Chapter 8, Problem 82GP

To determine

To Calculate: The ratio of angular speeds for a spinning skater with outstretched arms and with arms held tightly against her body.

Answer to Problem 82GP

  $0.54$

Explanation of Solution

Assumption:

Let the human has a mass of 60 kg. The total percentage of masses of arms and hands is 12.5 % of the total mass of the human. The total percentage of the mass of the rest of the body is about 87.5% of the total mass of the body.

The mass of the skater is 52 kg. Each arm of the skater has a mass of 4 kg. The height of the skater is 150 cm with a radius of 15 cm and the length of the arm is 50 cm.

Given :

  $\begin{array}{l}{M}_{total}=60\mathrm{kg}\\ M_{body}=52\mathrm{kg}\\ M_{arm}=4\mathrm{kg}\end{array}$

  

Formula used:

The moment of inertia of the cylinder is:

  $I_{cyl}=\frac{1}{2}{M}_{cyl}{R}^2$

The moment of inertia of the rod is:

  $I_{rod}=\frac{1}{2}{M}_{rod}{L}^2$

The moment of inertia of skater when the arms held tightly:

  $$

  $I_{in}=\frac{1}{2}{M}_{total}{R}^2$

The moment of inertia of the skater with stretched arms:

  $\begin{array}{l}{I}_{out}=I_{body}+2I_{arms}\\ =\frac{1}{2}{M}_{total}{R}^2+2\left(\frac{1}{3}{M}_{arm}{L}^2\right)\end{array}$

From the conservation of angular momentum:

  $\begin{array}{l}{L}_{in}=L_{out}\\ I_{in}{\omega }_{in}=I_{out}{\omega }_{out}\\ \frac{I_{in}}{I_{out}}=\frac{{\omega }_{out}}{{\omega }_{in}}\\ \frac{{\omega }_{out}}{{\omega }_{in}}=\frac{\frac{1}{2}{M}_{total}{R}^2}{\frac{1}{2}{M}_{total}{R}^2+2\left(\frac{1}{3}{M}_{arm}{L}^2\right)}\mathrm{.................................................}(1)\end{array}$

  $$

Calculation:

Putting the values in equation (1):

  $\begin{array}{l}\frac{{\omega }_{out}}{{\omega }_{in}}=\frac{\frac{1}{2}(60){(15)}^2}{\frac{1}{2}(52){(15)}^2+2\left(\frac{1}{3}(4){(50)}^2\right)}\\ =\frac{6750}{5850+6666.67}=\frac{6750}{12516.67}=0.54\end{array}$

Conclusion:

The ratio of angular speeds for a spinning skater with outstretched arms and with arms held tightly against her body is 0.54.