Question

A skater spins with an angular speed of 5.9 rad/s with her arms outstretched. She lowers her arms, decreasing her moment of inertia by a factor of 1.7. Ignoring friction on the skates, determine the ratio of her final kinetic energy to her initial kinetic energy.

Answer 1

the ratio of her final kinetic energy to her initial kinetic energy is 1.7.

Step-by-step explanation:

Given;

initial angular speed, ω₁ = 5.9 rad/s

let her initial moment of inertia = I₁

her final moment of inertia
`I_2 = (I_1)/(1.7)`

Apply the principle of conservation of angular momentum to determine the final angular speed of the girl;

`\omega_1I_1 = \omega_f I_2\\\\\omega_f = (\omega _1 I_1)/(I_2) \\\\\omega_f = (5.9 * I_1)/(I_1/1.7) \\\\\omega = 5.9 * 1.7 \\\\\omega_f = 10.03 \ rad/s`

The initial rotational kinetic energy is given as;

`K.E_I = (1)/(2)I_1 \omega_I ^2`

The final rotational kinetic energy is given as;

`K.E_f = (1)/(2)I_2 \omega_f ^2`

The ratio of her final kinetic energy to her initial kinetic energy is given as;

`(K.E_f)/(K.E_I)= ((1)/(2)I_2 \omega_f^2 )/((1)/(2) I_1\omega _1^2) \\\\(K.E_f)/(K.E_I)= (I_2 \omega_f^2)/( I_1\omega _1^2) \\\\(K.E_f)/(K.E_I)= (I_1/1.7 * \omega_f^2)/( I_1 * \omega _1^2) \\\\(K.E_f)/(K.E_I)= ( \omega_f^2)/( 1.7 \omega _1^2) \\\\(K.E_f)/(K.E_I)= ( (10.03)^2)/( 1.7(5.9)^2) = (17)/(10) = 1.7`

Therefore, the ratio of her final kinetic energy to her initial kinetic energy is 1.7.