Question

A circular disk of moment of inertia it is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega_i$. Another disk of moment of inertia $I_b$ is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $\omega_f$. The energy lost by the initially rotating disc to friction is

• A. $\frac{1}{2}\frac{I^2_b}{(I_t+I_b)}\omega^2_i$
• B. $\frac{1}{2}\frac{I^2_t}{2(I_t+I_b)}\omega^2_i$
• C. $\frac{I_b-I_t}{(I_t+I_b)}\omega^2_i$
• D. $\frac{1}{2}\frac{I_bI_t}{(I_t+I_b)}\omega^2_i$

Answer 1

The Correct Option is D

To find the energy lost by the initially rotating disk to friction, we must analyze the problem using the principles of angular momentum conservation and energy conservation.

1. Initial Scenario:
   • The initial angular momentum of the system is solely due to the rotating disk with moment of inertia \( I_t \) and angular velocity \( \omega_i \). Thus, the initial angular momentum \( L_i \) is given by: L_i = I_t \omega_i
2. Final Scenario:
   • After the second disk with moment of inertia \( I_b \) is dropped on the first, both disks rotate together with a common angular velocity \( \omega_f \). The final angular momentum \( L_f \) is: L_f = (I_t + I_b) \omega_f
   • Conserving angular momentum, we equate initial and final angular momenta: I_t \omega_i = (I_t + I_b) \omega_f
   • Solving for \( \omega_f \), we get: \omega_f = \frac{I_t \omega_i}{I_t + I_b}
3. Energy Analysis:
   • The initial rotational kinetic energy \( K_i \) of the system is: K_i = \frac{1}{2} I_t \omega_i^2
   • The final rotational kinetic energy \( K_f \) is: K_f = \frac{1}{2} (I_t + I_b) \omega_f^2
   • Substituting \( \omega_f \) from above, the expression becomes: K_f = \frac{1}{2} (I_t + I_b) \left( \frac{I_t \omega_i}{I_t + I_b} \right)^2 = \frac{1}{2} \frac{I_t^2 \omega_i^2}{I_t + I_b}
   • The energy lost \( \Delta E \) to friction is the difference between the initial and final kinetic energies: \Delta E = K_i - K_f = \frac{1}{2} I_t \omega_i^2 - \frac{1}{2} \frac{I_t^2 \omega_i^2}{I_t + I_b}
   • Simplifying this, we have: \Delta E = \frac{1}{2} \omega_i^2 \left( I_t - \frac{I_t^2}{I_t + I_b} \right)
   • This simplifies further to: \Delta E = \frac{1}{2} \omega_i^2 \frac{I_b I_t}{I_t + I_b}

Thus, the energy lost by the initially rotating disk to friction is \frac{1}{2}\frac{I_b I_t}{(I_t+I_b)}\omega^2_i.