(890) Problem 12: An ice skater is spinning at 6.4 revs and has a moment of iner

Question

(890) Problem 12: An ice skater is spinning at 6.4 revs and has a moment of inertia of 0.36 kg-m 33% Part (a) Calculate the angular momentum in kilogram meters squared per second, of the ice skater spinning at 6.4 revs. Grade Summary Potential Submissions 0% 100% Attempts remaining cotan asinacos0 atan0 acotan0 sinh0 cosh0 tanh0cotanh0 O Degrees Radians 6per attempt etailed view 1 23 END Submit Hint I give up! Hints: 5% deduction per hint. Hints remaining- Feedback: 2% deduction per feedback. 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 1.5 rev/s. - 33% Part (c) Suppose instead he keeps his arms in and allows friction ofthe ice to slow him to 3.75 revs. What is the magnitude of the average torque that was exerted, in N·m. if this takes 19 s?

Explanation / Answer

(a) = 6.4 rev/s = 2*6.4 rad/s = 40.21 rad/s

angular momentum = I* = 0.36kg.m^2 * 40.21rad/s = 14.48 kg.m^2/s

(b) = = 1.5 rev/s = 2*1.5 = 9.424 rad/s

moment of inertia I = 14.48 / 9.424 kg.m^2 = 1.536 kg.m^2

(c) f = = 3.75 rev/s = 2*3.75 rad/s = 23.56 rad/s

angular accel = (f - i) / t = (23.56 – 40.21) / 19 rad/s^2 = -0.876 rad/s^2

Av torque T = I. = 0.36kg.m^2 * -0.876 rad/s^2 = -0.315 N.m

Hope this helps :)

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