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urbano
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Homework Statement
A 60kg ice skater is spinning around at 300 degrees per second. His radius of gyration is .5m. As he pulls his arms in he does 300 J of work.
i.What is his initial kinetic energy?
ii.What is his final kinetic energy ?
iii.Assuming that his new radius of gyration is .32m, what is his final moment of inertia ?
iv.What is his final angular momentum ?
Homework Equations
1/2mv^2 Kinetic energy = 1/2 mass X velocity squared
I= mk^2 Inertia = mass X radius of gyration squared
I x ω Angular momentum = inertia X angular velocity
The Attempt at a Solution
i.I converted degrees to radians
1/2 X 60kg X 5.24 radians squared = 157.2 J
ii. I just added 300J on here as I assumed if the skater done 300J of work as he pulled the arms in I'd just add this on. 457.2 J
iii. Final moment of inertia
inertia = 60kg X .32m^2 = 6.144 kg.m
iV. final angular momentum:
= I X ω
= 6.144kg.m X ω
ω= ..?
or could I go
KE = 1/2m X V^2
so KE = 1/2 I x ω^2
457.2J = 1/2 I x ω^2
√ω = √(457.2/3.07)
ω= 12.2 rad/s