Respuesta :
Answer:
K = I₀ w₀²
Explanation:
This is a problem of conservation of angular momentum,
We take the counterclockwise turn as positive
Initial. Before the crash
L₀ = -I₀ 3w₀ + I₀ w₀
Final. The two albums together
[tex]L_{f}[/tex]= (I₀ + I₀) wf
The moment is preserved
L₀ = [tex]L_{f}[/tex]
-I₀ 2w₀ = (2I₀) [tex]w_{f}[/tex]
[tex]w_{f}[/tex] = -w₀
Let's look for kinetic energy
K = ½ I w²
initial
K₀ = ½ I₀ (-3w₀)² + ½ I₀ w₀²
K₀ = ½ I₀ w₀² (9-1)
K₀ = 4 I₀ w₀² = E₀
Final
K = ½ 2 I₀ (-w₀)²
K = I₀ w₀²
The kinetic energy of the system is [tex]I_0 \omega _0^2[/tex] object reaches a common angular speed.
What is Kinetic energy?
It is defined as the energy of an object or body due to the motion of the object.
From the conservation of angular momentum,
Assume the counterclockwise turn is positive,
Before the crash,
[tex]L_0 = - I_o \times 3\omega _0 + I_0\omega _0[/tex]
Final. The two disk together
= [tex](I_0 +I_0 ) \omega _f[/tex]
Since the moment is preserved
[tex]L_0 = L_f[/tex]
[tex]I_0 \times 2 \omega _0 = 2I_0 \times \omega _f[/tex]
[tex]\omega _f = \omega _0[/tex]
For kinetic energy
[tex]K = \dfrac 12 I\omega ^2[/tex]
initial Kinetic energy,
[tex]K_0= \dfrac 12 I_0 ( -3\omega _0)^2 + \dfrac 12 I_0 w_0^2\\\\\K_0 = \dfrac 12 I_0 w_0^2 (9-1)\\\\K_0 = 4 I_0 \omega ^2[/tex]
Final kinetic energy,
[tex]K = \dfrac 12 2I_0 (-\omega _0)^2\\\\K = I_0 \omega _0^2[/tex]
Therefore, the kinetic energy of the system is [tex]I_0 \omega _0^2[/tex]object reaches a common angular speed.
To know more about the disk's kinetic energy,
https://brainly.com/question/17063287
