Questions: An ice dancer with her arms stretched out starts into a spin with an angular velocity of 1.00 rad / s. Her moment of inertia with her arms stretched out is 248 kg m^2. What is the increase in her rotational kinetic energy when she pulls in her arms to make her moment of inertia 1.40 kg m^2?

Solution Steps

The initial rotational kinetic energy \( K_i \) can be calculated using the formula: \[ K_i = \frac{1}{2} I_i \omega_i^2 \] where \( I_i = 248 \, \mathrm{kg \, m^2} \) and \( \omega_i = 1.00 \, \mathrm{rad/s} \).

\[ K_i = \frac{1}{2} \times 248 \, \mathrm{kg \, m^2} \times (1.00 \, \mathrm{rad/s})^2 \] \[ K_i = 124 \, \mathrm{J} \]

Using the conservation of angular momentum, the initial angular momentum \( L_i \) is equal to the final angular momentum \( L_f \): \[ L_i = I_i \omega_i = I_f \omega_f \] where \( I_f = 1.40 \, \mathrm{kg \, m^2} \).

Solving for \( \omega_f \): \[ \omega_f = \frac{I_i \omega_i}{I_f} \] \[ \omega_f = \frac{248 \, \mathrm{kg \, m^2} \times 1.00 \, \mathrm{rad/s}}{1.40 \, \mathrm{kg \, m^2}} \] \[ \omega_f = 177.1429 \, \mathrm{rad/s} \]

The final rotational kinetic energy \( K_f \) can be calculated using the formula: \[ K_f = \frac{1}{2} I_f \omega_f^2 \]

\[ K_f = \frac{1}{2} \times 1.40 \, \mathrm{kg \, m^2} \times (177.1429 \, \mathrm{rad/s})^2 \] \[ K_f = \frac{1}{2} \times 1.40 \, \mathrm{kg \, m^2} \times 31383.6735 \, \mathrm{(rad/s)^2} \] \[ K_f = 21968.5714 \, \mathrm{J} \]

The increase in rotational kinetic energy \( \Delta K \) is: \[ \Delta K = K_f - K_i \]

\[ \Delta K = 21968.5714 \, \mathrm{J} - 124 \, \mathrm{J} \] \[ \Delta K = 21844.5714 \, \mathrm{J} \]

Final Answer