Question

(10%) Problem 10: An ice skater is spinning at 6.8 rev/s and has a moment of inertia of 0.24 kg-m 33% Part (a) Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 6.8 rev/s L 10.25 Correct > 33% Part (b) He reduces his rate of rotation by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kilogram meters squared) if his rate of rotation decreases to 1.75 rev/s. Grade Summary 5% 95% Potential sin0 cotan asin0 cos0 tan() π ( acos0 Attempts remaining: (5% per attempt) detailed view atan acotan sinh0 cosh tanh cotanhO Degrees Radians 5% Submit Hint Hints: 0% deduction per hint. Hints remaining: Feedback: 0% deduction per feedback. 33% Part (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.75 rev/s. What is the magnitude of the average torque that was exerted, in N·m, if this takes 11 s?

Answer 1

(a) L = mvN = Iw = 0.24 times (6.8 times 2 pi/s) L = 10.25 kg m^2/s (b) Moment of Inertia = (I_f) = L/W = 10.25/1.75 times 2 pi I_2 = 0.933 kg m^2 (c) Average Torque (T) = I alpha = I(w_f - w_i)/Delta t = 0.24 (3.75 times 2 pi - 6.8 times 2 pi)/11 T = -0.418 N.M