Question

A 51 kg skater twirls at 0.64 rev/s with her arms extended. She then brings her arms in and twirls at 0.95 rev/s. The skater's mass moment of inertia can be determined by approximating her to be a uniform cylinder with radius 17 cm when her arms are in.
a. Determine the mass moment of inertia of the skater with her arms extended.
b. Determine the percent of change in kinetic energy when the skater brings her arms in.

Answer 1

Answer:

(a) 1.09 kg m²

(b) 48.9 %

Explanation:

mass of skater, m = 51 kg

frequency when the arms are extended, f = 0.64 rev/s

frequency when the arms are in, f' = 0.95 rev/s

radius of cylinder when the arms in, r' = 17 cm

Moment of inertia of the body of skater when the arms in, I' = 0.5 x m x r'²

I' = 0.5 x 51 x 0.17 x 0.17 ² = 0.737 kg m²

(a)

By using the conservation of angular momentum

I x ω = I' x ω'

I x 2 x π x f = I' x 2 x π x f'

I x 0.64 = 0.737 x 0.95

I = 1.09 kg m²

Thus, the moment of inertia when the arms are in extended is 1.09 kg m².

(b)

Initial kinetic energy when the arms are extended, k = 0.5 x I x ω²

K = 0.5 x 1.09 x 4 x 3.14 x 3.14 x 0.64 x 0.64 = 8.8 J

Final kinetic energy when the arms are in, k' = 0.5 x I' x ω'²

K' = 0.5 x 0.737 x 4 x 3.14 x 3.14 x 0.95 x 0.95 = 13.1 J

Percentage change in kinetic energy

$\frac{K' - K}{K}\times 100=\frac{13.1 - 8.8 }{8.8 }\times 100$ = 48.9 %

Thus, the percentage change in kinetic energy is 48.9 %.