Question

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together. Upon ending, their arms extend outward, proclaiming their finish. Not quite as noticeably, one leg goes out as well. Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg*m^2 and for arms and legs in is 0.80 kg*m^2. If she starts out spinning at 5.0 rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?

Answer 1

Answer:

$w_f=1.143\frac{rev}{s}$

Explanation:

1) Notation

$I_{i}=0.8kgm^2$ (Inertia with arms and legs in)

$I_{f}=3.5kgm^2$ (Inertia with arms out and one leg extended)

$w_{i}=5\frac{rev}{s}$

$w_{f}=?$  (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.

$L_i =L_f$   (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:

$I_{i}w_{i}=I_{f}w_{f}$ (2)

And from equation (2) we can solve for $w_f$ like this:

$w_f=\frac{I_i w_i}{I_f}$   (3)

And replacing the values given into equation (3) we got:

$w_f=\frac{0.8kgm^2 x5\frac{rev}{s}}{3.5kgm^2}=1.143\frac{rev}{s}$

And that would be the final answer $w_f=1.143\frac{rev}{s}$.