Question

7) An Ice skater with her arms extended is rotating with initial kinetic energy Eki and has a moment of inertia I, then when she pulls her arms close to her body her moment of inertia goes down to 0.25 I. The final total kinetic energy (Ekf ) of the skater is A. Eki B. 2Eki C. 4Eki D. 16Eki E. Eki/4

Answer 1

C. 4Eki

Step-by-step explanation:

The kinetic energy due to the rotation is:

`K = (1)/(2)\cdot I \cdot \omega^(2)`

The motion of the ice skater is modelled after the Principle of Angular Momentum Conservation:

`I_(o)\cdot \omega_(o) = I_(f)\cdot \omega_(f)`

The initial angular speed is:

`\omega_(o) = \sqrt{(2\cdot E_(ki))/(I) }`

The final angular speed is:

`\omega_(f) = (I_(o))/(I_(f))\cdot \omega_(o)`

`\omega_(f) = 4\cdot \sqrt{(2\cdot E_(ki))/(I) }`

The final kinetic energy is:

`K = (1)/(2)\cdot \left((1)/(4)\cdot I \right)\cdot \left((32\cdot E_(ki))/(I) \right)`

`K = 4\cdot E_(ki)`

The correct answer is option C.