Question

Suppose an ice skater is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of 2.34 kg ⋅ m2 with her arms extended and of 0.363 kg ⋅ m2 with her arms close to her body. (These moments of inertia are based on reasonable assumptions about a 60.0-kg skater.) (a) What is her angular velocity in revolutions per second after she pulls in her arms? (b) What is her rotational kinetic energy before and after she does this?

Answer 1

`a) \:5.16\:\:rev/s`

`b)\:29.6\:\:J,\:\:191\:\:J`

Step-by-step explanation:

a) We will apply conservation of momentum,

`L = L'\\I\omega=I'\omega'`

`\omega'=(I)/(I')\omega=(2.34)/(0.363)=0.800=5.16\:\:rev/s`

b) We will use rotational kinetic energy,

Initial value of kinteric energy will be as follows:

`KE_(rot)=(1)/(2)I\omega^2\\KE_(rot)=0.5*2.34*(0.800*2\pi)^2=29.6 \:\:J`

Final value of kinteric energy will be as follows:

`KE_(rot)\:'=(1)/(2)I'\omega'^2\\KE_(rot)\:'=0.5*0.363*(5.16*2\pi)^2=191 \:\:J`

Answer 2

a) w₂ = 5.16 rev / s , b) K₀ = 29.55 J, K₂ = 190.88 J

Step-by-step explanation:

The skater forms an isolated system whereby its angular momentum is preserved

Initial. With outstretched arms

L₀ = I₀ w₀

Final. With arms stuck

L₂ = I₂W₂

L₀ = L₂

I₀ w₀ = I₂ w₂

w₂ = I₀ / I₂ w₀

Let's calculate

w₂ = 0.800 2.34 / 0.363

w₂ = 5.16 rev / s

b) The kinetic energy is

K = ½ I w²

Let's reduce to the SI system

w₀ = 0.800 rev / s (2π rad / 1 rev) = 5.026 rad / s

w₂ = 5.16 rev / s (2π rad / 1 rev) = 32.42 rad / s

The initial kinetic energy

K₀ = ½ I₀ w₀²

K₀ = ½ 2.34 5,026²

K₀ = 29.55 J

The final kinetic energy

K₂ = ½ I₂ w₂²

K₂ = ½ 0.363 32.42²

K₂ = 190.88 J