Use the parallel axis theorem to calculate the moment of inertia of a sprinter’s
ID: 3280884 • Letter: U
Question
Use the parallel axis theorem to calculate the moment of inertia of a sprinter’s leg about the hip axis, as in the figure. The moments of inertia of the upper leg (thigh), lower leg (calf), and foot about their respective centers of mass are 0.1052, 0.0504, and 0.0038kg.m2; these centers of mass are, respectively, 0.30, 0.45, and 0.53 m from the hip rotation axis, and these segments, respectively, have masses 7.21, 3.01, and 1.05 kg.
0.1052 kg-my g-m2 0.30 0.53 m 0.45 m 0.0038 kg-m? 0.0504 kg-m2Explanation / Answer
M.I. OF UPPER LEG ABOUT HIP AXIS =0.1052+7.21*(0.3)^2=0.7541 kgm^2 (using I=Ic+mh^2)
M.I. OF lower LEG ABOUT HIP AXIS =0.0504+3.01*(0.45)^2=0.6599kgm^2
M.I. OF foot ABOUT HIP AXIS =0.0038+1.05*(0.53)^2=0.2987 kgm^2
Total moment of inertia about the hip axis = 1.7127 kgm^2
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