Question

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 coinciding with each other. What is the loss in kinetic energy of the system in the process?

• A. $\frac{I_{1}I_{2}\left(\omega_{1} -\omega_{2}\right)^{2}}{2\left(I_{1}-I_{2}\right)}$
• B. $\frac{I_{1}I_{2}\left(\omega_{1} -\omega_{2}\right)^{2}}{\left(I_{1}+I_{2}\right)}$
• C. $\frac{I_{1}I_{2}\left(\omega_{1} +\omega_{2}\right)^{2}}{\left(I_{1}-I_{2}\right)}$
• D. $\frac{I_{1}I_{2}\left(\omega_{1} +\omega_{2}\right)^{2}}{2\left(I_{1}-I_{2}\right)}$

Answer 1

Answer: (A)

$\frac{I_{1}I_{2}\left(\omega_{1} -\omega_{2}\right)^{2}}{2\left(I_{1}-I_{2}\right)}$

Answer 2

Let the moment of inertia of disc $I$ be $I_1$ and the angular speed of disc $I$ be $\omega_1$. Let the moment of inertia of disc $II$ be $I_2$ and the angular speed of disc $II$ be $\omega_2$. Angular momentum of disc $I$ be $L _1 = I_1 \omega_1$ and angular momentum of disc $II$ be $L_2= I_2\omega_2$. $\therefore$ The initial angular momentum of two discs is given by $L_i = I_1 \omega_1 + I_2 \omega_2$ When two discs are brought into contact face to face (one on top of other) and their axis of rotation coincide, the moment of inertia $I$ of the system is equal to the sum of their individual moments of inertia i.e., $I = I_1 + I_2$. Let $\omega$ be the final angular speed of the system. $\therefore$ The final angular momentum of the system is given by $L_f = (I_1 +I_2)\omega$ According to law of conservation of angular momentum, $L_i =L_f$ $I_1 \omega _1 + I_2 \omega_2 = (I_1+I_2)\omega$ $\omega = \frac{I_1 \omega _1 + I_2 \omega_2 }{I_1 +I_2 } \quad ...(i)$ Kinetic energy of disc $I$ is given by $K_1 = \frac{1}{2} I_1\omega_1^2$ Kinetic energy of disc $II$ is given by $K_2 = \frac{1}{2} I_2 \omega_2$ $\therefore$ Initial kinetic energy of two discs is $K_i = \frac{1}{2} I_1\omega_1^2 + \frac{1}{2} I_2 \omega_2^2$ Final kinetic energy of the system is $K_f =\frac{1}{2} (I_1 +I_2)\omega^2$ $=\frac{1}{2}(I_1+I_2) (\frac{I_1\omega_1 + I_2\omega_2}{I_1+I_2})$ (Using $(i)$) $=\frac{1}{2} \frac{(I_1\omega_1 +I_2 \omega_2)^2}{(I_1 +I_2)}$ Loss in kinetic energy of the system $\Delta K = K_{i} -K_{f} = \frac{1}{2}\left(I_{1}\omega_{1}^{2} +I_{2}\omega_{2}^{2}\right) - \frac{\left(I_{1}\omega_{1} +I_{2}\omega_{2}\right)^{2}}{2\left(I_{1}+I_{2}\right)}$ $= \frac{1}{2}I_{1}\omega_{1}^{2} +\frac{1}{2}I_{2}\omega_{2}^{2} - \frac{1}{2}\frac{ I_{1}^{2}\omega_{1}^{2}}{\left(I_{1}+I_{2}\right)} - \frac{1}{2}\frac{I_{2}^{2}\omega_{2}^{2}}{\left(I_{1}+I_{2}\right)} -\frac{1}{2}\frac{2I_{1}I_{2}\omega_{1}\omega_{2}}{\left(I_{1}+I_{2}\right)}$ $= \frac{1}{\left(I_{1}+I_{2}\right)}[\frac{1}{2}I_{1}^{2}\omega_{1}^{2} +\frac{1}{2}I_{1}I_{2}\omega_{1}^{2}$ $+ \frac{1}{2}I_{1}I_{2}\omega_{2}^{2}+\frac{1}{2}I_{2}^{2}\omega_{2}^{2}-\frac{1}{2}I_{1}^{2}\omega_{1}^{2}$ $-\frac{1}{2}I_{2}^{2}\omega_{2}^{2} - I_{1}I_{2}\omega _{1}\omega_{2}]$ $= \frac{I_{1}I_{2}}{2\left(I_{1}+I_{2}\right)}\left(\omega_{1}^{2}+\omega_{2}^{2} -\omega_{1}\omega_{2}\right)$ $= \frac{I_{1}I_{2}\left(\omega_{1}-\omega_{2}\right)^{2}}{2\left(I_{1}+I_{2}\right)}$