91影视

Mass of the person, $m=75\text{鈥夆赌�}kg$

The velocity of the person, $v=\text{鈥�}5.0\text{鈥夆赌�}km/h$

Mass distribution of arms and hands together is 13%

Mass distribution of Legs and feet together is 37%

Length of the arm, $l_{arm}=\text{鈥�}70\text{鈥夆赌�}cm$

Length of legs, $l_{leg}=\text{鈥�}90\text{鈥夆赌�}cm$

Change in Angular moment of arms and legs$螖胃={60}^0=\frac{蟺}{3}\text{鈥夆赌�}rad$

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

The average angular velocity

${蝇}_{avg}=\frac{螖胃}{螖t}$

$螖胃$=change in angular coordinates, $螖t$=change in time

Rotational kinetic energy

$K_{rotational}=\frac{1}{2}I\text{鈥�}{蝇}^2$

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

Moment of inertia of the rod about one end.

$I=\frac{1}{3}M{L}^2$

Where M is mass and L is the length

The transitional kinetic energy

$K_{translational}=\frac{1}{2}m{v}^2$

Where m is mass and v is velocity

The average angular velocity can be given as,

${蝇}_{avg}=\frac{螖胃}{螖t}$

$螖胃$= change in angular coordinates, $螖t$=change in time

Since we know in a brisk walk, the arms and legs each move through an angle of about $卤{30}^0$ (a total of${60}^0$) from the vertical in approximately 1 second.

Therefore,

$\begin{array}{c}螖胃={60}^0\\ =\frac{蟺}{3}\text{鈥夆赌�}rad\end{array}$

And

$螖t=1\text{鈥夆赌�}s$

So, average angular velocity is

$\begin{array}{l}{蝇}_{avg}=\frac{蟺/3\text{鈥夆赌�}rad}{1\text{鈥夆赌�}s}\\ {蝇}_{avg}=蟺/3\text{鈥夆赌�}rad/s\end{array}$

Hence the angular velocity of arm and leg is $\frac{蟺}{3}rad/s$

Rotational kinetic energy can be given as

$K_{rotational}=\frac{1}{2}I\text{鈥�}{蝇}^2$

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

Moment of inertia can be calculated by using the formula for moment of inertia of rod about one end.

That is, $I=\frac{1}{3}M{L}^2$

Where M is mass and L is length

So, moment of inertia of arm and hand is

$\begin{array}{c}{I}_{arm+hands}=\frac{1}{3}M{L}^2\\ =\frac{1}{3}\left(0.13脳75\text{鈥夆赌�}kg\right){\left(0.7\text{鈥夆赌�}m\right)}^2\\ =1.5925\text{鈥夆赌�}kg鈭�m^2\end{array}$

Moment of inertia of leg and feet is

$\begin{array}{c}{I}_{leg+feets}=\frac{1}{3}M{L}^2\\ =\frac{1}{3}\left(0.37脳75\text{鈥夆赌�}kg\right){\left(0.9\text{鈥夆赌�}m\right)}^2\\ =7.4925\text{鈥夆赌�}kg鈭�m^2\end{array}$

Total rotational kinetic energy

$\begin{array}{l}{K}_{rotational}=\frac{1}{2}\left(9.085\text{鈥夆赌�}kg鈭�m^2\right){\left(蟺/3\text{鈥夆赌�}rad/s\right)}^2\\ K_{rotational}=4.9814\text{鈥夆赌�}J\end{array}$

Hence the rotational kinetic energy in person鈥檚 arms and legs is $4.9814J$

Total kinetic energy can be given as,

$K_{total}=K_{rotational}+K_{translational}$

The transitional kinetic energy

$K_{translational}=\frac{1}{2}m{v}^2$

Where m is mass and v is velocity

Therefore, transitional kinetic energy of person is

$\begin{array}{c}{K}_{translational}=\frac{1}{2}\left(75\text{鈥夆赌�}kg\right)\left(5\text{鈥�}\left(\frac{1000\text{鈥夆赌�}m}{3600\text{鈥夆赌�}s}\right)\text{鈥�}\right)\\ K_{translational}=\frac{1}{2}\left(75\text{鈥夆赌�}kg\right)\left(5\text{鈥�}\left(\frac{1000\text{鈥夆赌�}m}{3600\text{鈥夆赌�}s}\right)\text{鈥�}\right)\\ K_{translational}=72.338\text{鈥夆赌�}J\end{array}$

Therefore, total kinetic energy

$\begin{array}{c}{K}_{total}=K_{rotational}+K_{translational}\\ K_{total}=4.9814\text{鈥夆赌�}J+72.338\text{鈥夆赌�}J\\ =77.3194\text{鈥夆赌�}J\end{array}$

Hence the total kinetic energy due to both his forward motion and his rotation is$77.3194J$

.

We know total kinetic energy is

$K_{total}=77.3194\text{鈥夆赌�}J$

Energy due to rotation of arm and legs is

$K_{rotational}=4.9814\text{鈥夆赌�}J$

The percentage of kinetic energy due to rotation

$\begin{array}{c}{K}_{rotational\left(\%\right)}=\frac{K_{rotational}}{K_{total}}脳100\\ =\frac{4.9814\text{鈥夆赌�}J}{77.3194\text{鈥夆赌�}J}脳100\\ =6.44\%\end{array}$

Hence the percentage of the kinetic energy due to rotation of his arm and leg is 6.44%.