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
