Question

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)

Answer #1

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

so we can write I_{1}_{1} =
I_{2}_{2}

Intial angular speed _{1} = 15
rad/s

final angular speed _{2} =
?

Intial rotational inertia I_{1} = 1.72
kgm^{2}

Final rotational inertia I_{2} = 0.61
kgm^{2}

a) so we can substitute values in I_{1}_{1}
= I_{2}_{2}

then _{2} =
I_{1} * _{1}/I_{2}
= 1.72 * 15/0.61

= 42.3 rad/s

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

Intial kinetic energy K_{1} = 1/2 * I_{1} *
_{1}^{2}
= 1/2 * 1.72 * (15)^{2} = 193.5 J

Final kinetic energy K_{2} = 1/2 * I_{2} *
_{2}^{2}
= 1/2 * 0.61 * (42.3)^{2} = 545.73 J

Change in kinetic energy = K_{2} - K_{1} =
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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