Question

Two discs of moment of inertia \( I_1 = 4 \, kg \, m^2 \) and \( I_2 = 2 \, kg \, m^2 \) about their central axes and normal to their planes, rotating with angular speeds \( 10 \, rad/s \) and \( 4 \, rad/s \) respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is __________ J.

Answer 1

Correct Answer: 24

The kinetic energy of a rotating body is given by the formula:

\( KE = \frac{1}{2} I \omega^2 \),

where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

Kinetic Energy of Each Disc Before Contact: For disc 1:

\( KE_1 = \frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} (4 \, \text{kg m}^2)(10 \, \text{rad/s})^2 = \frac{1}{2} \times 4 \times 100 = 200 \, \text{J} \).

For disc 2:

\( KE_2 = \frac{1}{2} I_2 \omega_2^2 = \frac{1}{2} (2 \, \text{kg m}^2)(4 \, \text{rad/s})^2 = \frac{1}{2} \times 2 \times 16 = 16 \, \text{J} \).

Total Kinetic Energy Before Contact:

\( KE_{\text{total initial}} = KE_1 + KE_2 = 200 \, \text{J} + 16 \, \text{J} = 216 \, \text{J} \).

Finding Final Angular Velocity After Contact: When the discs come into contact, they will rotate together, and we can use the principle of conservation of angular momentum. Initial angular momentum \( L_{\text{initial}} \):

\( L_{\text{initial}} = I_1 \omega_1 + I_2 \omega_2 = (4 \, \text{kg m}^2)(10 \, \text{rad/s}) + (2 \, \text{kg m}^2)(4 \, \text{rad/s}) = 40 + 8 = 48 \, \text{kg m}^2/\text{s} \).

The total moment of inertia after they are in contact:

\( I_{\text{total}} = I_1 + I_2 = 4 \, \text{kg m}^2 + 2 \, \text{kg m}^2 = 6 \, \text{kg m}^2 \).

Final angular velocity \( \omega_f \):

\( L_{\text{final}} = I_{\text{total}} \omega_f \implies 48 = 6 \omega_f \implies \omega_f = 8 \, \text{rad/s} \).

Final Kinetic Energy After Contact:

\( KE_{\text{final}} = \frac{1}{2} I_{\text{total}} \omega_f^2 = \frac{1}{2} (6 \, \text{kg m}^2)(8 \, \text{rad/s})^2 = \frac{1}{2} \times 6 \times 64 = 192 \, \text{J} \).

Calculating Loss in Kinetic Energy:

\( \text{Loss in KE} = KE_{\text{total initial}} - KE_{\text{final}} = 216 \, \text{J} - 192 \, \text{J} = 24 \, \text{J} \).

Answer 2

The problem asks for the loss in kinetic energy when two rotating discs with given moments of inertia and angular speeds are brought into contact and rotate together as a single system.

Concept Used:

This problem is an example of an inelastic rotational collision. The key principles to be used are:

1. Conservation of Angular Momentum: When the two discs are brought into contact, the forces between them are internal to the system. Since there is no external torque acting on the system about the axis of rotation, the total angular momentum of the system is conserved.

\[ L_{\text{initial}} = L_{\text{final}} \] \[ I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_{\text{final}} \]

2. Rotational Kinetic Energy: The kinetic energy of a rotating body is given by \( K = \frac{1}{2} I \omega^2 \). The loss in kinetic energy is the difference between the total initial kinetic energy and the final kinetic energy of the combined system.

\[ \Delta K = K_{\text{initial}} - K_{\text{final}} \]

Step-by-Step Solution:

Step 1: List the given initial conditions.

Moment of inertia of the first disc, \( I_1 = 4 \, \text{kg} \, \text{m}^2 \)
Angular speed of the first disc, \( \omega_1 = 10 \, \text{rad/s} \)
Moment of inertia of the second disc, \( I_2 = 2 \, \text{kg} \, \text{m}^2 \)
Angular speed of the second disc, \( \omega_2 = 4 \, \text{rad/s} \)

Step 2: Apply the principle of conservation of angular momentum to find the final common angular speed (\(\omega_f\)).

The total angular momentum before contact is the sum of the angular momenta of the two discs:

\[ L_i = I_1 \omega_1 + I_2 \omega_2 = (4)(10) + (2)(4) = 40 + 8 = 48 \, \text{kg} \, \text{m}^2/\text{s} \]

After contact, the two discs rotate together. The total moment of inertia of the combined system is \( I_f = I_1 + I_2 = 4 + 2 = 6 \, \text{kg} \, \text{m}^2 \). The final angular momentum is:

\[ L_f = I_f \omega_f = (I_1 + I_2) \omega_f = 6 \omega_f \]

Equating initial and final angular momenta (\(L_i = L_f\)):

\[ 48 = 6 \omega_f \implies \omega_f = \frac{48}{6} = 8 \, \text{rad/s} \]

Step 3: Calculate the total initial kinetic energy (\(K_i\)) of the system.

\[ K_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 \] \[ K_i = \frac{1}{2} (4)(10)^2 + \frac{1}{2} (2)(4)^2 = 2(100) + 1(16) = 200 + 16 = 216 \, \text{J} \]

Step 4: Calculate the final kinetic energy (\(K_f\)) of the combined system.

\[ K_f = \frac{1}{2} (I_1 + I_2) \omega_f^2 \] \[ K_f = \frac{1}{2} (6)(8)^2 = 3(64) = 192 \, \text{J} \]

Final Computation & Result:

Step 5: Calculate the loss in kinetic energy (\(\Delta K\)).

The loss in kinetic energy is the difference between the initial and final kinetic energies.

\[ \Delta K = K_i - K_f = 216 \, \text{J} - 192 \, \text{J} \] \[ \Delta K = 24 \, \text{J} \]

The loss in kinetic energy of the system in the process is 24 J.