Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident.
- What is the angular speed of the two-disc system?
- Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take $\omega_1\neq\omega_2$
Exercise
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- Moment of inertia of dise $\text{I}=\text{I}_{1}$
Angular speed of disc $\text{II}=\text{I}_2$
Angular momentum of disc $\text{II}=\omega_1$
Angular momentum of disc $\text{I},\text{L}_1=\text{I}_1\omega_1$
Angular momentum of disc $\text{II},\text{L}_2=\text{I}_2\omega_2$
Total initial angular momentum, $\text{L}_{\text{i}}=\text{I}_1\omega_1+\text{I}_2+\omega_2$
When the two discs are joined together, their moments of inertia get added up.
Moment of inertia of the system of two discs $\text{I}=\text{I}_1+\text{I}_2$
Let $\omega$ be the angular speed of the system.
Total final angular momentum, $\text{L}_{\text{f}}=(\text{I}_1+\text{I}_2)\omega$
Using the law of conservation of angular momentum, we have,
$\text{L}_{\text{i}}=\text{L}_{\text{f}}$
$\text{I}_1\omega_1+\text{I}_2\omega_2=(\text{I}_1+\text{I}_2)\omega$
$\therefore\ \omega=\frac{\text{I}_1\omega_1+\text{I}_2\omega_2}{\text{I}_1+\text{I}_2}$
- Kinetic energy of disc I, $\text{E}_1=\frac{1}{2}\text{I}_1\omega^2_1$
Total initial kinetic energy, $\text{E}_{\text{i}}=\frac{1}{2}\big(\text{I}_1\omega^2_1+\text{I}_2\omega^2_2\big)$
When the discs are joined, their moments of inertia get added up.
Moment of inertia of the system, $\text{I}=\text{I}_1+\text{I}_2$
Angular speed of the system $=\omega$
Final kinetic energy $E_f$
$=\frac{1}{2}(\text{I}_1+\text{I}_2)\omega^2$
$=\frac{1}{2}(\text{I}_1+\text{I}_2)\Big(\frac{\text{I}_1\omega_1+\text{I}_2\omega_2}{\text{I}_1+\text{I}_2}\Big)^2=\frac{1}{2}\frac{(\text{I}_1\omega_1+\text{I}_2\omega_2)^2}{\text{I}_1+\text{I}_2}$
$\therefore\ \text{E}_{\text{i}}-\text{E}_{\text{f}}$
$=\frac{1}{2}\big(\text{I}_1\omega^2_1+\text{I}_2\omega^2_2\big)-\frac{(\text{I}_1\omega_1+\text{I}_2\omega_2)^2}{2(\text{I}_1+\text{I}_2)}$
$=\frac{1}{2}\text{I}_1\omega_1^2+\frac{1}{2}\text{I}_2\omega_2^2-\frac{1}{2}\frac{\text{I}^2_1\omega^2_1}{(\text{I}_1+\text{I}_2)}-\frac{1}{2}\frac{\text{I}^2_2\omega^2_2}{(\text{I}_1+\text{I}_2)}-\frac{1}{2}\frac{2\text{I}_1\text{I}_2\omega_1\omega_2}{(\text{I}_1+\text{I}_2)}$
$=\frac{1}{(\text{I}_1+\text{I}_2)}\Big[\frac{1}{2}\text{I}^2_1\omega^2_1+\frac{1}{2}\text{I}_1\text{I}_2\omega^2_1+\frac{1}{2}\text{I}_1\text{I}_2\omega^2_2+\frac{1}{2}\text{I}^2_2\omega^2-\frac{1}{2}\text{I}^2_1\omega^2_1-\frac{1}{2}\text{I}_2^2\omega^2_2-\text{I}_1\text{I}_2\omega_1\omega_2\Big]$
$=\frac{\text{I}_1\text{I}_2}{2(\text{I}_1+\text{I}_2)}\big[\omega^2_1+\omega^2_2-2\omega_1\omega_2\big]$
$=\frac{\text{I}_1\text{I}_2(\omega_1-\omega_2)^2}{2(\text{I}_1+\text{I}_2)}$
All the qauantities on RHS are positive.
$\therefore\ \text{E}_{\text{i}}-\text{E}_{\text{f}}>0$
$\text{E}_{\text{i}}>\text{E}_{\text{f}}$
The loss of KE can be attributed to the frictional force that comes into play when the two discs come in contact with each other.
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