Question

Suppose we want to calculate the moment of inertia of a 67 kg skater, relative to a vertical axis through their center of mass. Part (a) First calculate the moment of inertia (in kg⋅m2) when the skater has their arms pulled inward by assuming they are cylinder of radius 0.11 m. $sig.gif?tid=8O75-6C-7E-4A-961B-22885$ Part (b) Now calculate the moment of inertia of the skater (in kg⋅m2) with their arms extended by assuming that each arm is 5% of the mass of their body. Assume the body is a cylinder of the same size, and the arms are 0.775 m long rods extending straight out from the center of their body being rotated at the ends. $sig.gif?tid=8O75-6C-7E-4A-961B-22885$ I =

Answer 1

Jiven, The man of skater m = 67kg Purta: Sadis, Raoull m The moment of Inertia , I = 1 / 2 MR? I = 1/2 * 67 * (0.1² I = 0.4054 kgoma - Part bi The length of aims I = 0775m 54. of mass, mint 67*0,05 = 3135 kg 2 I = me? + [ite m'g* + m'( +)# 2 = 6+coon" + [2:2 3:35* 60-7+5)* + 3:26(0729 +0.1)**2 [ = 0.4054+[0:167670+ 0.829153 & 2 I z 2139 kg.m? . I z 2