BIO Human Rotational Energy. A dancer is spinning at 72 rpm about an axis through her center with her arms outstretched

(Fig. P9.83). From biomedical measurements, the typical distribution

of mass in a human body is as follows:

Head: 7.0%

Arms: 13% (for both)

Trunk and legs: 80.0%

Suppose you are this dancer. Using this information plus length

measurements on your own body, calculate (a) your moment of inertia about your spin axis and (b) your rotational kinetic energy. Use Table 9.2 to model reasonable approximations for the pertinent parts of your body.

Angular velocity $\omega =72\text{\hspace{0.17em}\hspace{0.17em}rpm}$

Mass distribution

head 7.0%

Arms 13%

Trunk and legs 80%

Personal data

Mass $m=70\text{\hspace{0.17em}\hspace{0.17em}kg}$

Length of the arm, $l_{arm}=0.5m$

The radius of the head,$r_{head}=0.7m$

The radius of the (truck and leg),$r_{truck+head}=0.12m$

Rotational Motion

If the motion of an object is around a circular path, in a fixed orbit, then that motion is known as rotational motion.

Since our hands and legs are rotating, we must apply the formula of rotational motion

$K_{rotational}=\frac{1}{2}I{\omega }^2$

Where I is the moment of inertia of the rigid body andis angular velocity.

$I=\frac{2}{5}M{R}^2$

Where M is mass and R is the radius

$I=\frac{1}{2}M{R}^2$

Where M is mass and R is the radius

$I_{cm}=\frac{1}{12}M{L}^2$

Where M is mass and L is the length

The total moment of inertia of the body is the sum moment of inertia of all the parts of the body.

$I_{total}=I_{arm}+I_{head}+I_{trunk+leg}$

Considering the head as a sphere, its moment of inertia can be given by using the formula for the moment of inertia of a solid sphere.

$\begin{array}{c}{I}_{head}=\frac{2}{5}m{r}_h^2\\ =\frac{2}{5}\left(0.07\left(70\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\right){\left(0.07\right)}^2\\ =0.0096\text{\hspace{0.17em}\hspace{0.17em}kg}\end{array}$

Considering the trunk and leg together as a cylinder, their moment of inertia can be given by using the formula for the moment of inertia of a solid cylinder.

$\begin{array}{c}{I}_{head}=\frac{1}{2}m{r}_{trunk+leg}^2\\ =\frac{1}{2}\left(0.8\left(70\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\right){\left(0.12\right)}^2\\ =0.4032\text{\hspace{0.17em}\hspace{0.17em}kg}\end{array}$

Considering the hands as a rod, its moment of inertia can be given by using the formula for the moment of inertia of a rod.

To find the moment of inertia of the arm about the center axis of the body we must use the parallel axis theorem,

That is

$I=I_{cm}+m{d}^2$

Where is a moment of inertia about the center of mass, m is mass d is the distance between two axes.

Therefore

Therefore, the total moment of Inertia is

$\begin{array}{c}{I}_{total}=I_{arm}+I_{head}+I_{trunk+leg}\\ =1.4353{\text{\hspace{0.17em}\hspace{0.17em}kg.m}}^{\text{2}}+0.0096\text{\hspace{0.17em}\hspace{0.17em}kg}+0.4032\text{\hspace{0.17em}\hspace{0.17em}kg}\\ \text{=\hspace{0.17em}}1.500{\text{\hspace{0.17em}kg.m}}^{\text{2}}\end{array}$

Hence the total moment of inertia of the body is $1.500{\text{\hspace{0.17em}kg.m}}^{\text{2}}$

Rotational kinetic energy can be given as

$K_{rotational}=\frac{1}{2}I{\omega }^2$

Where I is a moment of inertia of the rigid body and is angular velocity.

We know

$I_{total}\text{=}1.500{\text{\hspace{0.17em}kg.m}}^{\text{2}}$

$\omega =72\text{\hspace{0.17em}\hspace{0.17em}rpm}$

Converting angular velocity into rad/s

$\omega =72\left(\frac{2\pi }{60}\right)\text{rad}/\text{s}$

Total rotational kinetic energy

$\begin{array}{l}{K}_{rotational}=\frac{1}{2}\left(1.500{\text{\hspace{0.17em}kg.m}}^{\text{2}}\right){\left(72\left(\frac{2\pi }{60}\right)\text{rad}/\text{s}\right)}^2\\ K_{rotational}=42.59J\end{array}$

Hence the rotational kinetic energy is $42.59J$