Question

Calculate the moment of inertia (in kg·m2) of a skater given the following information.
(a)The 76.0 kg skater is approximated as a cylinder that has a 0.140 m radius.

(b)The skater with arms extended is approximately a cylinder that is 68.0 kg, has a 0.140 m radius, and has two 0.950 m long arms which are 4.00 kg each and extend straight out from the cylinder like rods rotated about their ends.

Answer 1

Use the moment of inertia for a solid cylinder about its central axis, I_cyl = (1/2) M R^2, and for a rod about one end, I_rod,end = (1/3) m L^2.

(a) M = 76.0 kg, R = 0.140 m
I = 1/2 · 76.0 · (0.140)^2
(0.140)^2 = 0.0196 → 1/2 · 76.0 · 0.0196 = 0.7448 kg·m^2
I ≈ 0.745 kg·m^2

(b) Cylinder: M = 68.0 kg, R = 0.140 m
I_cyl = 1/2 · 68.0 · (0.140)^2 = 0.6664 kg·m^2

Each arm: m = 4.00 kg, L = 0.950 m
I_arm = 1/3 · 4.00 · (0.950)^2
(0.950)^2 = 0.9025 → I_arm = (1/3) · 4.00 · 0.9025 = 1.2033 kg·m^2

Two arms: 2 · 1.2033 = 2.4067 kg·m^2

Total I = I_cyl + I_arms = 0.6664 + 2.4067 = 3.0731 kg·m^2

I ≈ 3.07 kg·m^2