# Chapter 8: Angular momentum

## 8.3: Collisions

In addition to describing the rotational motion of a single object while its moment of inertia changes, we can also look at what happens when two rotating objects collide.

#### Example 8.2

Two disks are spinning counter-clockwise. One disk has a mass of 1 kg, a radius of 2 cm, and is spinning counterclockwise at 3 rad/s. The other has a mass of 0.5 kg, a radius of 1 cm, and is rotating the same direction at 6 rad/s. The smaller disk is moved towards the larger disk so that they contact each other face-to-face. Friction between the two disks causes them to rotate together afterwards; what is their angular speed?

\begin{align*} L_i &= L_f \\ L_{1,i} + L_{2,i} &= L_{1,f} + L_{2,f} \\ I_1\omega_1 + I_2\omega_2 &= I_1\omega_f + I_2\omega_f \end{align*}

Note that both disks are rotating at the same angular velocity after the collision; this can be factored out and we have

$I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega_f$

Now, this looks very similar to what we have seen with completely inelastic collisions when dealing with linear momentum! However, instead of adding the masses of the two objects, we add their moments of inertia. Substituting the appropriate expression for the moment of inertia of a disk, we continue

\begin{align*} \frac{1}{2}m_1r_1^2\omega_1 + \frac{1}{2}m_2r_2^2 &= \left(\frac{1}{2}m_1r_1^2 + \frac{1}{2}m_2r_2^2\right)\omega_f \\ \hookrightarrow \omega_f &= \frac{m_1r_1^2\omega_1 + m_2r_2^2\omega_2}{m_1r_1^2 + m_2r_2^2} \\ &= \frac{(1\ \textrm{kg})(2\ \textrm{cm})^2(3\ \textrm{rad/s}) + (0.5\ \textrm{kg})(1\ \textrm{cm})^2(6\ \textrm{rad/s})}{(1\ \textrm{kg})(2\ \textrm{cm})^2 + (0.5\ \textrm{kg})(1\ \textrm{cm})^2} \\ &= 3.33\ \textrm{rad/s} \end{align*}

#### Example 8.3

A Lazy Susan is a rotating platform placed in the center of a table and used for easily passing dishes of food from one person to another. Consider a Lazy Susan with a mass of 2 kg and a radius of 12 cm. You give it a push so that it rotates at 3 rad/s. Then, you place a 0.5 kg dish of food on it halfway from the center to the edge. How fast will the Lazy Susan be rotating after you set the dish down?

A Lazy Susan is essentially a disk, so it’s moment of inertia is $$I_s = \frac{1}{2}m_sr^2$$. The dish of food can be treated as a point mass, traveling in a circle with a radius of $$r_f = \frac{r}{2}$$. Note that even though a dish can be disk-shaped, we do not treat it as a disk in this case. This is because the dish is not rotating around it’s own center; it is instead traveling around a circular path. Angular momentum is conserved.

\begin{align*} L_i &= L_f \\ I_s\omega_i &= (I_s + I_d)\omega_f \\ \frac{1}{2}m_sr^2\omega_i &= \left(\frac{1}{2}m_sr^2 + m_dr_d^2\right)\omega_f \\ \frac{1}{2}m_sr^2\omega_i &= \left(\frac{1}{2}m_sr^2 + m_d\left(\frac{r}{2}\right)^2\right)\omega_f \\ \frac{1}{2}m_sr^2\omega_i &= \left(\frac{1}{2}m_sr^2 + m_d\frac{r^2}{4}\right)\omega_f \\ \frac{1}{2}m_sr^2\omega_i &= \frac{r^2}{2}\left(m_s + \frac{m_d}{2}\right)\omega_f \\ \hookrightarrow \omega_f &= \frac{m_s\omega_i}{m_s + \frac{m_d}{2}} \end{align*}

It is common practice to eliminate fractions-within-fractions. Multiply both the numerator and denominator by two:

\begin{align*} \omega_f &= \frac{2(m_s\omega_i)}{2\left(m_s + \frac{m_d}{2}\right)} \\ &= \frac{2m_s\omega_i}{2m_s + m_d} \\ &= \frac{2(2\ \textrm{kg})(3\ \textrm{rad/s})}{2(2\ \textrm{kg})+(0.5\ \textrm{kg})} \\ &= 2.7\ \textrm{rad/s} \end{align*}