91Ó°ÊÓ

Angular velocity $\mathrm{Ó\neg }=72\text{ â¶Ä‰r±è³¾}$

Mass distribution

head 7.0%

Arms 13%

Trunk and legs 80%

Personal data

Mass $m=70\text{ â¶Ä‰k²µ}$

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

The radius of the head,$r_{head}=\text{ }0.7\text{ â¶Ä‰}m$

The radius of the (truck and leg),$r_{truck+head}=\text{ }0.12\text{ â¶Ä‰}m$

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

$K_{rotational}=\frac{1}{2}I\text{ }{\mathrm{Ó\neg }}^2$

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.

$\begin{array}{c}{I}_{head}=\frac{2}{5}m{r}_h^2\\ =\frac{2}{5}\left(0.07\left(70\text{ â¶Ä‰k²µ}\right)\right){\left(0.07\right)}^2\\ =0.0096\text{ â¶Ä‰k²µ}\end{array}$

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.

$\begin{array}{c}{I}_{head}=\frac{1}{2}m{r}_{trunk+leg}^2\\ =\frac{1}{2}\left(0.8\left(70\text{ â¶Ä‰k²µ}\right)\right){\left(0.12\right)}^2\\ =0.4032\text{ â¶Ä‰k²µ}\end{array}$

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

$\begin{array}{c}{I}_{total}=I_{arm}+I_{head}+I_{trunk+leg}\\ =1.4353{\text{ â¶Ä‰k²µ.³¾}}^{\text{2}}+0.0096\text{ â¶Ä‰k²µ}+0.4032\text{ â¶Ä‰k²µ}\\ \text{= }1.500{\text{ k²µ.³¾}}^{\text{2}}\end{array}$

Hence the total moment of inertia of the body is $1.500{\text{ k²µ.³¾}}^{\text{2}}$

Rotational kinetic energy can be given as

$K_{rotational}=\frac{1}{2}I\text{ }{\mathrm{Ó\neg }}^2$

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

We know

$I_{total}\text{=}1.500{\text{ k²µ.³¾}}^{\text{2}}$

$\mathrm{Ó\neg }=72\text{ â¶Ä‰r±è³¾}$

Converting angular velocity into rad/s

$\mathrm{Ó\neg }=72\text{ â¶Ä‰}\left(\frac{2\mathrm{Ï€}}{60}\right)\text{rad}/\text{s}$

Total rotational kinetic energy

$\begin{array}{l}{K}_{rotational}=\frac{1}{2}\left(1.500{\text{ k²µ.³¾}}^{\text{2}}\right){\left(72\text{ â¶Ä‰}\left(\frac{2\mathrm{Ï€}}{60}\right)\text{rad}/\text{s}\right)}^2\\ K_{rotational}=42.59\text{ â¶Ä‰â¶Ä‰}J\end{array}$

Hence the rotational kinetic energy is $42.59J$