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 speed $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident.
- Does the law of conservation of angular momentum apply to the situation? why?
- Find the angular speed of the two-disc system.
- Calculate the loss in kinetic energy of the system in the process.
- Account for this loss.
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Key concept: Law of Conservation of Angular Momentum. For a syatem of particles, $\frac{\text{d}\vec{\text{L}}}{\text{dt}}=\vec{\tau}_\text{ex},\text{ i.e, }\frac{\text{dL}_\text{x}}{\text{dt}}=\tau_{\text{ex(x)}}$ $\frac{\text{dL}_\text{y}}{\text{dt}}=\tau_{\text{ex(y)}}=\frac{\text{dL}_\text{z}}{\text{dt}}=\tau_{\text{ex(z)}}$ $\text{If }\vec{\tau}_\text{ex}=0,\vec{\text{L}}=\text{Constant}$ $\text{If}\ \tau_{\text{ex(x)}}=0,\text{ L}_\text{x}=\text{Constant}, \text{ if}\ \tau_{\text{ex(y)}}=0,\text{ L}_\text{y}=\text{Constant}$ $\text{If}\ \tau_{\text{ex(x)}}=0,\text{ L}_\text{z}=\text{constant},$ i.e., the angular momentum of a system of particles about an axis is zero. This is called conservation of angular momentum principle. The two discs will acquire common angular speed after some time due to friction between them. According to the diagram,
Applying principle of conservation of angular momentum,
$\text{L}_\text{f}=\text{L}_\text{i}$
$\Rightarrow\text{ I}\omega=\text{I}_1\omega_1+\text{I}_2\omega_2$
$\Rightarrow\ \omega=\frac{\text{I}_1\omega_1+\text{I}_2\omega_2}{\text{I}}=\frac{\text{I}_1\omega_1+\text{I}_2\omega_2}{\text{I}_1+\text{I}_2}$ $(\because\ \text{I}=\text{I}_1+\text{I}_2)$
$=\frac{1}{2}\frac{(\text{I}_1\omega_1+\text{I}_2\omega_2)}{(\text{I}_1+\text{I}_2)}$
Initial KE of two discs,
$\text{K}_\text{i}=\frac{1}{2}\big(\text{I}_1\omega^2_1+\text{I}_2\omega_2^2\big)$
$\Delta\text{K}=\text{K}_\text{f}-\text{K}_\text{i}$
$\Delta\text{K}=\frac{\text{I}_1\text{I}_2}{2(\text{I}_1+\text{I}_2)}(\omega_1-\omega_2)^2<0$
- Yes, the law of conservation of angular momentum can be applied. Since gravitational force and normal reaction (external forces) are passing through the axis of rotation hence produce no torque. As there is no external torque, the law of conservation of angular momentum is applicable.
- External forces, gravitation and normal reaction act through the axis of rotation, hence, produce no torque. Let the common angular velocity of the system is Applying principle of conservation of angular momentum.
Applying principle of conservation of angular momentum,
$\text{L}_\text{f}=\text{L}_\text{i}$
$\Rightarrow\text{ I}\omega=\text{I}_1\omega_1+\text{I}_2\omega_2$
$\Rightarrow\ \omega=\frac{\text{I}_1\omega_1+\text{I}_2\omega_2}{\text{I}}=\frac{\text{I}_1\omega_1+\text{I}_2\omega_2}{\text{I}_1+\text{I}_2}$ $(\because\ \text{I}=\text{I}_1+\text{I}_2)$
- Final KE of the system,
$=\frac{1}{2}\frac{(\text{I}_1\omega_1+\text{I}_2\omega_2)}{(\text{I}_1+\text{I}_2)}$
Initial KE of two discs,
$\text{K}_\text{i}=\frac{1}{2}\big(\text{I}_1\omega^2_1+\text{I}_2\omega_2^2\big)$
$\Delta\text{K}=\text{K}_\text{f}-\text{K}_\text{i}$
$\Delta\text{K}=\frac{\text{I}_1\text{I}_2}{2(\text{I}_1+\text{I}_2)}(\omega_1-\omega_2)^2<0$
- Hence, there is a loss in KE of the system. The loss in kinetic energy is mainly due to the work against the friction between the two discs.
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