Question

Two discs have moments of inertia I\(_1\) and I\(_2\) about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, \(\omega_1\) and \(\omega_2\) respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:

• A. \( \frac{I_1I_2}{2(I_1 + I_2)}(\omega_1 - \omega_2)^2 \)
• B. \( \frac{I_1I_2}{(I_1 + I_2)}(\omega_1 - \omega_2)^2 \)
• C. \( \frac{(\omega_1 - \omega_2)^2}{2(I_1 + I_2)} \)
• D. \( \frac{(I_1 - I_2)^2 \omega_1 \omega_2}{2(I_1 + I_2)} \)

Hint:

This formula for loss of kinetic energy in a perfectly inelastic rotational collision is analogous to the formula for loss of kinetic energy in a one-dimensional perfectly inelastic linear collision: \( \Delta K = \frac{1}{2}\frac{m_1m_2}{m_1+m_2}(v_1-v_2)^2 \). The term \( \frac{m_1m_2}{m_1+m_2} \) is the reduced mass. Similarly, \( \frac{I_1I_2}{I_1+I_2} \) is the reduced moment of inertia. Remembering this analogy can help you recall the formula quickly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have two rotating discs that are brought into contact. Due to frictional forces between them, they eventually rotate with a common angular velocity. Since this is an inelastic collision, some kinetic energy will be lost. We need to calculate this loss.
Step 2: Key Formula or Approach:
1. Conservation of Angular Momentum: Since there is no external torque on the two-disc system, the total angular momentum before and after contact is conserved.
2. Kinetic Energy of Rotation: The rotational kinetic energy of a body is given by \( K = \frac{1}{2}I\omega^2 \).
3. Loss in KE: \( \Delta K = K_{\text{initial}} - K_{\text{final}} \).
Step 3: Detailed Explanation:
Initial State:
Initial angular momentum of the system, \( L_i = I_1\omega_1 + I_2\omega_2 \).
Initial kinetic energy of the system, \( K_i = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 \).
Final State:
When the discs are brought into contact, they rotate together with a common final angular velocity, \( \omega_f \).
The total moment of inertia of the combined system is \( I_f = I_1 + I_2 \).
Final angular momentum of the system, \( L_f = (I_1 + I_2)\omega_f \).
Applying Conservation of Angular Momentum:
\( L_i = L_f \)
\( I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega_f \)
Solving for the final angular velocity:
\[ \omega_f = \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2} \] Calculating Final Kinetic Energy:
\( K_f = \frac{1}{2}I_f\omega_f^2 = \frac{1}{2}(I_1 + I_2)\left(\frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}\right)^2 \)
\[ K_f = \frac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)} \] Calculating Loss in Kinetic Energy:
\( \Delta K = K_i - K_f \)
\( \Delta K = \left(\frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2\right) - \frac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)} \)
Taking a common denominator of \( 2(I_1 + I_2) \):
\( \Delta K = \frac{(I_1\omega_1^2 + I_2\omega_2^2)(I_1 + I_2) - (I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)} \)
Expanding the numerator:
Numerator = \( (I_1^2\omega_1^2 + I_1I_2\omega_1^2 + I_1I_2\omega_2^2 + I_2^2\omega_2^2) - (I_1^2\omega_1^2 + I_2^2\omega_2^2 + 2I_1I_2\omega_1\omega_2) \)
Numerator = \( I_1I_2\omega_1^2 + I_1I_2\omega_2^2 - 2I_1I_2\omega_1\omega_2 \)
Numerator = \( I_1I_2(\omega_1^2 + \omega_2^2 - 2\omega_1\omega_2) \)
Numerator = \( I_1I_2(\omega_1 - \omega_2)^2 \)
Substituting back into the expression for \( \Delta K \):
\[ \Delta K = \frac{I_1I_2(\omega_1 - \omega_2)^2}{2(I_1 + I_2)} \] Step 4: Final Answer:
The loss in kinetic energy is \( \frac{I_1I_2}{2(I_1 + I_2)}(\omega_1 - \omega_2)^2 \).