An ice skater is spinning at 6.8 rev/s and has a moment of inertia of 0.52 kg ⋅ m2.

a) Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 6.8 rev/s.b) 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 0.75 rev/s.

c) 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 14 s?

Relevance

angular momentum L = J*ω = 0.52*6.8*2*3.1416 = 22.217 kg*m^2/sec

Reduction of his rate of rotation by extending arms is done at constant angular momentum L, therefore :

Value of his moment of inertia J' = L/ω' = 22.217/(0.75*2*3.1416) = 4.715 kgm^2

angular acceleration α = 2*3.1416*(3.25-6.8)/14 = -1.593 rad/sec^2

braking torque Tb = J*α = 0.52*-1.593 = -0.828 N*m

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• a)

Angular momentum = Inertia x angular velocityL = IωL = I*(2π* rev/s)L = 0.52 *2π * 6.8 =  22 kg m²/sb)

I = L/(2π* rev/s) = 22/(2π*0.75) = 4.7 kg m²

c)

average torque, τ = Inertia * angular acceleration

τ = Iα

τ = I(Δω/ Δt) = 0.52(2π*6.8 - 2π*3.25)/14 = 0.83 N⋅m

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