# The principle of riding a swing

They know how, but they can’t tell the principle.

“How can you go so high?”
In the world, there are some principles are difficult to describe in words even if everyone knows it.
I think one of them is how to ride a swing.
Can you explain how to ride a swing with physics equation?

Standing or sitting at the midpoint of a moving swing changes the length of the pendulum.
In other words, a swing can be thought of as a pendulum that can change its length.
Consider the pendulum that moves as shown in the picture below. When you going down from A to O

The reduced potential energy equals the increased kinetic energy.
$mg r_1 (1 – cos \theta_1 ) = \frac{1}{2}mv^2 \\ v^2 = 2gr_1 (1 – cos \theta_1 )$

When you going from O to B.

Reduced kinetic energy equals increased potential energy.
$\frac{1}{2}mv^2 = mg r_2 (1 – cos \theta_2 ) \\ v^2 = 2gr_2 (1 – cos \theta_2 )$

Change in angle while moving from A to B

The speed at ‘O’ does not changed,
$2g r_1 (1 – cos \theta_1 ) = 2g r_2 (1 – cos \theta_2 ) \\ r_1 (1 – cos \theta_1 ) = r_2 (1 – cos \theta_2 )$ $\therefore \theta_2 = cos^{-1} ({\frac{r_2 – r_1(1 – cos \theta_1)}{r_2}})$

How to ride a swing well?

When going down, we need to increase the length of ‘r’. (= sit down.)
Conversely, when you go up, you just have to stand up.
For example, let’s calculate the increase in angle by substituting the following condition.

• $$r_1 = 3m$$
• $$r_2 = 2m$$
• $$\theta_1 = 45˚$$ (Angle of starting point)

\begin{align} \theta_2 &= cos^{-1} ({\frac{r_2 – r_1(1 – cos \theta_1)}{r_2}}) \\ &= cos^{-1} ({\frac{2 – 3(1 – cos 45˚)}{2}}) \\ &\approx 55.9˚ \end{align}

In this case, the angle is increased by about 1.24 times.