A disk with moment of inertia I₁ = 40 kg  m² and angular velocity ₁ = 20 rad/s is dropped on to a stationary second disk along the axis of rotation. The second disk 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

When the first disk is dropped onto the second disk, the total angular momentum is conserved, but some rotational kinetic energy is lost due to friction. The total initial angular momentum is given by:
L = I1ω1 + I2ω2
where I1 and ω1 are the moment of inertia and angular velocity of the first disk, and I2 and ω2 are the same for the second disk. Since the second disk is initially stationary, ω2 = 0, and we can solve for the initial angular velocity of the first disk:
ω1 = L / I1 = (I1 + I2)ωf / I1
where ωf is the final common angular velocity of the two disks after they have reached their final positions. Since the disks are rotating about the same axis, we can assume that ωf is the same for both disks. Using conservation of energy, we can relate the initial and final kinetic energy:
K1i + K2i = K1f + K2f + ΔK
where Ki and Kf are the initial and final kinetic energies of the two disks, and ΔK is the change in kinetic energy due to friction. Since the second disk is initially stationary, K2i = 0, and we can solve for ΔK:
ΔK = K1i - K1f = (1/2)I1ω1^2 - (1/2)(I1 + I2)ωf^2
Substituting for ω1 and simplifying, we get:
ΔK = (1/2)(I1I2/(I1 + I2))(ωf - ω1)^2
Plugging in the numbers, we get:
ΔK = (1/2)(40 kg·m^2)(60 kg·m^2/(40 kg·m^2 + 60 kg·m^2))(0 - 20 rad/s)^2 = 4800 J
Therefore, the rotational kinetic energy lost is 4800 J, which corresponds to answer choice B.

A Disk with Moment of Inertia I1 = 40 Kg

A Disk with Moment of Inertia I₁ = 40 Kg