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

Answer:

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

Explanation:

Given;

initial angular speed, ω₁ = 5.9 rad/s

let her initial moment of inertia = I₁

her final moment of inertia $I_2 = \frac{I_1}{1.7}$

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

$\omega_1I_1 = \omega_f I_2\\\\\omega_f = \frac{\omega _1 I_1}{I_2} \\\\\omega_f = \frac{5.9 \times I_1}{I_1/1.7} \\\\\omega = 5.9 \times 1.7 \\\\\omega_f = 10.03 \ rad/s$

The initial rotational kinetic energy is given as;

$K.E_I = \frac{1}{2}I_1 \omega_I ^2$

The final rotational kinetic energy is given as;

$K.E_f = \frac{1}{2}I_2 \omega_f ^2$

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

$\frac{K.E_f}{K.E_I}= \frac{\frac{1}{2}I_2 \omega_f^2 }{\frac{1}{2} I_1\omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{I_2 \omega_f^2}{ I_1\omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{I_1/1.7 \times \omega_f^2}{ I_1 \times \omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{ \omega_f^2}{ 1.7 \omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{ (10.03)^2}{ 1.7(5.9)^2} = \frac{17}{10} = 1.7$

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