An ice skater whirls with her arms and one leg stuck out as shown on the left in Figure 11.93, making one complete turn inThen she quickly moves her arms up above her head and pulls her leg in as shown at the right in Figure 11.93.

(a) Estimate how long it now takes for her to make one complete turn. Explain your calculations, and state clearly what approximations and estimates you make. (b) Estimate the minimum amount of chemical energy she must expended to change her configuration.

An ice skater spins vertically on the tip of one skate with her arms and one leg outstretched, and then pull her leg in and her arms to a vertical position over her head, so she can spin much faster. In this situation, the total momentum of the system is conserved and in which there is no external torque, even if the moment of inertia of the system does change due to change in shape of the system.

Estimate a skater of mass around 50 kg that about 40 kg is in her torso plus one leg as a cylinder of radius r_torso=0.1 m. Then the moment of inertia of the cylinder is

$$\begin{gathered} I_{torso}=\frac{1}{2}{M}_{torso}{r_{torso}}^2 \\ =\frac{1}{2}\left(40kg\right)(0.1m{)}^2 \\ =0.2kg.m^2 \end{gathered}$$

Estimate the mass of her leg of length from axis of rotation$r_{leg}=1.0m$, is $m_{leg}=5kg$and mass of arm of length $0.6m$from axis of rotation is with these estimates the moment of inertia of the skater is

$$\begin{gathered} I_1=I_{torso}+\frac{1}{2}\left(2m_{arm}\right){r_{arm}}^2+\frac{1}{2}{m}_{leg}{r_{leg}}^2 \\ =0.2kg.m^2+(2.5kg)(0.6m{)}^2+\frac{1}{2}\left(5kg\right)(1m{)}^2 \\ =3.6kg.m^2 \end{gathered}$$

If the skater pulls her arms and leg inside then the moment of inertia of the skater is

$$\begin{gathered} I_1=I_{torso}+m{r}^2 \\ =0.2kg.m^2+\left(10kg\right)(0.1{)}^2 \\ =0.3kg.m^2 \end{gathered}$$

(a)Let us consider a skater spin initially with an angular speed${\omega }_1,$after she changed her configuration the angular speed increases to

From the law of conservation of angular momentum,

$I_1{\omega }_1=I_2{\omega }_2$

Thus, the angular speed of the skater is

${\omega }_2=\frac{I_1{\omega }_1}{I_2}$

Given that, the she makes one revolution in one second before pulls her arms.

So,

Therefore, the time taken to completer one revolution is

$\frac{1rev}{T}=\frac{I_1(1rev/s)}{I_2}$

Rearrange the above equation and isolate, T

$T=\frac{I_2}{I_1}$

$=\frac{0.3kg.m^2}{3.6kg.m^2}$

$=0.083kg.m^2$

(b)The skater has expended his chemical energy in order to increase her kinetic energy. The initial kinetic energy of the system is

$$\begin{gathered} K_1=\frac{1}{2}{I}_1{{\omega }_1}^2 \\ =\frac{1}{2}(3.6kg.m^2)(2\pi rad/1\sec {)}^2 \\ =71.0J \end{gathered}$$

The final kinetic energy of the system is

$$\begin{gathered} K_2=\frac{1}{2}{I}_2{{\omega }_2}^2 \\ =\frac{1}{2}{I}_2{\left(\frac{I_1}{I_2}{\omega }_1\right)}^2 \\ =\frac{1}{2}\frac{{I_1}^2}{I_2}{{\omega }_1}^2 \\ =\frac{1}{2}\frac{{(3.6kg.m^2)}^2}{0.3kg.m^2}{\left(\frac{2\pi rad}{1\sec }\right)}^2 \\ =852.7J \end{gathered}$$

Thus, the amount of chemical energy needed to change her configuration is

$$\begin{gathered} \Delta K=K_2-K_1 \\ =852.7J-71.0J \\ =781.7J \end{gathered}$$