Question

a cylinder of inertia 1.72 kgm2 rotates about a vertical frictionless axle with angular velocity 39 radians/second. a second cylinder of inertia 2 kgm2 is initially not rotating and it drops onto the first cylinder. the two surfaces are rough and they eventually come to the same angular velocity. calculate the magnitude of the loss of kinetic energy (in j)

Answer 1

The magnitude of the loss of kinetic energy is 1246.10 J.

Step-by-step explanation:

To calculate the magnitude of the loss of kinetic energy, we need to find the initial and final angular velocities of the two cylinders after they come to the same angular velocity.

Using the law of conservation of angular momentum, we can set up the following equation:

I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω

Where I₁, I₂ are the moments of inertia of the two cylinders, ω₁, ω₂ are their initial angular velocities, and ω is the final angular velocity.

Substituting the given values, we have:

1.72*(39) + 2*(0) = (1.72+2)*ω

Solving for ω:

ω = 28.37 rad/s

Next, to calculate the loss of kinetic energy, we can use the equation:

ΔKE = ½(I₁ω₁² + I₂ω₂²) - ½(I₁ + I₂)ω²

Substituting the given values, we have:

ΔKE = ½(1.72*(39)² + 2*(0)²) - ½(1.72+2)*(28.37)²

ΔKE = -1246.10 J

Therefore, the magnitude of the loss of kinetic energy is 1246.10 J.