1592 A 53 kg skater twirls at 0.77 rev/s with her arms extended. She then brings

Question

1592 A 53 kg skater twirls at 0.77 rev/s with her arms extended. She then brings her arms in and twirls at 1.06 rev/s. The skater's mass moment of inertia can be determined by approximating her to be a uniform cylinder with radius 15 cm when her arms are in NOTES IMAGES DISCUSS UNITS STATSHELP Part Description Answer Chk History Determine the mass moment of inertia of the skater with her arms extended. (include units with answer) 22.43 pts,90% | # tries: 0 Format Check Show Details 290 try penalty Hints: 1,0 Determine the percent of change in kinetic energy when the skater brings her arms in. 22.43 pts.909 # tries : B. 0 Show Details 2% try penalty Hints: 1,0

Explanation / Answer

here,

initial moment of inertia , I0 = 0.5 * m * r^2

I0 = 0.5 * 53 * 0.15^2 kg.m^2 = 0.596 kg.m^2

w0 = 0.77 rev/s

w = 1.06 rev/s

a)

let the moment of inertia of skater when arms are extended be I

using conservation of angular momentum

I * w = I0 * w0

I * 1.06 = 0.596 * 0.77

I = 0.43 kg.m^2

the moment of inertia of skater when arms are extended is 0.43 kg.m^2

b)

the percentage of change in kinetic energy , % = ( 0.5 * I * w^2 - 0.5 * I0 * w0^2)/(0.5 * I0 * w0^2) * 100

% = ( 0.43 * 1.06^2 - 0.596 * 0.77^2) /(0.596 * 0.77^2) * 100

% = 36.7 %

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