Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω₁ and ω₂. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is

Question

A thin circular disc of mass \( M \) and radius \( R \) is rotating in a horizontal plane about an axis passing through its center and perpendicular to its plane with angular velocity \( \omega \). If another disc of the same dimensions but of mass \( M/2 \) is placed gently on the first disc co-axially, then the new angular velocity of the system is:

Answer 1

To solve this problem, we need to understand the concept of rotational motion and how angular momentum is conserved. We are given two discs with the same moment of inertia rotating with angular velocities \(\omega_1\) and \(\omega_2\). When these discs are brought into contact face-to-face, their axes of rotation coincide, leading to a change in energy due to the conservation of angular momentum.

Let's denote the moment of inertia of each disc as \(I\). The initial angular momentum (\(L_i\)) of the system is the sum of the angular momenta of the two discs:

L_i = I \omega_1 + I \omega_2 = I (\omega_1 + \omega_2)

When the discs are brought together, they will rotate with a common angular velocity (\(\omega'\)) due to conservation of angular momentum, which is:

L_f = 2I \omega'

Equating initial and final angular momentum:

I (\omega_1 + \omega_2) = 2I \omega'

Solving for \(\omega'\):

\omega' = \frac{(\omega_1 + \omega_2)}{2}

Next, let's analyze the initial and final kinetic energies to find the loss in energy. The initial kinetic energy (\(KE_i\)) is:

KE_i = \frac{1}{2}I \omega_1^2 + \frac{1}{2}I \omega_2^2

The final kinetic energy (\(KE_f\)) with common angular velocity is:

KE_f = \frac{1}{2}(2I)(\frac{(\omega_1 + \omega_2)}{2})^2 = I \frac{(\omega_1 + \omega_2)^2}{4}

Therefore, the loss of energy (\(\Delta KE\)) is:

\Delta KE = KE_i - KE_f = \left[ \frac{1}{2}I \omega_1^2 + \frac{1}{2}I \omega_2^2 \right] - I \frac{(\omega_1 + \omega_2)^2}{4}

Simplifying this expression:

\Delta KE = \frac{1}{2}I (\omega_1^2 + \omega_2^2) - \frac{1}{4}I (\omega_1^2 + 2\omega_1\omega_2 + \omega_2^2)

\Delta KE = \frac{1}{2}I \omega_1^2 + \frac{1}{2}I \omega_2^2 - \frac{1}{4}I \omega_1^2 - \frac{1}{2}I \omega_1\omega_2 - \frac{1}{4}I \omega_2^2

\Delta KE = \frac{1}{4}I (\omega_1^2 + \omega_2^2 - 2\omega_1\omega_2)

\Delta KE = \frac{1}{4}I (\omega_1 - \omega_2)^2

Therefore, the loss of energy during the process when the discs are brought in contact is given by the expression: \(\frac{1}{4}I (\omega_1 - \omega_2)^2\).

Answer 2