# Answer to Question #7918 in Mechanics | Relativity for Bhawana

Question #7918

An ice skater spins about a vertical axis at an angular speed of 15 rad s−1 when her arms are

outstretched. She then quickly pulls her arms into her sides in a very small time interval so

that the frictional forces due to ice are negligible. Her initial moment of inertia about the

axis of rotation is 1.72 kg m2 and her final moment of inertia is 0.61 kg m2. What is the

change in her angular speed? What is the change in her kinetic energy? Explain this change

in kinetic energy.

outstretched. She then quickly pulls her arms into her sides in a very small time interval so

that the frictional forces due to ice are negligible. Her initial moment of inertia about the

axis of rotation is 1.72 kg m2 and her final moment of inertia is 0.61 kg m2. What is the

change in her angular speed? What is the change in her kinetic energy? Explain this change

in kinetic energy.

Expert's answer

Let moments of inertia be I1 and I2 and angular speeds be w1 and w2

Angular momentum is conserved

I1 x w1 = I2 x w2

w2 = 1.72 x 15 / 0.61

= 43 rad/s (approximately)

Therefore change in angular speed = 43 - 15 = 28rad/s increase

Change in KE = ( I2 x w2^2 - I1 x w1^2 ) / 2

=( 0.61 x 43^2 - 1.72 x 15^2) / 2

= 363.45J

That is KE has actually increased .This is because of external work done by the skater while pulling her hands closer

Angular momentum is conserved

I1 x w1 = I2 x w2

w2 = 1.72 x 15 / 0.61

= 43 rad/s (approximately)

Therefore change in angular speed = 43 - 15 = 28rad/s increase

Change in KE = ( I2 x w2^2 - I1 x w1^2 ) / 2

=( 0.61 x 43^2 - 1.72 x 15^2) / 2

= 363.45J

That is KE has actually increased .This is because of external work done by the skater while pulling her hands closer

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