Two disks are mounted (like a merry-go-round) on low-friction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with rotational inertia $\mathbf{3}\mathbf{.30}\mathbf{}\mathbf{kg}\mathbf{.}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$about its central axis, is set spinning counter clockwise at $\mathbf{450}\mathbf{}\mathbf{rev}\mathbf{/}\mathbf{min}$. The second disk, with rotational inertia $\mathbf{6}\mathbf{.60}\mathbf{}\mathbf{kg}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$about its central axis, is set spinning counter clockwise at $\mathbf{900}\mathbf{}\mathbf{rev}\mathbf{/}\mathbf{min}\mathbf{.}\mathbf{}$They then couple together.

(a) What is their angular speed after coupling? If instead the second disk is set spinning clockwise at 900 rev/min,

(b) what are their angular speed?

(c) What are their direction of rotation after they couple together?

1. The rotational inertia of first disk is,${\text{I}}_1=3.3\text{kg}.{\text{m}}^2$
2. The rotational inertia of second disk is$,{\text{I}}_2=6.6\text{kg}.{\text{m}}^2$
3. The angular speed of the first disk,${\text{ω}}_1=450\mathrm{rev}/\min$
4. The angular speed of the second disk is,${\text{ω}}_2=900\mathrm{rev}/\min$

Using the conservation law of the angular momentum we can find the angular speed of the system after coupling. Then by using the sign convention, we can find the angular speed of the system when the second disk is spinning clockwise.

Formula:

The law of conservation of angular momentum,${\mathbf{L}}_{\mathbf{i}}\mathbf{}\mathbf{=}{\mathbf{L}}_{\mathbf{f}}$

(a)

The law of conservation of angular momentum gives,${\text{L}}_{\text{i}}={\text{L}}_{\text{f}}$.

Angular momentum of the system before coupling = Angular momentum of the system after coupling

$\begin{array}{c}\Rightarrow {\text{I}}_1{\text{ω}}_1+{\text{I}}_2{\text{ω}}_2=({\text{I}}_1+{\text{I}}_2)\text{ω}\\ \Rightarrow \text{ω}=\frac{{\text{I}}_1{\text{ω}}_1+{\text{I}}_2{\text{ω}}_2}{{\text{I}}_1+{\text{I}}_2}\\ \Rightarrow \text{ω}=\frac{\left(3.3\right)\left(450\right)+\left(6.6\right)\left(900\right)}{3.3+6.6}\\ \Rightarrow \mathbf{\omega }=750\mathbf{rev}/\mathbf{min}\end{array}$

(b)

If the second disk is spinning clockwise, its angular speed would be $-900\mathrm{rev}/\min$.

So, the angular speed of the system is

$\begin{array}{l}\text{ω}=\frac{\left(3.3\right)\left(450\right)+\left(6.6\right)(-900)}{3.3+6.6}\\ \mathbf{\omega }=-450\mathbf{rev}/\mathbf{min}\end{array}$

(c)

The minus sign indicates that$\overrightarrow{\omega }$ is clockwise, that is, in the direction of the second disk’s initial angular velocity.