Chapter 8, Problem 55P

To determine

To Find: The final moment of inertia.

The way skater accomplished the change.

Answer to Problem 55P

The final moment of inertia is $0.7666{\text{kg·m}}^{\text{2}}$ .

Explanation of Solution

Given Information:

Initial spin rotation, ${\omega }_i=\frac{1.0}{2}\text{rev/s}$

Final spin rotation, ${\omega }_f=3\text{rev/s}$

Initial moment of inertia, ${\text{I}}_i=4.6\text{kg}\cdot {\text{m}}^{\text{2}}$

Formula used:

Law of conservation of momentum

  $\begin{array}{l}{I}_i{\omega }_i=I_f{\omega }_f\\ I_f=\frac{I_i{\omega }_i}{{\omega }_f}\end{array}$

Here, $I_i$ is the initial moment of inertia,

  ${\omega }_i$ is the initial spin rotation,

  ${\omega }_f$ is the final spin rotation.

  $I_f$ is the final moment of inertia.

Calculation: As,

  $\begin{array}{l}{I}_f=\frac{I_i{\omega }_i}{{\omega }_f}\\ I_f=\frac{\left(4.6\right).\left(\frac{1}{2}\right)}{3}\\ I_f=0.7666{\text{kg·m}}^{\text{2}}\end{array}$

The skater can physically accomplish this small moment of inertia by pullingher arms inwards. The closer the mass is to the rotational axis, the smaller is the moment of inertia.

Conclusion: The final moment of inertia is $0.7666\text{kg}·{\text{m}}^{\text{2}}$ .