Question

Part A 

What is the angular momentum of a figure skater spinning at 2.8 rev/s with arms in close to her body, assuming her to be a uniform cylinder with a height of 1.5 m. a radius of 15 cm. and a mass of 48 kg ? Express your answer using two significant figures. 

Part B 

How much torque (in magnitude) is required to slow her to a stop in 4.8 s. assuming she does not move her arms? 

Express your answer using two significant figures.

Answer 1

Given

Angular velocity of skater ω = 2.8 rev /s

Height of cylinder h = 1.5

Mass of skater   m = 48 kg

radius of cylinder   r = 15 cm

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

                                = 0.15 m

a)

Angular momentum of skater is

                L = I ω

Moment of inertia of cylinder is

                I = m r ² / 2

                   = ( 0.5 ) ( 48 kg ) ( 0.15 m )²

                   = 0.54 kg m²

There fore Angular momentum of the skater is

                L = ( 0.54 kg m² )  ( 2.8 rev /s ) ( 2π rad / rev )

                    = 9.5 kg m² / s    

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b)

Torque is required to slow her to a stop

                τ = L / t

                   = (9.5 kg m² s ) / ( 4.8 s )

                    = 2.0 Nm

Feel free to drop comment if you have any doubt or any part of solution is wrong.