Final answer:
Using the conservation of angular momentum, the common angular speed of the ice skaters after joining hands is found to be 7.50 rad/s.
Step-by-step explanation:
The student is dealing with a rotational dynamics problem, specifically involving conservation of angular momentum. When the two ice skaters of mass 68 kg, moving at 6.0 m/s, join hands while maintaining a 1.6 m separation, they will begin rotating about a common axis due to the conservation of angular momentum. Since they were originally moving linearly, we need to equate the initial linear momentum with the final angular momentum to find the angular speed after joining hands.
Let's assume they grab hands and rotate around the center point between them. The initial linear momentum L for each skater can be calculated by:
L = m * r * v
Where m is the mass, r is the distance to the rotation axis (half the distance between them, which is 0.8 m), and v is the initial velocity.
So for one skater, L = 68 kg * 0.8 m * 6.0 m/s = 326.4 kg*m^2/s.
Since there are two skaters and they both contribute to the angular momentum, the total initial angular momentum is:
L_total = 2 * 326.4 kg*m^2/s = 652.8 kg*m^2/s.
After they join hands, we use the conservation of angular momentum:
L_initial = I_total * ω_final
Where I_total is the total moment of inertia and ω_final is the final angular speed. The moment of inertia for each skater is I = m * r^2, and for both skaters:
I_total = 2 * m * r^2 = 2 * 68 kg * (0.8 m)^2 = 87.04 kg*m^2.
Finally, we solve for the final angular speed ω_final:
ω_final = L_total / I_total = 652.8 kg*m^2/s / 87.04 kg*m^2 = 7.50 rad/s.
Therefore, the common angular speed of the ice skaters after joining hands is 7.50 rad/s