velocity vector) is toward the center of the circle. Last two sentences important.
To accelerate something takes a force: $F=ma$. Given which way it’s accelerating you know the force it needs.
