# Classical Rotations

By

Tee Kai Ze

,

Guan Xin

,

Jensie Low

,

Seow Ryan

●

Published 2022-08-27

Consider a particle of mass m rotating around an axis. The inertia towards linear motion is the particle's mass m. For rotational motion, mass is insufficient to quantify the amount of inertia for rotational motion. The distance away from the axis of rotation plays a role too. One way to demonstrate this is to rotate a rod with a weight attached. The further the weight is away from the axis of rotation, the harder it is to rotate. The reluctance to rotate is termed as moment of inertia.

The particle's motion is characterised by:

• r : distance between the particle and axis
• I : moment of inertia
• v : velocity
• \omega : angular velocity
• p : linear momentum
• L : angular momentum
• T : kinetic energy

The above parameters are closely related by the following equations:

\begin{equation}
I=mr^{2}
\end{equation}

\begin{equation}
v=r\omega
\end{equation}

\begin{equation}
p=mv
\end{equation}

\begin{equation}
L=I\omega
\end{equation}

\begin{equation}
T=\frac{1}{2}mv^{2}=\frac{1}{2}mr^{2}\omega^{2}=\frac{1}{2}I\omega^{2}
\end{equation}

Let us now consider the case of two particles with masses m_{1} and m_{2} joined by a rigid (massless) rod of length R. The 2-body system has a center of mass. Let us set the center of mass to be at the origin, mass m_{1} to be located at distance r_{1} away from the center of mass, and mass m_{2} at r_{2} away. Since the 2 masses are at opposite sides of the center of mass,

\begin{equation}
r_{1}+r_{2}=R
\end{equation}

By principle of moments¹

\begin{equation}
m_{1}r_{1}=m_{2}r_{2}
\end{equation}

Now set these masses on the xy plane, have the z axis to cut through the center of mass and allow the masses to rotate around the z axis. This is the rigid rotor which we will use to model a rotating diatomic molecule.

\begin{equation}
r_{1}=\frac{m_{2}}{m_{1}+m_{2}}R
\end{equation}

\begin{equation}
r_{2}=\frac{m_{1}}{m_{1}+m_{2}}R
\end{equation}

The moment of inertia of the system is given by

\begin{align}
I & =m_{1}r_{1}^{2}+m_{2}r_{2}^{2}\nonumber \\
 & =m_{1}\left(\frac{m_{2}}{m_{1}+m_{2}}R\right)^{2}+m_{2}\left(\frac{m_{1}}{m_{1}+m_{2}}R\right)^{2}\nonumber \\
 & =\frac{m_{1}m_{2}R^{2}}{(m_{1}+m_{2})^{2}}(m_{2}+m_{1})\nonumber \\
 & =\frac{m_{1}m_{2}}{m_{1}+m_{2}}R^{2}\nonumber \\
 & =\mu R^{2}
\end{align}

\mu is the called "effective mass". By expressing the moment of inertia in Eq.(10) allows us to treat the 2-body system like a single particle.

Since the rotor is "rigid", the two masses share the same angular velocity \omega. The angular momentum of the system is

\begin{equation}
L=I\omega
\end{equation}

The kinetic energy is

\begin{equation}
T=\frac{1}{2}m_{1}r_{1}^{2}\omega^{2}+\frac{1}{2}m_{2}r_{2}^{2}\omega^{2}=\frac{1}{2}I\omega^{2}=\frac{L^{2}}{2I}
\end{equation}

What is the potential energy of the system?

\begin{equation}
V=0
\end{equation}

No potential energy, the rigid rotor here rotates "freely".

In chapter 2, we learned about the case of one particle attached to a massless spring. We shall now extend the idea to two particles connected by a spring.

The equations of motion of the two masses are

\begin{equation}
m_{1}\ddot{x}_{1}=k(x_{2}-x_{1}-x_{0})
\end{equation}

\begin{equation}
m_{2}\ddot{x}_{2}=-k(x_{2}-x_{1}-x_{0})
\end{equation}

Similar to how the 2-masses rigid rotor can be re-modeled as a rotation of a particle with an effective mass, the 2-masses-connected-by-a-spring system can be re-modeled as a single particle undergoing simple harmonic motion:

\begin{equation}
\ddot{x}=-\omega^{2}x
\end{equation}

where \omega is the angular frequency

\begin{equation}
\omega=\sqrt{\frac{k}{\mu}}
\end{equation}

k is the spring constant, and \mu is the effective mass given by

\begin{equation}
{\displaystyle \mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}}
\end{equation}

The kinetic energy of the system is

\begin{equation}
T=\frac{1}{2}\mu\dot{x}^{2}=\frac{p^{2}}{2\mu}
\end{equation}

The potential energy of the system is

\begin{equation}
V=\frac{1}{2}kx^{2}=\frac{1}{2}\mu\omega^{2}x^{2}
\end{equation}

For the interested reader, the detailed derivation of the above equations is given in Appendix.