## 11.2 Newton’s laws, applied to rotation

As we have seen throughout this entire course, physical quantities dealing with translational motion tend to have analogous counterparts for rotation. We can think of two of Newton’s laws in terms of rotation:

- Inertia: an object’s rotational velocity will only change if it is acted on by some torque.
- The total torque acting on an object is equal to the rate of change of its momentum: $$\tau_\textit{net} = \frac{\Delta L}{\Delta t} \tag{11.2}$$ If the moment of inertia of the system does not change, the total torque acting is equal to the object’s rotational inertia (moment of inertia) times its angular acceleration: $$\tau_\textit{net} = I\alpha \tag{11.3}$$

Newton’s third law does not directly apply to rotation. Consider that you need to apply a force to cause rotation; whatever object you are applying a force to is applying the same force back on you (Newton’s third law). However, the *torque* you exert on the object is not necessarily the same as the *torque* the object is exerting on you.

For example, say you exert some force \(F\) on a door by pushing a distance \(r\) from the door’s hinges. Your arm makes a 90° angle with the door. By Newton’s third law, the door exerts the same force \(F\) on you. However, the force the door exerts on you makes a 180° angle with your arm: you are keeping your arm straight, and the door exerts the force directly into your hand. So while you exert some torque on the door, the door does not exert any torque on you (because \(\sin(180^\circ) = 0\)).

Newton’s third law is not being violated here—the door is exerting the same force on you that you exert on it—the third law just does not apply directly to torque.

#### Example 11.2

A windmill is at rest. Two seconds later, it has an angular momentum of 500 kg·m^{2}/s. What was the net torque on the windmill?

\[

\begin{align*}

\tau_\textit{net} &= \frac{\Delta L}{\Delta t} \\

&= \frac{L_f – L_i}{\Delta t} \\

&= \frac{(500 – 0)\ \textrm{kg}\cdot\textrm{m}^2\textrm{/s} }{2\ \textrm{s}} \\

&= 250\ \textrm{N}\cdot\textrm{m}

\end{align*}

\]

#### Example 11.3

You have two pendulums that consist of a heavy mass attached to a light rod. On one pendulum, the rod has a length \(\ell\). On the other, the rod has a length of \(\frac{\ell}{2}\). Both pendulums start out in a horizontal position, and have equal masses. At the moment they are released, which pendulum has a larger angular acceleration? By how much?

Since the rod is light, we only need to consider the mass at the end, which may be treated as a point mass. Torque is exerted on each pendulum by the force of gravity pulling on the mass at the end of the rod. We can use Newton’s second law for rotation to determine the angular acceleration of each rod.

##### The longer rod

\[

\begin{align*}

\tau_\textit{net} &= I\alpha \\

\tau^G &= m\ell^2\alpha \\

mg\ell &= m\ell^2\alpha \\

\hookrightarrow \alpha &= \frac{g}{\ell}

\end{align*}

\]

##### The shorter rod

\[

\begin{align*}

\tau_\textit{net} &= I\alpha \\

\tau^G &= m\left(\frac{\ell}{2}\right)^2\alpha \\

mg\frac{\ell}{2} &= m\frac{\ell^2}{4}\alpha \\

\hookrightarrow \alpha &= \frac{2g}{\ell}

\end{align*}

\]

The shorter rod has twice the angular acceleration of the longer rod.