# 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):

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

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