 Objectives:

• To understand and be able to calculate the acceleration, $g$ of free fall
• To identify use the techniques and procedures used to determine the acceleration of free fall using a trapdoor and electromagnet arrangement or light gates and timer

If you drop a ball or stone, it falls to the ground. The ball’s velocity will increase as it falls, which means that the ball has an acceleration. Usually they fall too quickly for our eye to notice the acceleration, to us it seems like they just go very fast. The force which causes the ball to accelerate is the pull of the Earth’s gravity. Another name for this force is the weight of the ball.

The centre of gravity The force is shown as an arrow pointing vertically downwards and it comes from what it’s called the centre of gravity. The centre of gravity of an object is defined as the point where its entire weight appear to act. It’s different from the centre of mass, which is the point where its entire mass appear to act. These points may be at the same place, but not always.

A large rock has a greater weight than a small rock, but if you push them off a cliff, they will have the same acceleration, which means that they will fall at the same rate. It was Galileo who dropped a large cannon ball and a small cannon ball and showed that they landed simultaneously. We wouldn’t normally think that a feather can fall at the same speed as a cannon ball, but we are misled from the presence of air resistance. The force of air resistance has a large effect on the falling feather, more than on the ball. On the moon, the feather and cannon ball would fall side by side because of the absence of air resistance. Check out the following video presented by Bryan Cox which proves the above paragraph:

If we measure the acceleration of freely falling objects on the surface of the Earth, we find a value to be $9.81 \ ms^{-1}$. The value of $g$ however does depend on where you are on the Earth’s surface as it varies from place to place – this change is negligible and differs depending on distance from the centre of Earth.

Using the equation $F = ma$, where a is the acceleration due to gravity $a = g$, the weight of an object (the force acting downwards) can be given as: $F = ma \rightarrow F = mg \rightarrow W = mg$

where; $W$ is the weight $m$ is the mass of an object $g$ is the acceleration due to gravity.

Determining the acceleration due to gravity

One way to measure the acceleration of free fall $g$ would be to try bungee-jumping. You would need to carry a stopwatch and measure the time between jumping from the platform and the moment when the elastic rope begins to slow your fall. If you knew the length of the unstretched rope, you could calculate $g$. For example, lets say the unstretched rope has a length of $50 \ m$ and you begin a stopwacth and jump from the platform simultaneously. If you notice the stopwatches time as $3.1 \ s$ before the rope begins to stretch we can use the equations of motion to calculate the acceleration: $s = 50 \ m$ $u = 0 \ ms^{-1}$ $v = \text{unknown}$ $a = \text{?}$ $t = 3.1 \ s$

The equation $s = ut + \frac{1}{2}at^{2}$ is then used: $50 = 0 \ times 3.1 + \frac{1}{2} \times a \times 3.1^{2}$ $50 = \frac{1}{2} \times a \times 3.1^{2}$ $50 = 4.805 \times a$ $a = 10.4 \ ms^{-2}$ Easier methods are usually used, but the principles are the same. Usually a steel ball-bearing is used and it is attached to an electromagnet. When the current to the magnet is switched off, the ball begins to fall and an electronic timer starts. The ball falls through a trapdoor and this breaks a circuit to stop the timer. So now we know the time taken for the ball to fall from rest through the distance between the bottom of the ball and the trapdoor. Given $s = ut + \frac{1}{2}at^{2}$, we know that the initial velocity $u = 0 ms^{-1}$, so we can now write; $s = \frac{1}{2}at^2$

Since $s = h$ as it is a falling through a height, $h$ and $a = g$ the acceleration due to gravity, so finally we can write; $h = \frac{1}{2}gt^{2}$

If we record the values of $t$ varying the height from which the ball drops, a graph can then be plotted with $h$ on the y-axis and $t^{2}$ on the x-axis, which will lead us to calculate $g$ from the gradient. The reason for plotting $h$ on the y-axis and $t^{2}$ on the x-axis is because the equation $h = \frac{1}{2}gt^{2}$ is then in the form $y = mx + c$ where; $y = h$ $x = t^{2}$ $c = 0$
Therefore the gradient is equivalent to the half times the acceleration, $m = \frac{1}{2}g$

Since the gradient is equal to the acceleration, the acceleration due to gravity can be determine and using the graph above; $m = \frac{\Delta y}{\Delta x} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{0.78-0.10}{0.16-0.02} = 4.85$ $m = \frac{1}{2}g$ $4.85 = \frac{1}{2}g$ $g = 9.71 ms^{-2}$

As you can see, this setup (and by just taking approximate values from the graph) gives a much more accurate value that compares to the true value to $g = 9.81 ms^{-2}$.

Sources of uncertainty

The electromagnet may retain some magnetism when it is switched off, and this may tend to slow the ball’s fall – you will learn more about this in the electromagnetism topic in A2. Consequently, the time $t$ recorded by the timer may be longer than if the ball were to fall completely freely. From the equation above, if $t$ is too great, then $g$ will be too small. This is an example of a systematic error, all the results are systematically distorted so that they are too great (or too small) as a consequence of the experimental equipment. Measuring the height is awkward as well. You can probably only find the value of $h$ to within one millimetre at best. So there is a random error in the value of the height and this will result in a slight scatter of the points on the graph, and a degree of uncertainty in the final value of $g$. These are only some of the sources but there may be more you may think about – have a thought and leave a comment below if you can think of them.