A disc with moment of inertia I₁ = 40 kg.m² and angular velocity $\omega$₁ = 20 rad/s is dropped on to a stationary second disc along the axis of rotation.The second disc has moment of inertia I₂ = 60 kg.m².How much rotational kinetic energy is lost? A)3200 J B)4800 J C)8000 J D)11300 J E)15000 J

Question

Quizplus · Accepted Answer

The initial rotational kinetic energy of the first disc can be calculated as:
K_i = (1/2)I_1*omega^2 = (1/2)*40*20^2 = 16000 J

When the first disc drops onto the stationary second disc, the final angular velocity can be calculated using conservation of angular momentum:
I_1*omega_i = (I_1 + I_2)*omega_f
20*40 = (40+60)*omega_f
omega_f = 8 rad/s

The final rotational kinetic energy of the combined system can be calculated as:
K_f = (1/2)*(I_1 + I_2)*omega_f^2 = (1/2)*100*8^2 = 3200 J

The rotational kinetic energy lost in the collision is:
delta_K = K_i - K_f = 16000 - 3200 = 12800 J

Therefore, the answer is B) 4800 J.

A Disc with Moment of Inertia I1 = 40 Kg ω\omegaω

A Disc with Moment of Inertia I₁ = 40 Kg $\omega$