# Proof 24: Newton’s 2nd law in rotation

Get some friends to help you push the stuck carousel with your daughter sitting in it. Each applies a force on her. Friction and other force may counteract. Many forces could be involved – let’s denote them $F_1$, $F_2$, $F_3$ and so on – and we use Newton’s 2nd law (see the Newton’s 2nd law skill) on all these forces that are acting at this position to find the acceleration of your daughter.

\begin{align} \sum F&=ma\quad\Leftrightarrow\quad \\ \underbrace{F_1+F_2+F_3+\cdots}_\text{all forces}&=ma \quad\Leftrightarrow\quad \\ \frac{\tau_1}r+\frac{\tau_2}r+\frac{\tau_3}r+\cdots&=ma \quad\Leftrightarrow\quad \qquad\leftarrow \text{Torque }\tau=Fr\Leftrightarrow F=\tau/r\\ \underbrace{\tau_1+\tau_2+\tau_3+\cdots}_\text{all torques}&=rma\\ \sum \tau &=rma \quad\Leftrightarrow\quad \\ \sum \tau &=rm\underbrace{r\alpha}_a \quad\Leftrightarrow\quad \qquad \leftarrow \text{Geometric bond } a=r\alpha\\ \sum \tau &=mr^2\alpha \end{align}

The mass $m$ is just the daughter. But actually not only her is being set in motion, so not only her mass must be involved. The entire carousel structure is being set in motion and the mass of every “particle” of it is involved. The farther away from the centre, the longer it must be moved per second, so the “tougher”.

Each “particle” can be accounted for my splitting the carousel arm (and other parts of the structure) into infinitely many “particles” with each “particle” being pushes/pulled in by its neighbour particles. Due to Newton’s 3rd law, the force from one neighbour “particle” is also felt by that neighbour “particle” itself, but oppositely. So, when summing up all forces for the entire situation, all these “particle”-“particle” force-pairs cancel out. We end with only all the “toughnesses”, all the inertias” $mr^2$ adding up. So, let’s add a sum symbol to the “toughness” group of parameters:

\begin{align} \sum \tau &=\underbrace{\sum (mr^2)}_I\alpha\quad\Leftrightarrow\\ \sum \tau &=I\alpha \qquad\blacksquare\\ \end{align}

This “toughness” chunk is called the moment-of-inertia:

$$I=\sum mr^2$$

Note, this result is also often written as an integral in case of an infinite amount of “particles” being infinitely small (a continuous body):

$$I=\int r^2\;\mathrm dm$$

## Proof 14: Motion equations

In motion along a path, $a$ is acceleration, $v_0$ and $v$ are initial and final (current) speed, and $t_0$ and $t$ are initial and final…

## Proof 31: Elastic-collision speed “conservation”

We call a collision elastic if the total kinetic energy is conserved (same before as after the collision) and there are no other energies involved.…

## Proof 22: Geometric bonds

Angle $\theta$ measured in radians is ‘the number of radius-lengths along the periphery’. Multiply that number with the radius $r$ to get the total length…

## Proof 35: Kinetic energy

Linear version of kinetic energy A spaceship in outer space starts its rocket engines to propel itself forwards with a rocket force. That force does…

## Proof 32: Momentum conservation law

The moment conservation law and Newton’s 3rd law as well as the idea of impulse seem equivalent (they all describe impacts/collisions). And in fact the…