Question

An ice skater is spinning at 6.8 rev/s and has a moment of inertia of 0.24 kg ⋅ m2.

Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 6.8 rev/s.

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 1.25 rev/s.

Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.25 rev/s. What is the magnitude of the average torque that was exerted, in N ⋅ m, if this takes 11 s?

Answer 1

Dale Part Ii= 0.24 kg i m² wie - 6.8 rev's - 6.8x 2 x 3.14 – 42.7 rad /see rad see ular Momentum (Li) = Ii Wr - 0.24x 42.7 =110.24 kg m² /sec .part-2 Wh= 1.25 rew's Is = ? y Conservation of Angular Moment was Licht In wi= If Wt 0.24 X 6.8= It x 1.25 It = 1.30 kg.m Moment of Inentia = 1.30kgom? part-3 3225-608 Wi= 42.7 rad/see Ws - 3.25rev see = 3.25x2x3.17 - Ws = 20.4 rad/ see x = wf-Wi" - 20.4-4217 Torque = 2a-2.02. yad (see Lixa = 10.24x -2.02 Torque = 20.68 Nom. * PLEASE GIVE POSITVE RATINGS