A uniform disk of mass $\mathbf{10}\mathbf{m}$ and radius $\mathbf{3}\mathbf{.0}\mathbf{r}$can rotate freely about its fixed centre like a merry-go-round. A smaller uniform disk of mass $\mathbf{m}$and radius $\mathbf{r}$lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of $\mathbf{20}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}\mathbf{.}$Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding).

(a) What then is their angular velocity about the centre of the larger disk?

(b) What is the ratio of$\mathbf{K}\mathbf{/}{\mathbf{K}}_{\mathbf{0}}$ the new kinetic energy of the two-disk system to the system’s initial kinetic energy?

1. Mass of the bigger disk is$10\text{m}$
2. Radius of the larger disk is$3\mathrm{r}$
3. Mass of the smaller disk is$\mathrm{m}$
4. Radius of the smaller disk is$\mathrm{r}$
5. ${\mathrm{I}}_{\mathrm{disk}}=30{\text{0kg.m}}^{\text{2}}$

Calculate the initial and final moment of inertia by using the formula for the rotational inertia of the disk. Use law of conservation of momentum to find the final angular speed. Then, calculate the ratio of final to initial kinetic energies.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula are as follow:

$\begin{array}{l}\mathbf{I}\mathbf{=}\frac{{\mathbf{MR}}^{\mathbf{2}}}{\mathbf{2}}\\ \mathbf{Initial}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}\mathbf{=}\mathbf{Final}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}\end{array}$

Where,Iis moment of inertia, M is mass and R is radius.

(a)

Using the formula for the rotational inertia of the disk, find the initial rotational inertia of the system,

$\begin{array}{l}{\mathrm{I}}_{\mathrm{i}}={\mathrm{I}}_{\mathrm{bigdisk}}+{\mathrm{I}}_{\mathrm{smalldisk}}\\ \mathrm{I}=\frac{{\mathrm{MR}}^2}{2}+\frac{{\mathrm{mr}}^2}{2}=\frac{10\mathrm{m}\times 9{\mathrm{r}}^2}{2}+\frac{{\mathrm{mr}}^2}{2}=\frac{91{\mathrm{mr}}^2}{2}\end{array}$

The new rotational inertia of system is,

${\text{I}}_{\text{f}}=\frac{{\mathrm{MR}}^2}{2}+\frac{{\mathrm{mr}}^2}{2}+\mathrm{m}{(\mathrm{R}-\mathrm{r})}^2=\frac{99}{2}{\mathrm{mr}}^2$

By law of conservation of angular momentum,

$\begin{array}{c}{\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}\\ {\mathrm{I}}_{\mathrm{i}}{\mathrm{\omega }}_{\mathrm{i}}={\mathrm{I}}_{\mathrm{f}}{\mathrm{\omega }}_{\mathrm{f}}\\ \frac{91{\mathrm{mr}}^2}{2}\times 20=\frac{99}{2}{\mathrm{mr}}^2\times {\mathrm{\omega }}_{\mathrm{f}}\\ {\mathrm{\omega }}_{\mathrm{f}}=18\text{rad/s}\end{array}$

Hence, angular velocity about the center of the larger disk is, $\mathrm{\omega }=18\frac{\text{rad}}{\text{s}}.$

(b)

Now,

$\frac{{\mathrm{I}}_{\mathrm{f}}}{{\mathrm{I}}_{\mathrm{i}}}=\frac{99}{91}$

And,

$\frac{{\mathrm{\omega }}_{\mathrm{f}}}{{\mathrm{\omega }}_{\mathrm{i}}}=\frac{91}{99}$

Thus,

$\frac{{\mathrm{KE}}_{\mathrm{f}}}{{\mathrm{KE}}_{\mathrm{i}}}=\frac{{\mathrm{I}}_{\mathrm{f}}{\mathrm{\omega }}_{\mathrm{f}}^2}{{\mathrm{I}}_{\mathrm{i}}{\mathrm{\omega }}_{\mathrm{i}}^2}=\left(\frac{99}{91}\right)\times {\left(\frac{91}{99}\right)}^2=0.92$

Hence, the ratio of new kinetic energy to initial kinetic energy is $\mathrm{0.92.}$

Therefore, using the formula for rotational inertia and conservation of angular momentum, the ratio of kinetic energies can be found.