Respuesta :
Given:
I₁ = 0.70 kg-m², the moment of inertia with arms and legs in
I₂ = 3.5 kg-m², the moment of inertia with arms and a leg out.
ω₁ = 4.8 rev/s, the angular speed with arms and legs in.
That is,
ω₁ = (4.8 rev/s)*(2π rad/rev) = 30.159 rad/s
Let ω₂ = the angular speed with arms and a leg out.
Because momentum is conserved, therefore
I₂ω₂ = I₁ω₁
ω₂ = (I₁/I₂)ω₁
= (0.7/3.5)*(30.159)
= 6.032 rad/s
ω₂ = (6.032 rad/s)*(1/(2π) rev/rad) = 0.96 rev/s
Answer: 0.96 rev/s
I₁ = 0.70 kg-m², the moment of inertia with arms and legs in
I₂ = 3.5 kg-m², the moment of inertia with arms and a leg out.
ω₁ = 4.8 rev/s, the angular speed with arms and legs in.
That is,
ω₁ = (4.8 rev/s)*(2π rad/rev) = 30.159 rad/s
Let ω₂ = the angular speed with arms and a leg out.
Because momentum is conserved, therefore
I₂ω₂ = I₁ω₁
ω₂ = (I₁/I₂)ω₁
= (0.7/3.5)*(30.159)
= 6.032 rad/s
ω₂ = (6.032 rad/s)*(1/(2π) rev/rad) = 0.96 rev/s
Answer: 0.96 rev/s
