In this video, we will observe how induced eddy currents in a copper plate slow down a magnetic pendulum. 

When the pendulum is set in motion, it usually oscillates for quite a while. This pendulum consists of a strong magnet.

If we slide a copper plate underneath the magnet while it is in motion, the magnet comes to a stop quickly. Note that copper is not a ferromagnetic material, which means it does not get attracted to a stationary magnet.

As the magnet moves across an area on the copper plate, the change in magnetic flux induces eddy currents on the plate. These eddy currents flow in such a way as to repel the magnet as it approaches the plate and attracts the magnet as it leaves the plate, therefore slowing the magnetic pendulum.

When we pull the copper sheet out from under a stationary magnetic pendulum, the eddy currents will flow in such a way that it becomes attracted to the copper sheet.

Moving the copper sheet to and fro at a certain frequency (the pendulum's natural frequency), the magnetic pendulum can be made to oscillate again.

Using a chain of rubber bands, I swung a ball around in a vertical loop. This demonstration shows how the tension in an elastic band changes according to the position of the ball, by referring to the length of the elastic band.

When the ball of mass $m$ is at the bottom of the loop, the centripetal force is given by the difference between tension $T_{bottom}$ and weight $mg$, where $T_{bottom}$ varies depending on the speed of the ball $v_{bottom}$ and the radius of the curvature $r_{bottom}$.

$T_{bottom} - mg = \dfrac{mv_{bottom}^2}{r_{bottom}}$

When the ball is at the top of the path, it is given by

$T_{top} + mg = \dfrac{mv_{top}^2}{r_{top}}$

As the weight is acting in the same direction to tension when the ball is at the top, a smaller tension is exerted by the elastic band to maintain a centripetal force. Therefore , $T_{bottom} > T_{top}$.

The GeoGebra app below shows a simpler version of a vertical loop - a circular path with a fixed radius $r$. Consider a ball sliding around a smooth circular loop. The normal contact force varies such that

$N_{bottom} = \dfrac{mv_{bottom}^2}{r} + mg$

$N_{top} = \dfrac{mv_{top}^2}{r} - mg$

It can be shown that the minimum height at which the ball must be released in order for it to complete the loop without losing contact with the track is 2.5 times the radius of the frictionless circular track.

If we were to consider the rotational kinetic energy required for the ball to roll, the required initial height will have to be 2.7 times the radius, as shown in the video below:

Many thanks to Dr Darren Tan for his input. Do check out his EJSS simulation of a mass-spring motion in a vertical plane, which comes with many more features such as the ability to vary the initial velocity of the mass, graphs showing the variation of energy and velocity, as well as an option for a mass-string motion as well.

This is the best video I have come across so far that is used to introduce the idea of perpetual motion engines and why it violates the principle of conservation of energy.

Source: https://arxiv.org/pdf/2007.00926.pdf?fbclid=IwAR2Y6CkFXeCrJH7nxf624To8RcpdRSs0e3_COto9ev6MK-Z_wnObeRHvVlg

Here's a quick video to demonstrate the movement of a ball initially moving in a circle before its centripetal force (contact force by the circular wall) is removed. The ball is observed to move in a tangent to the circle, in accordance with Newton's 1st Law, since there is no longer a net force acting on it.