Question

What is the angular momentum of a figure skater spinning at 3.5 rev/s with arms in close to her body, assuming her to be a uniform cylinder with a height of 1.6 m , a radius of 13 cm, and a mass of 60 kg?

B.) How much torque is required to slow her to a stop in 5.8 s, assuming she does not move her arms?

Answer 1

Given

Angular velocity of skater ω = 3.5 rev /s

Heigth of cylinder h = 1.6

Mass of skater   m = 60 kg

radius of cylinder   r = 13 cm

                                = ( 13 cm ) ( 0.01 m / 1 cm )

                                = 0.13 m

a)

Angular momentum of skater is

                L = I ω

Moment of inertia of cylinder is

                I = m r ² / 2

                   = ( 0.5 ) ( 60 kg ) ( 0.13 m )²

                   = 0.507 kg m²

There fore Angular momentum of the skater is

                L = ( 0.507 kg m² )  ( 3.5 rev /s ) ( 2π rad / rev )

                    =   11.149 kg m² / s    

                           or    

                    ˜   11.15 kg m² / s

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b)

Torque is required to slow her to a stop

                τ = L / t

                   = (11.149 kg m² s ) / ( 5.8 s )

                    = 1.922 Nm