Respuesta :
Answer:
[tex]w_f=1.143\frac{rev}{s}[/tex]
Explanation:
1) Notation
[tex]I_{i}=0.8kgm^2[/tex] (Inertia with arms and legs in)
[tex]I_{f}=3.5kgm^2[/tex] (Inertia with arms out and one leg extended)
[tex]w_{i}=5\frac{rev}{s}[/tex]
[tex]w_{f}=?[/tex] (variable of interest)
2) Analysis of the situation
For this case we can assume that there are no external forces acting on the skater, so based on this assumption we don’t have any torque from outside acting on the system. And for this reason, we can consider the angular momentum constant throughout the movement.
On math terms then the initial angular momentum would be equal to the final angular momentum.
[tex]L_i =L_f[/tex] (1)
The angular momentum of a rigid object is defined "as the product of the moment of inertia and the angular velocity and is a vector quantity"
3) Formulas to use
Using this definition we can rewrite the equation (1) like this:
[tex]I_{i}w_{i}=I_{f}w_{f}[/tex] (2)
And from equation (2) we can solve for [tex]w_f[/tex] like this:
[tex]w_f=\frac{I_i w_i}{I_f}[/tex] (3)
And replacing the values given into equation (3) we got:
[tex]w_f=\frac{0.8kgm^2 x5\frac{rev}{s}}{3.5kgm^2}=1.143\frac{rev}{s}[/tex]
And that would be the final answer [tex]w_f=1.143\frac{rev}{s}[/tex].
Angular speed ( in rev/s ) when arms out and one leg open extended outward is 1.14 rev/s
What is moment of Inertia of a rotating body?
moment of inertia, I is the measure of distibution of the mass of the body along the axis of rotatiton.
I = angular momentum, L / angular velocity, ω
L = I * ω
L1 = L2
L1 = angular momentum arms and leg in
L2 = angular momentum arms out and one leg extended
I1 * ω1 = I2 * ω2 conservation of angular momentum
0.8 * 5 = 3.5 * ω2
ω2 = 4 / 3.5
ω2 = 1.14 rev/s
Read more on moment of inertia here: brainly.com/question/3406242