91Ó°ÊÓ

Torque is the measure of the force that can cause an object to rotate about an axis. Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes an angular acceleration. Hence, torque can be defined as the rotational equivalent of linear force.

In the absence of external torque, the total angular momentum of the rotating stool-barbells hold in your hand system remains constant:

$$\begin{gathered} L=\text{constant} \\ I\mathrm{Ó\neg }=\text{constant} \end{gathered}$$

Apply this equation for the case of arms holding barbells extended fully. In the case of barbells pulls in close to your body, we have

$I\mathrm{Ó\neg }=I\text{'}\mathrm{Ó\neg }\text{'}......\left(1\right)$

The amount of rational kinetic energy expended in this process

$E=\frac{1}{2}I\text{'}\mathrm{Ó\neg }{\text{'}}^2\frac{1}{2}\mathrm{Ó\neg }......\left(2\right)$

Here,

$I→$Moment of inertia of the system in case (1)

$I\text{'}→$Moment of inertia of the system in case (2)

$\mathrm{Ó\neg }→$Angular speed of the system in case (1)

$\mathrm{Ó\neg }\text{'}→$Angular speed of the system in case (2)

(a)In order to solve this problem, it is essential to consider some required numerical data. Let us consider the numerical data arbitrarily (on your own choice) as follows.

Assume the rotation axis passes through center of the turntable table and the person.

Moment of inertia of the system in case (1), $I=1500kg.m^2$

Magnitude of the each holding mass, $m=10kg$

Distance between centers of axis to any one of the arm in case (1), $r=8m$

After pulling his arms close to the body, the distance between the centers of axis to any one of the arms, $r\text{'}=1m$

Time to complete one rotation (time period of the system) in case (1), $T=2s$

So, angular momentum of the system in case (1), $\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$

$$\begin{gathered} =\frac{2\mathrm{Ï€}}{2s} \\ =\mathrm{Ï€}rad/s \\ =3.14rad/s \end{gathered}$$

Moment of inertia of the system in case (2) becomes very less when his arms pulled very close to his body.

$$\begin{gathered} I\text{'}=I-(2m{r}^2-2mr{\text{'}}^2) \\ =I-2m(r^2-r{\text{'}}^2) \\ =(1500kg.m^2)-2\left(10kg\right)\left[{\left(8m\right)}^2-{\left(1m\right)}^2\right] \\ =(1500kg.m^2)-1260kg.m^2 \\ =240kg.m^2 \end{gathered}$$

By using equation (1), the angular velocity of the system in case (2) becomes

$$\begin{gathered} \mathrm{Ó\neg }\text{'}=\frac{I\mathrm{Ó\neg }}{I\text{'}} \\ =\frac{(1500kg.m^2)(\mathrm{Ï€}rad/s)}{(240kg.m^2)} \\ =19.6rad/s \end{gathered}$$

The corresponding time period (time taken to complete one rotation) is given by

$$\begin{gathered} T\text{'}=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }\text{'}} \\ =\frac{2(3.14)rad}{(19.6rad/s)} \\ =0.32s \end{gathered}$$

Finally, it is clear that, when his arms pulled close to his body, moment of inertia of the system decreases. This decreases of moment of inertia causes to increase the angular velocity of the system. As angular velocity inversely proportional to time period (time complete to turn one complete rotation), the time period is decreased drastically.

Using equation (2), the amount of energy expend during this process given by,

$E=\frac{1}{2}I\text{'}\mathrm{Ó\neg }{\text{'}}^2\frac{1}{2}I{\mathrm{Ó\neg }}^2$

Plug the given data in the above equation,

$$\begin{gathered} E=\frac{1}{2}(240kg.m^2)(19.6rad/s{)}^2\frac{1}{2}(1500kg.m^2)(3.14rad/s^2) \\ =46099.2J-7394.7 \\ =38704.5J \end{gathered}$$

$$\begin{gathered} E=\frac{1}{2}(240kg.m^2)(19.6rad/s{)}^2\frac{1}{2}(1500kg.m^2)(3.14rad/s^2) \\ =46099.2J-7394.7 \\ =38704.5J \end{gathered}$$