A-level Topics

16 July

Non-Uniform Vertical Circular Motion

Seng Kwang Tan 07 Circular Motion, 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.

15 July

Perpetual Motion Engines

Seng Kwang Tan 04 Energy and Fields, IP3 06 Work, Energy and Power 0 Comments

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.

12 July

Pressure and Projectile – Penguin’s Poop

Seng Kwang Tan 05 Projectile Motion, 12 Temperature and Ideal Gases 0 Comments

An interesting paper on the range of penguin’s poop. Their motivation? “Such information is useful for keepers to avoid the direct hitting of faeceses.”

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

12 July

Centripetal force on and off

Seng Kwang Tan 07 Circular Motion, Demonstrations 0 Comments

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.

12 July

Escape Velocity

Seng Kwang Tan 08 Gravitational Fields, 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, $v = \sqrt{\dfrac{2GM}{R_P}}$.

24 May

Angular velocity

Seng Kwang Tan 07 Circular Motion, GeoGebra 0 Comments

This GeoGebra app shows how angular velocity ω is the rate of change of angular displacement (i.e. $\omega=\dfrac{\theta}{t}$) and is dependent on the speed and radius of the object in circular motion (i.e. $v=r\omega$).

Students can explore the relationships by doing the following:

Keeping r constant and varying ω.

Keeping ω constant and varying r.

Keeping v constant by varying r and ω.