Acceleration and free fall

In a nutshell

When an object is dropped from a point on Earth, it accelerates towards the centre of the Earth. When an object is accelerating only due to gravity, it is in free fall. The acceleration due to gravity is $$g$$, which can be determined experimentally using an electromagnet and a trapdoor, or a camera. 

Equations

Constants

Variables

Acceleration due to gravity

When an object is released on Earth, it will accelerate vertically towards the centre of the Earth. When an object is accelerating only due to gravity, it is said to be in free fall, as gravity is constant, the equations of motion can be used, and the substitution $$a = g$$ made. The acceleration due to gravity in free fall is  $$g = 9.81 \ ms^{-2}$$. 

Electromagnet and trap door

The acceleration due to gravity can be calculated using multiple different methods. The most common method uses an electromagnet and trapdoor. 

1. Electromagnet and steel ball   2. Circuit connecting to switch  and timer 3. Height of fall 4. Trapdoor 5. Ruler

An electromagnet holds a steel ball which is suspended over a trapdoor. The distance between the ball and trapdoor is known. As soon as the electromagnet is deactivated, and the ball begins to fall, a timer begins, which stops as soon as the trapdoor is hit, as this breaks the circuit. The ball begins at rest, so its initial velocity, $$u$$ is $$0 \ ms^{-1}$$, and the distance is known and time measured, the acceleration due to gravity can be calculated using: 

$$s = ut + \dfrac{1}{2}at^2$$

When $$u = 0 \ ms^{-1}$$ and $$a = g$$:

$$s = \dfrac{1}{2} gt^2$$

​Rearranging for $$g$$​:

​$$g = \dfrac{2s}{t^2}$$​​

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Using a camera

The acceleration due to gravity can be calculated by filming the fall of a ball against a ruler. If the number of frames per second of the camera is known, then the time taken to travel a known distance can be calculated, and used to find $$g$$.

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Example: 

A ball drops $$47 \ cm$$​ from an electromagnet to a trapdoor in $$0.31 \ s$$. Use this information to determine a value for g.

State variables:

​$$s = 47 \ cm = 47 \times 10^{-2} \ m, \ \ t = 0.31 \ s$$

State equation and rearrange:

​$$s = \dfrac{1}{2}gt^2 \newline \\[0.1in] g = \dfrac{2s}{t^2}$$

Substitute in and solve:

​$$g = \dfrac{2 \times 47 \times 10^{-2}}{(0.31)^2}\newline \\[0.1in] g = 9.78 \ ms^{-2}$$

The acceleration due to gravity here is calculated as $$\underline{g = 9.78 \ ms^{-2}}$$.​

This value of $$g$$ is not the same as the theoretical value due to uncertainties, which occur in every measurement. The measurements are not very precise, and are only taken to a minimum of 2sf, this means the value of $$g$$ is only correct to 2sf, where it would equal the theoretical value of $$9.8 \ ms^{-2}$$.​

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