An ice skater doing a toe spin with outstretched arms has an aangular speed of 4

Question

An ice skater doing a toe spin with outstretched arms has an aangular speed of 4.75 rad/s. The skater then tucks in herarms, decreasing her moment of inertia by 8.5%. (a)what isthe resulting angular speed? (b) By what factor does theskater's kinetic energy change? (neglect any frictionaleffects) my answer is 5.15 rad/s. does that seem right? An ice skater doing a toe spin with outstretched arms has an aangular speed of 4.75 rad/s. The skater then tucks in herarms, decreasing her moment of inertia by 8.5%. (a)what isthe resulting angular speed? (b) By what factor does theskater's kinetic energy change? (neglect any frictionaleffects) my answer is 5.15 rad/s. does that seem right?

Explanation / Answer

    Initial angular speed, 1 = 4.74 rad/s     Change in Moment of inertia, I = 8.5 %     Let Initial moment of inertia , I1 = 100            Finalmoment of inertia, I2 = 100 - 8.5                                                    = 91.5 (a)     From law of conservation of angularmomentum,     I1 1 = I2 2     100 * 4.75 = 91.5 * 2     Resulting angular speed, 2 = 5.19rad/s (b)     Fractional increase in kinetic energy,     KE / E = [ ( 1/2 ) I2 22 - ( 1/2 ) I1 1 2 ] / (1/2 ) I1 1 2                   = ( I2 22 - I1 1 2 )/ I1 1 2                         = ( 91.5 * 5.192 - 100 * 4.752 )/ ( 100 * 4.752 )                   = ( 2464.65 - 2256.25 ) / 2256.25                   = 0.0924     % Increase in kinetic energy = 9.24 %

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