Respuesta :
Answer:
a) w₂ = 5.16 rev / s , b) K₀ = 29.55 J, K₂ = 190.88 J
Explanation:
The skater forms an isolated system whereby its angular momentum is preserved
Initial. With outstretched arms
L₀ = I₀ w₀
Final. With arms stuck
L₂ = I₂W₂
L₀ = L₂
I₀ w₀ = I₂ w₂
w₂ = I₀ / I₂ w₀
Let's calculate
w₂ = 0.800 2.34 / 0.363
w₂ = 5.16 rev / s
b) The kinetic energy is
K = ½ I w²
Let's reduce to the SI system
w₀ = 0.800 rev / s (2π rad / 1 rev) = 5.026 rad / s
w₂ = 5.16 rev / s (2π rad / 1 rev) = 32.42 rad / s
The initial kinetic energy
K₀ = ½ I₀ w₀²
K₀ = ½ 2.34 5,026²
K₀ = 29.55 J
The final kinetic energy
K₂ = ½ I₂ w₂²
K₂ = ½ 0.363 32.42²
K₂ = 190.88 J
Answer:
[tex]a) \:5.16\:\:rev/s[/tex]
[tex]b)\:29.6\:\:J,\:\:191\:\:J[/tex]
Explanation:
a) We will apply conservation of momentum,
[tex]L = L'\\I\omega=I'\omega'[/tex]
[tex]\omega'=\frac{I}{I'}\omega=\frac{2.34}{0.363}=0.800=5.16\:\:rev/s[/tex]
b) We will use rotational kinetic energy,
Initial value of kinteric energy will be as follows:
[tex]KE_{rot}=\frac{1}{2}I\omega^2\\KE_{rot}=0.5*2.34*(0.800*2\pi)^2=29.6 \:\:J[/tex]
Final value of kinteric energy will be as follows:
[tex]KE_{rot}\:'=\frac{1}{2}I'\omega'^2\\KE_{rot}\:'=0.5*0.363*(5.16*2\pi)^2=191 \:\:J[/tex]
