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:

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