10 ac 10 Problem 10 An ice skater is spinning at 64 revs and

#10 (a-c)

(10%) Problem 10: An ice skater is spinning at 6.4 rev/s and has a moment of inertia of 0.56 kg·m eis - 33% Part (a) Calculate the angular momentum, in kilogram meters squared per second of the ice s inertia (in kilogram meters squared) if his rate of rotation decreases to 1.25 rev/s average torque that was exerted, in N·m, if this takes 14 s? ter spinnin art (b) He reduces his rate of rotation by extending his arms and increasing his moment of inertia. Find the value of his moment of 33% Part c) Suppose instead he keeps his arms in and allows friction of the ce to low hi n to 3.75 rev/s what is the magnitude of the Grade Summary Deductions Potential 100% ave | | (171819 Submissions Attempts remaining: 7 (1% per attempt) detailed view | tan() sin() cos() cotan sinacos0 atan aco sinh0 cosh tan cotan) Degrees Radians NO

Solution

a) angular momentum = I w

= 0.56 * 6.4 * 2 pi = 22.52 kg.m²/s

b) moment of inertia = 22.52 / (1.25 * 2pi)

= 2.87 kg.m²

c) a = 2 pi (6.4 - 3.75) / 14 = 1.19 rad/s²

magnitude of average torque = I * alpha = 0.56 * 1.19

= 0.666 N.m