Final Answer:
(a) The skater's angular speed when she first grabs the pole is approximately 3.91 rad/s.
(b) The skater's angular speed after she pulls her center of mass to a distance of r2 = 0.23 m from the pole is approximately 8.55 rad/s.
Step-by-step explanation:
Step 1: Define the given values:
Mass of the skater (m) = 60 kg
Initial speed of the skater (v) = 1.33 m/s
Initial distance from the pole (r1) = 0.34 m
Final distance from the pole (r2) = 0.23 m
Step 2: Calculate the initial angular momentum:
Angular momentum is conserved in this case, as there is no external torque acting on the system. Therefore, the initial angular momentum is equal to the final angular momentum.
L_initial = m * v * r1
L_initial = 60 kg * 1.33 m/s * 0.34 m
L_initial ≈ 27.84 kg m^2/s
Step 3: Calculate the initial angular speed:
The initial angular speed (ω_initial) can be calculated using the following equation:
ω_initial = L_initial / (m * r1^2)
ω_initial = 27.84 kg m^2/s / (60 kg * 0.34 m^2)
ω_initial ≈ 3.91 rad/s
Step 4: Calculate the final angular speed:
The final angular speed (ω_final) can be calculated using the same equation, but with the final distance (r2):
ω_final = L_initial / (m * r2^2)
ω_final = 27.84 kg m^2/s / (60 kg * 0.23 m^2)
ω_final ≈ 8.55 rad/s
Therefore, the skater's angular speed is approximately 3.91 rad/s when she first grabs the pole and 8.55 rad/s after she pulls her center of mass closer to the pole.