Chapter 8, Problem 111QAP

To determine

The angular velocity of the skater when he pulls his parallel to his trunk.

Answer to Problem 111QAP

The angular velocity of the skater is, $642.786rpm$.

Explanation of Solution

Given information:

The mass distribution of a human body can be considered as, $80\%$ for the trunk and legs $13\%$, for the arms and hands, and $7\%$ for the head.

The spinning skater with hands outstretched can be modeled as a vertical cylinder with two uniform rods attached to it.

The mass of the skater is, $62kg$ and he is, $1.8m$ tall.

His trunk can be modeled as a cylinder with $35cm$ diameter.

The length of both arm and a hand is, $65cm$.

The spinning speed of the skater with his arms outstretched is, $70rpm$.

Formula used:

  $I={\displaystyle \int m.r^2}$

  $I_1{\omega }_1=I_2{\omega }_2$

Calculation:

Let us first consider the hand outstretched position.

  

The moment of inertia of the arms can be calculated as follows.

The mass of an arm and a hand is,

  $62kg\times \frac{13}{100}\times \frac{1}{2}$

  $4.03kg$

Since the arms are modeled as uniform rods, let us assume that they have a uniform linear density $\lambda$.

  $\lambda =\frac{4.03}{0.65}$

  $\lambda =6.2$

Let us consider the moment of inertia of small particle of the rod, which is $x$ length away from its center of rotation.

Using the given formula, its moment of inertia can be written as,

  $I={\displaystyle \int x^2}dm$

  $I={\displaystyle \int x^2}d\left(\lambda x\right)$

  $I=\lambda {\displaystyle \int x^2}dx$

Applying the limits,

  $I=\lambda {\displaystyle {\int }_{0.35}^1{x}^2}dx$

  $I=\frac{6.2}{3}\left(1^3-{0.35}^3\right)$

  $I=1.978$

The moment of inertia for the both arms is, $3.956kg.m^2$.

Let us now consider the head and the trunk.

The mass of the trunk and head is,

  $62kg\times \frac{87}{100}$

  $53.94kg$

Assuming the cylinder to be uniform, let us denote its density by $\rho$.

  $\rho =\frac{53.94}{\pi \times {0.35}^2\times 1.8}$

  $\rho =77.8668$

It can be modeled as a solid cylinder.

  

Using the given formula, its moment of inertia can be written as,

  $I={\displaystyle \int r^2}dm$

  $I={\displaystyle \int r^2}d\left(\rho V\right)$

  $I=\rho {\displaystyle \int r^2}dV$

But, $dV=2\pi rLdr$.

  $I=2\pi \rho L{\displaystyle \int r^{3}dr}$

Applying the limits;

  $I=2\pi \rho L{\displaystyle \underset{0}{\overset{0.175}{\int }}{r}^{3}dr}$

  $I=\left(2\pi \times 77.8668\times 1.8\right){\displaystyle \underset{0}{\overset{0.175}{\int }}{r}^{3}dr}$

  $I=0.2065kg.m^2$

Considering the whole system, the moment of inertia when the hands are outstretched is,

  $I_1=3.956kg.m^2+0.2065kg.m^2$

  $I_1=4.1625kg.m^2$

Let us now consider the case when skater is pulling his arms towards him.

The moment of inertia of the head and the trunk remains the same.

But moment of inertia of the hands and arms change.

Since there is no mass distribution perpendicular to the axis of rotation, the moment of inertia can be directly calculated.

  $\begin{array}{l}I=m{r}^2+m{r}^2\\ I=2m{r}^2\\ I=2\times 4.03\times {0.175}^2\\ I=0.2468kg.m^2\end{array}$

Therefore, the moment of inertia for the whole system will be,

  $I_2=0.2468kg.m^2+0.2065kg.m^2$

  $I_2=0.4533kg.m^2$

Now let us consider the conservation of angular momentum;

  $I_1{\omega }_1=I_2{\omega }_2$

  $\begin{array}{l}4.1625\times 70=0.4533\times {\omega }_2\\ {\omega }_2=642.786rpm\end{array}$

Conclusion:

The angular speed of the skater when he pulls his arms towards him is, $642.786rpm$.