16 July

Braking of a Magnetic Pendulum with Copper Plate

Seng Kwang Tan 18 Electromagnetic Induction, Demonstrations 0 Comments

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.

Eddy currents repels the magnet as it approaches
Eddy currents attracts the magnet as it leaves

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.

16 July

Non-Uniform Vertical Circular Motion

Seng Kwang Tan 06 Motion in a Circle, Demonstrations, GeoGebra 0 Comments

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.

Securing the elastic band to the ball with a shoelace

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.

12 July

Escape Velocity

Seng Kwang Tan 07 Gravitation, GeoGebra 0 Comments

Using the GeoGebra app above, I intend to demonstrate the relationship between total energy, kinetic energy and gravitational potential energy in a rocket trying to escape a planet’s gravitational field.

By changing the total energy of the rocket, you will increase the initial kinetic energy, thus allowing it to fly further from the surface of the planet. The furthest point to which the rocket can fly can be observed by moving the slider for “distance”. You will notice that the furthest point is where kinetic energy would have depleted.

Gravitational potential energy of an object is taken as zero at an infinite distance away from the source of the gravitational field. This means gravitational potential energy anywhere else takes on a negative value of $\dfrac{-GMm}{r}$. Therefore, the total energy of the object may be negative, even after taking into account its positive kinetic energy as total energy = kinetic energy + gravitational potential energy.

The minimum total energy needed for the rocket to leave the planet’s gravitational field is zero, as that will mean that the minimum initial kinetic energy will be equal to the increase in gravitational potential energy needed, according to the equation $\Delta U = 0 – (-\dfrac{GMm}{R_P})$, where $R_P$ is the radius of the planet.

Since $\dfrac{1}{2}mv^2 = \dfrac{GMm}{R_P}$, escape velocity, $v = \sqrt{\dfrac{2GM}{R_P}}$.