Answer:
a) the moment of inertia is 0.784 Kg*m²
b) the moment of inertia is with arms extended is 1.187 Kg*m²
c) the angular velocity in scenario (b) is 4.45 rad/s
Explanation:
The moment of inertia is calculated as
I = ∫ r² dm
since
I = Ix + Iy
and since the cylinder rotates around the y-axis then Iy=0 and
I = Ix = ∫ x² dm
if we assume the cylinder has constant density then
m = ρ * V = ρ * π R²*L = ρ * π x²*L
therefore
dm = 2ρπL* x dx
and
I = ∫ x² dm = ∫ x² 2ρπL* x dx = 2ρπL∫ x³ dx = 2ρπL (R⁴/4 - 0⁴/4) = ρπL R⁴ /2 = mR² /2
therefore
I skater = mR² /2 = 80 Kg * (0.140m)²/2 = 0.784 Kg*m²
b) since the arms can be seen as a thin rod
m = ρ * V = ρ * π R²*L = ρ * π R²*x
dm =ρ * π R² dx
I1 = ∫ x² dm = ∫ x² * ρ * π R² dx = ρ * π R²*∫ x² dx = ρ * π R²* ((L/2)³/3 - (-L/2)³/3)
= ρ * π R²*2*L³/24 = mL²/12
therefore
I skater 2 = I1 + I skater = mL²/12 + mR² /2= 8 Kg* (0.85m)²/12 +(80-8) Kg * (0.140m)²/2 = 1.187 Kg*m²
c) from angular momentum conservation
I s2 * ω s2 = I s1 * ω s1
thus
ω s2 = (I s1 / I s2 )* ω s1 /= (0.784 Kg*m²/1.187 Kg*m²) * 6.75 rad/s = 4.45 rad/s
