Question

# An ice skater spins about a vertical axis with an angular speed of 15 rad/s with...

An ice skater spins about a vertical axis with an angular speed of 15 rad/s with arms fully extended horizontally. Then the arms are pulled in quickly with no friction. Suppose the initial rotational inertia is 1.72 kg*m^2 and the final is .61 kg*m^2.

a) what is the final angular velocity of the skater?

b) what is the change in the skater's kinetic energy?

c) where does the additional kinetic engery come from? What is being done when the arms are pulled in? (Hint; this isn't a result of change of moment of inertia)

Here no friction is acting i.e. no torque is acting so angular momentum L = I is conserved

so we can write I11 = I22

Intial angular speed 1 = 15 rad/s

final angular speed 2 = ?

Intial rotational inertia I1 = 1.72 kgm2

Final rotational inertia I2 = 0.61 kgm2

a) so we can substitute values in I11 = I22

then 2 = I1 * 1/I2 = 1.72 * 15/0.61

b) to find change in skater's kinetic energy we need to first find intial and final kinetic energies

Intial kinetic energy K1 = 1/2 * I1 * 12 = 1/2 * 1.72 * (15)2 = 193.5 J

Final kinetic energy K2 = 1/2 * I2 * 22 = 1/2 * 0.61 * (42.3)2 = 545.73 J

Change in kinetic energy = K2 - K1 = 545.73 - 193.5 = 352.23 J

c) Here we observe that kinetic energy has increased.This is because when the skater pulls his arms he has done some work.This work is responsible for increase in kinetic energy

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