5.5 Dynamics of Rigid Bodies

As stated in chapter 4, rigid bodies are distinguished by the fact that all mass elements have the same angular velocity $\omega$ . Each mass element is therefore moving in circles with speed $v_{i}=R_{i}\omega$ , where $R_{i}$ is the distance from the axis of rotation. The smart choice for a coordinate system would be to align the $z$ -axis with the axis of rotation and use cylindrical coordinates $(R,\phi,z)$ , but let’s look at the angular momentum in cartesian coordinates. The velocity of a mass element (assuming that the rigid body has no linear motion) is $\boldsymbol{v}_{i}=v_{x,i}\hat{\boldsymbol{\imath}}+v_{y,i}\hat{\boldsymbol{% \jmath}}=-\omega y_{i}\hat{\boldsymbol{\imath}}+\omega x_{i}\hat{\boldsymbol{% \jmath}}$ . Then the total angular momentum is

$\displaystyle\boldsymbol{L}=\sum_{i}\boldsymbol{L}_{i}$	$\displaystyle=\sum_{i}\boldsymbol{r}_{i}\times m_{i}\boldsymbol{v_{i}}$	(5.58)
	$\displaystyle=\sum_{i}m_{i}\begin{vmatrix}\hat{\boldsymbol{\imath}}&\hat{% \boldsymbol{\jmath}}&\hat{\boldsymbol{k}}\\ x_{i}&y_{i}&z_{i}\\ -\omega y_{i}&\omega x_{i}&0\end{vmatrix}$	(5.59)
	$\displaystyle=\sum_{i}m_{i}x_{i}z_{i}\omega\hat{\boldsymbol{\imath}}+\sum_{i}m% _{i}y_{i}z_{i}\omega\hat{\boldsymbol{\jmath}}+\sum_{i}m_{i}(x_{i}^{2}+y_{i}^{2% })\omega\hat{\boldsymbol{k}}.$	(5.60)

The last term can be simplified to $\sum_{i}m_{i}R_{i}^{2}\omega\hat{\boldsymbol{k}}=I_{z}\omega\hat{\boldsymbol{k}}$ where $I_{z}$ is the moment of inertia about the $z$ axis. The first two terms are the products of inertia multiplied by $\omega$ . In the case where the products of inertia are zero, we have

\boldsymbol{L}=I_{z}\omega\hat{\boldsymbol{k}},

(5.61)

and the object is rotating around a principle axis.

Assuming this angular momentum is measured from the centre of mass of the rigid body, we can use $\boldsymbol{\tau}=\dot{\boldsymbol{L}}$ to get

\boldsymbol{\tau}=\dot{\boldsymbol{L}}=I\dot{\omega}\hat{\boldsymbol{k}}=I% \alpha\hat{\boldsymbol{k}}.

(5.62)

This is an angular analogue of Newton II where moment of inertia takes the place of mass. This derivation has also implicitly assumed that the torque is being applied around the axis of rotation and hence the direction of $\boldsymbol{L}$ does not change. A torque in another direction would change the direction of angular momentum, which leads to precession In the case of constant torque around the axis of rotation, the equation above implies that $\alpha$ is constant, which means we can use the angular SUVAT equations from chapter 4 to describe the motion.

Example 5.5.

Consider two connected masses on a massless pulley with $m_{1}>m_{2}$ . Suppose the system starts from rest and assume the string is massless, inextensible, and lies vertically. Find an expression for the magnitude of acceleration of the masses. To do this, we have to analyse the forces acting on the blocks and also the torques acting on the pulley.