Question

Two ice skaters, both of mass 68 kg, approach on parallel paths 1.6 m apart. Both are moving at 4.5 m/s with their arms outstretched. They join hands as they pass, still maintaining their 1.6 m separation, and begin rotating about one another. Treat the skaters as particles with regard to their rotational inertia.

(C) If they now pull on each other’s hands, reducing their radius to half its original value, what is their common angular speed after reducing their radius?

(D) Calculate the change in kinetic energy for the process described in part C.

Answer 1

To calculate the common angular speed after reducing their radius, we can use the conservation of angular momentum and the equation L = I * w. The initial and final moments of inertia can be calculated using the mass and radius. The final angular speed is determined by the equation w' = (r^2 * w_0) / (0.8)^2.

Step-by-step explanation:

To calculate the common angular speed after reducing their radius, we can use the conservation of angular momentum. The initial angular momentum of the system will be equal to the final angular momentum. The initial radius is 1.6 m and the final radius is half of that, 0.8 m. The initial moment of inertia is given by I = m * r^2, where m is the mass and r is the radius. The final moment of inertia can be calculated by considering two masses rotating about the new center of mass at a distance of 0.8 m, so I' = 2 * (m/2) * (0.8)^2.

Using the equation L = I * w, where L is the angular momentum and w is the angular speed, the initial angular momentum is L = I * w_0, and the final angular momentum is L' = I' * w'. Since the masses are equal, we can set L = L' and solve for w':

L = L'

I * w_0 = I' * w'

m * r^2 * w_0 = (m/2) * (0.8)^2 * w'

w' = (r^2 * w_0) / (0.8)^2