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

To solve this problem, we need to calculate the loss in kinetic energy when the two discs are brought into contact with their axes of rotation coaxial.

Initially, the kinetic energies of the two discs are given by:

• Kinetic energy of the first disc: K_1 = \frac{1}{2} I_1 \omega_1^2
• Kinetic energy of the second disc: K_2 = \frac{1}{2} I_2 \omega_2^2

When the discs are combined, they rotate together with a common angular velocity \omega_f

Step 1: Calculate the final angular velocity

The total angular momentum before contact is equal to the total angular momentum after contact:

I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega_f

Solve for the final angular velocity:

\omega_f = \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}

Step 2: Calculate the kinetic energy after combining

The kinetic energy of the system after combining is:

K_f = \frac{1}{2} (I_1 + I_2) \omega_f^2

Substitute the expression for \omega_f:

K_f = \frac{1}{2} (I_1 + I_2) \left(\frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}\right)^2

Step 3: Calculate the loss in kinetic energy

The loss in kinetic energy is given by the difference between the initial and the final kinetic energies:

\Delta K = (K_1 + K_2) - K_f

Substitute the expressions for K_1, K_2, and K_f:

• \Delta K = \left(\frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2\right) - \frac{1}{2} \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{(I_1 + I_2)}

Simplify and calculate:

• Use the identity: (a - b)^2 = a^2 + b^2 - 2ab
• \Delta K = \frac{I_1I_2}{2(I_1 + I_2)}(\omega_1 - \omega_2)^2

This gives us the final expression for the loss in kinetic energy. Therefore, the correct answer is:

\( \frac{I_1I_2}{2(I_1 + I_2)}(\omega_1 - \omega_2)^2 \)