Question

A flat solid disk of radius R and mass M is around vertical axle going through a center of mass of a disk. Suddenly a hoop of mass M and radius R is falling coaxially on the disk. Hoop did not rotate initially. If rotational energy of the system was initially RE, what is rotational energy of the system after hoop has landed?

Answer 1

Answer:

Kf = 2/9 RE

Explanation:

The initial mechanical energy of the system is the sum of the kinetic and potential energy of the two bodies as the height does not change we can take the zero in the position of the disk, the ring is still so it has no energy and the energy disk is energy rotation kinetics

      K = ½ m₁ w₁²

In the final position the disc and the ring rotate together, so calluses contribute energy

      Kf = K₁ + K₂ = ½ (m₁ + m₂) w₂²

Where m1 is the mass of the disk, m2 the mass of the ring and w is the initial and final angular velocity

To find the final angular velocity, we treat the case as an inelastic shock, where the kinetic moment (L) is preserved, the system is formed by the two bodies.

       L₀ = Lf

       L₀ = I₁ w₁ + 0

       Lf = (I₁ + I₂) w₂

       I₁ w₁ = (I₁ + I₂) w ₂

       w₂ = I₁ / (I₁ + I₂) w₁

We take the kinetic moments of the bodies

Disk     I₁ = ½ m₁ R₁²

Hoop   I₂ = m₂ R₂²

Let's calculate the final angular velocity

     w₂ = ½ m₁ R² / (1/2 m₁ R² + m₂ R²) w₁

     w₂ = ½ m₁ / (m₁/2 + m₂) w₁

With this value we can substitute and calculate the final kinetic energy

     Kf = ½ (m₁ + m₂) [½ m₁ / (m₁ /2 + m₂) w₁]²  

     Kf = 1/8 [(m₁ + m₂) m₁² / (m₁/2 + m₂)²] w₁²

Let's substitute the values ​​that the mass and radius of the disc and ring give us are the same (M, R)

    Kf = 1/8 [2M M² / (3M/2)²] w₁² = ¼ M³ / (9M² /4) w₁²

    Kf = 1/9 M w₁²

This is the final kinetic energy, let's say it based on the initial (RE)

    Ko = RE = ½ M w₁²

    Kf / Ko = (1/9 M w₁²) / (1/2 M w₁²)

    Kf / RE = 2/9

    Kf = 2/9 RE

This loss of kinetic energy is transformed into internal energy during the crash