A cockroach of mass $\mathbf{m}$ lies on the rim of a uniform disk of mass$\mathbf{4.00}\mathbf{}\mathbf{m}$ that can rotate freely about its centre like a merry- go-round. Initially the cockroach and disk rotate together with an angular velocity of$\mathbf{0.260}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}\mathbf{.}$Then the cockroach walks halfway to the centre of the disk.

(a) What then is the angular velocity of the cockroach – disk system?

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

(c) What accounts for the change in the kinetic energy?

1. The mass of the cockroach is$,{\mathrm{m}}_{\mathrm{c}}=\mathrm{m}.$
2. The mass of the disk is$,{\mathrm{m}}_{\mathrm{d}}=4.00\text{m}.$
3. The initial angular velocity of the system of the cockroach and disk is ${\mathrm{\omega }}_{{}_{\mathrm{i}}}=0.260\mathrm{rad}/\mathrm{s}.$,

Using the law of conservation of angular momentum, find the new angular speed of the cockroach-disk system. Then using this, find the new rotational kinetic energy of the system. Find the ratio of initial and new K.E of the system from these calculations.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

1. $\mathbf{K}\mathbf{.}\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}\mathbf{=}\mathbf{L\omega }$
   
2. The law of conservation of angular momentum,

${\mathbf{L}}_{\mathbf{i}}\mathbf{}\mathbf{=}{\mathbf{L}}_{\mathbf{f}}$

Where, L_i, L_f are initial and finalmomentums, I is moment of inertia, L is angular momentum and$\mathbf{\omega }$is angular frequency.

(a)

The initial angular momentum of the system is,

$\begin{array}{l}{\mathrm{L}}_{\mathrm{i}}={\mathrm{I}}_{\mathrm{i}}{\mathrm{\omega }}_{\mathrm{i}}\\ {\mathrm{L}}_{\mathrm{i}}={\mathrm{m}}_{\mathrm{c}}{\mathrm{R}}^2{\mathrm{\omega }}_{\mathrm{i}}+\frac{1}{2}{\mathrm{m}}_{\mathrm{d}}{\mathrm{R}}^2{\mathrm{\omega }}_{\mathrm{i}}\end{array}$

The final angular momentum of the system is,

${\mathrm{L}}_{\mathrm{f}}={\mathrm{m}}_{\mathrm{c}}{\left(\frac{\mathrm{R}}{2}\right)}^2{\mathrm{\omega }}_{\mathrm{f}}+\frac{1}{2}{\mathrm{m}}_{\mathrm{d}}{\mathrm{R}}^2{\mathrm{\omega }}_{\mathrm{f}}$

According to the law of conservation of angular momentum,

$\begin{array}{c}{\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}\\ {\mathrm{m}}_{\mathrm{c}}{\mathrm{R}}^2{\mathrm{\omega }}_{\mathrm{i}}+\frac{1}{2}{\mathrm{m}}_{\mathrm{d}}{\mathrm{R}}^2{\mathrm{\omega }}_{\mathrm{i}}={\mathrm{m}}_{\mathrm{c}}{\left(\frac{\mathrm{R}}{2}\right)}^2{\mathrm{\omega }}_{\mathrm{f}}+\frac{1}{2}{\mathrm{m}}_{\mathrm{d}}{\mathrm{R}}^2{\mathrm{\omega }}_{\mathrm{f}}\\ \left({\mathrm{m}}_{\mathrm{c}}{\mathrm{R}}^2+\frac{1}{2}{\mathrm{m}}_{\mathrm{d}}{\mathrm{R}}^2\right){\mathrm{\omega }}_{\mathrm{i}}=\left(\frac{{\mathrm{m}}_{\mathrm{c}}{\mathrm{R}}^2}{4}+\frac{1}{2}{\mathrm{m}}_{\mathrm{d}}{\mathrm{R}}^2\right){\mathrm{\omega }}_{\mathrm{f}}\end{array}$

$\begin{array}{l}{\mathrm{\omega }}_{\mathrm{f}}=\frac{\left[{\mathrm{m}}_{\mathrm{c}}{\mathrm{R}}^2+\frac{1}{2}{\mathrm{m}}_{\mathrm{d}}{\mathrm{R}}^2\right]}{\left[\frac{{\mathrm{m}}_{\mathrm{c}}{\mathrm{R}}^2}{4}+\frac{1}{2}{\mathrm{m}}_{\mathrm{d}}{\mathrm{R}}^2\right]}{\mathrm{\omega }}_{\mathrm{i}}\\ {\mathrm{\omega }}_{\mathrm{f}}=\frac{\left[{\mathrm{mR}}^2+\frac{1}{2}\left(4\mathrm{m}\right){\mathrm{R}}^2\right]}{\left[\frac{{\mathrm{mR}}^2}{4}+\frac{1}{2}\left(4\mathrm{m}\right){\mathrm{R}}^2\right]}{\mathrm{\omega }}_{\mathrm{i}}\\ {\mathrm{\omega }}_{\mathrm{f}}=\frac{3{\mathrm{mR}}^2}{\frac{9}{4{\mathrm{mR}}^2}}{\mathrm{\omega }}_{\mathrm{i}}\\ {\mathrm{\omega }}_{\mathrm{f}}=1.33\times 0.260\end{array}$

${\mathrm{\omega }}_{\mathrm{f}}=0.347\text{rad/s}$

Therefore, the final angular velocity of the cockroach disk system is $0.347\text{rad}/\text{s}$.

(b)

The ratio of the new kinetic energy to the initial K.E of the system is,

$\begin{array}{l}\frac{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}}=\frac{\frac{1}{2}{\mathrm{L\omega }}_{\mathrm{f}}}{\frac{1}{2}{\mathrm{L\omega }}_{\mathrm{i}}}\\ \frac{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}}=\frac{{\mathrm{\omega }}_{\mathrm{f}}}{{\mathrm{\omega }}_{\mathrm{i}}}\\ \frac{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}}=\frac{0.347}{0.260}\\ \frac{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}}=1.33\end{array}$

Hence,the ratio of the new kinetic energy to the initial kinetic energy of the system is$1.33$ .

(c)

Work done by the cockroach while walking towards the center of the disk accounts for the change in the K.E..

Therefore, using the law of conservation of angular momentum, the initial or final angular speed of the system can be found. By using this angular momentum, the rotational K.E of the system can be found.