A 45 kg figure skater is spinning on the toes of her skates at 1.0 rev/s. Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 kg, 20 cm average diameter, 160 cm tall) plus two rod-like arms (2.5 kg each, 66 cm long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 kg, 20-cm-diameter, 200-cm-tall cylinder. What is her new angular velocity, in rev/s?

Short Answer

Expert verified

Her new angular velocity = 3.3 rev/sec

Step by step solution

01

Given information

Mass of the skater 45 kg
Angular velocity of the skater= 1 rev/sec

Now details about the model

Mass of model =40 kg
Average diameter of model =20cm =0.2 m
so average radius of=0.1m
Length of the model =1.6 m
The specifications for the rod :
Mass of the rod =2.5 kg each
Length of arm =0.66 m

02

Explanation

First consider when arms are stretched,

The torso acts like a solid cylindrical shape, and the arms stretched acts like a rod

Total moment of inertia = Moment of inertia of the solid cylinder + Moment of inertia of cylindrical rod

Itotal=12Mr2+2Mr2Itotal=(40kg)(0.10m)22+2(2.5kg)(0.33m)2=0.744kgm2

angular momentum = Iω,

substitute the values we get

L=(0.744kg.m2)(1revolution/sec)=0.744kg.m2/s

Now consider when her arms are above head:

Total mass is on center and total mass is 45 kg.

So the moment of inertia is

I'=Mr22

Susbstitute the values

I'=(45kg)(0.10m)22=0.225kgm2

Calculate the momentum

L'=I'ω=0.225kg.m2ω..................................(1)

From the momentum conservation, equate equation(1) and equation(2)

L=L'Iωo=I'ωω=IωoI'

ω=0.744kg.m2/s0.225kg.m2=3.3rev/sec

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